# Brilliant Integration Contest - Season 2 (Part-1)

Hi Brilliant! Just like what Anastasiya Romanova conducted last year, this year I would also like to conduct an integration contest.

The aims of the Integration contest are to improve skills in the computation of integrals, to learn from each other as much as possible, and of course to have fun. Anyone here may participate in this contest.

The rules are as follows:

• I will start by posting the first problem. If there is a user solves it, then they must post a new one.

• You may only post a solution of the problem below the thread of problem and post your proposed problem in a new thread. Put them separately.

• Only make substantial comment that will contribute to the discussion.

• Make sure you know how to solve your own problem before posting it in case there is no one can answer it within 48 hours, then you must post the solution and you have a right to post another problem.

• If the one who solves the last problem does not post his/her own problem after solving it within a day, then the one who has a right to post a problem is the last solver before him/her.

• The scope of questions is only computation of integrals either definite or indefinite integrals.

• You are NOT allowed to post a multiple integrals problem.

• It is NOT compulsory to post original problems. But make sure it has not been posted on brilliant.

• Do not copy questions from last year's contest. If anyone found to do so he/she will be banned from taking further part in this contest

• You are also NOT allowed to post a solution using a contour integration or residue method.

The final answer can ONLY contain the following special functions: gamma function, beta function, Riemann zeta function, Dirichlet eta function, dilogarithm, digamma function, trigonometric integral, Wallis' integral, Bessel function, contour integration and Ramanujan's Master Theorem (including Mellin Transform).

Format your post is as follows:

 1 2 3 4 5 6 7 **SOLUTION OF PROBLEM xxx (number of problem) :** **[Post your solution here]** **PROBLEM xxx (number of problem) :** **[Post your problem here]** 

View Part 2

2 years, 5 months ago

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Problem 1:

Evaluate: $\int _{ 0 }^{ 1 }{\frac { x \ln\left( x \right) }{ \sqrt { 1-{ x }^{ 2 } } } \, dx }$

##### This problem has been solved by Surya Prakash.

- 2 years, 5 months ago

Another solution: Let $$\sqrt{(1-x^2)}=t$$

$$\frac{-x}{\sqrt{(1-x^2)}}dx=dt$$

So integration becomes

$$- \int_1^0 ln\sqrt{(1-t^2)}dt$$

$$-\frac{1}{2} \int_1^0 ln(1-t^2)dt$$

$$\frac{1}{2} \int_0^1 ln(1-t^2)dt$$

$$\frac{1}{2} \int_0^1 (ln(1-t)+ln(1+t))dt$$

This can be easily evaluated and comes out to be $$ln2-1$$

- 2 years, 5 months ago

$$\Large \text{Solution to Problem 1:}$$

$\int_{0}^{1} \dfrac{x \ln(x)}{\sqrt{1-x^2}} dx$ Take $$t = \arcsin(x) \iff dt = \dfrac{dx}{\sqrt{1-x^2}}$$ and the limits changes to $$t=0$$ to $$t=\dfrac{\pi}{2}$$. $\int_{0}^{\dfrac{\pi}{2}} \sin (t) \ln(\sin(t)) dt$

Consider,

$B(x,y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} dt$

Taking the transformation $$t=\sin^2 (z) \iff dt = 2\sin(z) \cos (z)dz$$, the integral transforms into

$B(x,y) = 2 \int_{0}^{\pi /2} \sin ^{2x-1} (z) \cos^ {2y-1} (z) dz$

But $$B(x,y) = \dfrac{\Gamma (x) \Gamma(y) }{\Gamma (x+y)}$$

So, $\int_{0}^{\pi /2} \sin ^{2x-1} (z) \cos^ {2y-1} (z) dz = \dfrac{1}{2} \dfrac{\Gamma (x) \Gamma(y) }{\Gamma (x+y)}$

Differentiate it w.r.t $$x$$

$2 \int_{0}^{\pi /2} \sin ^{2x-1} (z) \cos^ {2y-1} (z) \ln(\sin(z)) dz = \dfrac{1}{2} \dfrac{\partial}{\partial x} \dfrac{\Gamma (x) \Gamma(y) }{\Gamma (x+y)} \\ \int_{0}^{\pi /2} \sin ^{2x-1} (z) \cos^ {2y-1} (z) \ln(\sin(z)) dz = \dfrac{1}{4} \dfrac{\Gamma (x) \Gamma(y) }{\Gamma (x+y)} (\psi (x) - \psi(x+y))$

Take $$x= 1$$ and $$y= \dfrac{1}{2}$$.

$\int_{0}^{\pi / 2} \sin(z) \ln (\sin (z)) dz = \dfrac{1}{4} \dfrac{\Gamma (1) \Gamma(1/2) }{\Gamma (3/2)} (\psi (1) - \psi(3/2)) = \ln (2) -1$

- 2 years, 5 months ago

Problem 23:

Evaluate:$\int _{ 0 }^{ 1 }{ \log\left( 1+x \right) \log\left( 1-{ x }^{ 3 } \right) dx }$

###### Due to time constraint, Aditya Kumar decided to post a solution himself.

- 2 years, 5 months ago

Solution to Problem 23:

Substitute: $$x\longrightarrow \frac { 1-x }{ 1+x }$$

$I=\int_0^1 \ln(1+x)\ln(1-x^3)dx=2\int_0^1 \ln\left(\frac{2 x (3+x^2)}{(1+x)^3}\right)\ln\left(\frac{2}{1+x}\right)\frac{dx}{(1+x)^2}$

On separating the integrands,

$$\ln\left(\frac{2 x (3+x^2)}{(1+x)^3}\right)\ln\left(\frac{2}{1+x}\right) \\\small =\ln^22-\ln2\ln x+\ln2\ln(3+x^2)-4\ln2\ln(1+x)-\ln x\ln(1+x)+3\ln^2(1+x)-\ln(1+x)\ln(3+x^2)$$

1st integral:

$\int_0^1 \frac{dx}{(1+x)^2}=-\frac1{1+x}\Bigg{|}_0^1=\frac12$

2nd integral:

$\int_0^1 \frac{\ln x}{(1+x)^2}dx=\sum_{n\geq1} \frac{(-1)^n}{n}=-\ln2$

3rd integral:

$\int_0^1 \frac{\ln(1+x)}{(1+x)^2}dx=-\frac{\ln(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1\frac{dx}{(1+x)^2}dx=\frac12-\frac{\ln2}{2}$

4th integral:

$\int_0^1 \frac{\ln(3+x^2)}{(1+x)^2}dx=-\frac{\ln(3+x^2)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac{2x}{(1+x)(3+x^2)}dx \\=-\ln2+\ln3+2\operatorname{Re} \int_0^1 \frac{dx}{(1+x)(x+i\sqrt{3})}=-\ln2+\frac34\ln3+\frac{\pi}{4\sqrt{3}}$

5th integral:

$\int_0^1 \frac{\ln x \ln(1+x)}{(1+x)^2}dx=-\frac{\ln x\ln(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1\frac1{1+x}\left(\frac{\ln(1+x)}{x}+\frac{\ln x}{1+x}\right)dx \\=-\ln2+\int_0^1 \frac{\ln(1+x)}{x}dx-\int_0^1\frac{\ln(1+x)}{1+x}dx=\frac{\pi^2}{12}-\frac12\ln^22-\ln2$

6th integral:

$\int_0^1 \frac{\ln^2(1+x)}{(1+x)^2}dx=-\frac{\ln^2(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac{2\ln(1+x)}{(1+x)^2}dx \\=1-\ln2-\frac12\ln^22$

7th integral:

$\int_0^1 \frac{\ln(1+x)\ln(3+x^2)}{(1+x)^2}dx=-\frac{\ln(1+x)\ln(3+x^2)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac1{1+x}\left(\frac{2x\ln(1+x)}{3+x^2}+\frac{\ln(3+x^2)}{1+x}\right)dx \\=-\ln^22-\ln2+\frac34\ln3+\frac{\pi}{4\sqrt{3}}+2J$

where,

$J=\int_0^1\frac{x\ln(1+x)}{(1+x)(3+x^2)}dx=\operatorname{Re} \int_0^1 \frac{\ln(1+x)}{(1+x)(x+i\sqrt{3})}dx \\=-\frac18\ln^22-\operatorname{Re} \,\frac1{i\sqrt{3}-1}\int_0^1 \frac{\ln(1+x)}{x+i\sqrt{3}}dx$

Now on using the formula:

$\int_0^1 \frac{\ln(1+x)}{x+a}dx=\ln2 \ln\frac{a+1}{a-1}+\operatorname{Li}_2\left(\frac2{1-a}\right)-\operatorname{Li}_2\left(\frac1{1-a}\right)$

Putting $$a=i\sqrt{3}$$ then gives, after taking the real part, and assuming the principle value of the logarithm,

$\small J=-\frac18\ln^22+\frac{\pi\sqrt{3}}{12}\ln2+\frac14\operatorname{Re}\text{Li}_2(e^{i\pi/3})-\frac{\sqrt{3}}{4}\operatorname{Im}\text{Li}_2(e^{i\pi/3})-\frac14\operatorname{Re}\text{Li}_2(\frac12 e^{i\pi/3})+\frac{\sqrt{3}}{4}\operatorname{Im}\text{Li}_2(\frac12 e^{i\pi/3})$

Using:

$\displaystyle \,\, \operatorname{Re}\text{Li}_2(e^{i\pi/3})=\frac{\pi^2}{36}$,

$\displaystyle \,\, \operatorname{Im}\text{Li}_2(e^{i\pi/3})=\frac1{2\sqrt{3}}\psi_1\left(\frac13\right)-\frac{\pi^2}{3\sqrt{3}}$,

$\displaystyle \,\,\text{Li}_2(\frac12 e^{i\pi/3})=-\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right)-\frac12\ln^2\left(\frac34-i\frac{\sqrt{3}}{4}\right)$,

and

$\displaystyle \,\, \operatorname{Re}\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right)=\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n)^2\, 3^n}=\frac14\text{Li}_2\left(-\frac13\right)$

we have $\operatorname{Re}\text{Li}_2(\frac12 e^{i\pi/3})=\frac{\pi^2}{72}-\frac18\ln^2\left(\frac34\right)-\frac14\text{Li}_2\left(-\frac13\right)$

and $\operatorname{Im}\text{Li}_2(\frac12 e^{i\pi/3})=\frac{\pi}{12}\ln\left(\frac34\right)-\operatorname{Im}\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right).$

Finally, putting everything together gives: $I=\ln^22 - \frac18\ln^23 + 2\ln2\ln3 - \frac32\ln3 - 6\ln2 + 6-\frac{\pi}{4\sqrt{3}}(2 + \ln3) - \frac{37\pi^2}{72} + \frac12\psi_1\left(\frac13 \right ) - \frac14\text{Li}_2\left( -\frac13 \right ) + \sqrt{3}\Im\text{Li}_2\left( -\frac{i}{\sqrt{3}} \right )$

$...$

This was not original. I found it very awesome, so I thought to share it here.

- 2 years, 5 months ago

I don't know what's wrong with the rendering of the Latex, since it sometimes render and sometimes it doesn't. So I'm going to post in pictures the one's that didn't render. I think it's better to move this discussion to another note.

1st Integral:

3rd Integral:

4th Integral:

5th Integral:

6th Integral:

7th Integral:

- 2 years, 5 months ago

Problem 2:

Evaluate:

$\Large \int_{0} ^{\infty} \dfrac{x \cos (x^3)}{\exp (x^3)} \, dx$

###### This problem has been solved by Tanishq Varshney.

- 2 years, 5 months ago

$$\large{Solution~to~Problem~ 2}$$

Firstly I would add a proof of my solution

$$\large{\displaystyle \int _{0}^{\infty} e^{-ax}x^{n-1} dx=\frac{\Gamma (n)}{a^{n}}}$$ where $$\Gamma (n)$$ is Gamma function

Replace $$a \rightarrow a+ib$$ where $$i =\sqrt{-1}$$

$$\large{\displaystyle \int _{0}^{\infty} e^{-ax}e^{-ibx}x^{n-1} dx=\frac{\Gamma (n)}{(a+ib)^{n}}}$$

Put $$a=r \cos y ~ and ~ b=r \sin y$$.

So that $$r^2=a^2+b^2$$ and $$y=\arctan \left(\frac{b}{a}\right)$$

Using de moviers theorem

$$\large{\displaystyle \int _{0}^{\infty} e^{-ax}(\cos bx-i \sin bx)x^{n-1} dx=\frac{\Gamma (n)}{r^{n}}(\cos ny+\sin ny)^{-1}}$$

Comparing the real and imaginary parts we finally get

$$\large{\displaystyle \int _{0}^{\infty}x^{n-1} e^{-ax}(\cos bx) dx=\frac{\Gamma (n)}{r^{n}}(\cos ny)}$$

now in the given integral put $$x^{3}=t$$

The integral now becomes

$$\large{\frac{1}{3} \displaystyle \int^{\infty}_{0} t^{-\frac{1}{3}}e^{-t} \cos (t) dt}$$

here $$n=\frac{2}{3}$$;$$a=1$$ and $$b=1$$

Thus we get $$\large{\frac{1}{3}\frac{\Gamma \left(\frac{2}{3}\right) \cos \left(\frac{\pi}{6}\right)}{2^{\frac{1}{3}}}}$$

$$\large{\boxed{\frac{\sqrt{3} \Gamma \left(\frac{2}{3}\right)}{3.2^{\frac{4}{3}}}}}$$

- 2 years, 5 months ago

Problem 24:

Prove that

$\large \int_0^{\pi /2} \dfrac1{\sin^8 x + \cos^8 x } \, dx = \dfrac{\sqrt{10-\sqrt2}}{2} \pi .$

###### This problem has been solved by Tanishq Varshney.

- 2 years, 5 months ago

$$\text{Solution to Problem 24}$$

$$\large{\displaystyle \int ^{\infty}_{0} \frac{x^{m-1}}{1+x}dx=\frac{\pi}{\sin m \pi} \quad \quad 0<m<1}$$

Refer to problem 18 for this step.

Now in the integral $$\tan x \rightarrow t$$

$$\large{\displaystyle \int ^{\infty}_{0} \frac{(1+t^2)^3}{1+t^{8}}dt}$$

$$\large{t^8 \rightarrow z}$$

$$\large{\frac{1}{8} \left(\displaystyle \int ^{\infty}_{0} \frac{z^{\frac{7}{8}-1}}{1+z}+ \frac{3z^{\frac{5}{8}-1}}{1+z}+ \frac{3z^{\frac{3}{8}-1}}{1+z}+ \frac{z^{\frac{1}{8}-1}}{1+z} dz \right)}$$

$$\large{\rightarrow \frac{\pi}{4} \left(\frac{1}{\sin \left(\frac{\pi}{8} \right)}+\frac{3}{\cos \left(\frac{\pi}{8} \right)} \right)}$$

$$\large{=\frac{\sqrt{10-\sqrt{2}}}{2} \pi}$$

- 2 years, 5 months ago

Whoops, I accidentally deleted Tanishq's comment:

Problem 5:

$\int_0^{\pi /2} \tan^{-1} (1729\cos^2 x) \, dx$

###### This problem has been solved by Sudeep Salgia.

- 2 years, 5 months ago

I'll post the outline of the solution as it is a bit cumbersome. Let
$$\displaystyle I(a) = \int_0^{\frac{\pi}{2}} \arctan (a \cos^2 x) dx \Rightarrow I'(a) = \frac{\cos^2 x}{1 + a^2 \cos^4 x } dx = \frac{\sec^2 x}{\sec^4 x + a^2 } dx$$ Substitute $$\tan x = t$$, to get $$\displaystyle I'(a) = \int_0^{\infty} \frac{1}{t^4 + 2t^2 + (a^2 + 1)} dt$$ . Put $$a^2 + 1 = k^2$$ and with some simple rearrangement, we get,
$$\displaystyle 2k I'(a) = \int_0^{\infty} \frac{ 1+ \frac{k}{t^2}}{\left( t - \frac{k}{t} \right)^2 + 2(k+1)} - \int_0^{\infty} \frac{ 1- \frac{k}{t^2}}{\left( t + \frac{k}{t} \right)^2 + 2(-k+1)}$$ Substitute $$\displaystyle t - \frac{k}{t} = z$$ and $$\displaystyle t + \frac{k}{t} = y$$ in the respective integrals and evaluate them easily (They come out to be $$\arctan$$ of something ). After all this we can write,
$$\displaystyle I(a) = \frac{\pi}{2\sqrt{2}} \int \frac{1}{\sqrt{a^2 + 1}} \frac{1}{\sqrt{\sqrt{a^2 + 1} +1}} da$$
This can be rearranged to be written as
$$\displaystyle I(a) = \frac{\pi}{2\sqrt{2}} \int \frac{1}{1+ \frac{\sqrt{a^2 + 1} - 1 }{2}} \times \frac{1}{2\sqrt{ \frac{\sqrt{a^2 + 1} - 1 }{2} }} \times \frac{1}{2} \times \frac{a}{\sqrt{a^2 +1}} da$$

This evaluates to $$\displaystyle \frac{\pi}{2} \arctan \left( \sqrt{ \frac{\sqrt{a^2+1} - 1}{2}} \right)$$

The constant of integration is $$0$$ from $$I(0) = 0$$. Put $$a =1729$$ to get the answer.

- 2 years, 5 months ago

Problem 4:

Let $$P(x)$$ be a polynomial such that $$\frac { P\left( x+1 \right) }{ P\left( x \right) } =\frac { { x }^{ 2 }+x+1 }{ { x }^{ 2 }-x+1 }$$ and $$P(2)=3$$, evaluate$\int _{ 0 }^{ 1 }{ \left( { \tan }^{ -1 }\left( P(x) \right) .{ \tan }^{ -1 }\left( \sqrt { \frac { x }{ 1-x } } \right) \right) \, dx }.$

###### This problem has been solved by Tanishq Varshney.

- 2 years, 5 months ago

$$\text{Solution to Problem 4}$$

Clearly $$P(x)=x^2-x+1$$ and $$P(1-x)=P(x)$$

Also $$\tan^{-1} (x)=\cot^{-1} \left(\frac{1}{x}\right)$$

$$\tan^{-1}(x)+\cot^{-1}(x)=\frac{\pi}{2}$$

$$\large{I=\displaystyle \int^{1}_{0} \left( \tan^{-1}(P(x)) \cot^{-1} \left( \sqrt{\frac{x}{1-x}}\right) \right) dx}$$

$$\large{2I=\frac{\pi}{2} \displaystyle \int^{1}_{0}\tan^{-1} (x^2-x+1) dx}$$

$$\large{2I=\frac{\pi}{2}\left(\frac{\pi}{2} - \displaystyle \int^{1}_{0} \tan^{-1}\left(\frac{1}{1+x(x-1)}\right)\right)}$$

$$\large{2I=\frac{\pi}{2}\left(\frac{\pi}{2} - \displaystyle \int^{1}_{0} \tan^{-1}\left(\frac{x-(x-1)}{1+x(x-1)}\right) dx \right)}$$

$$\large{2I=\frac{\pi}{2}\left(\frac{\pi}{2} -( \displaystyle \int^{1}_{0} \tan^{-1} (x) dx- \displaystyle \int^{1}_{0} \tan^{-1} (x-1) dx)\right)}$$

$$\large{2I=\frac{\pi}{2}\left(\frac{\pi}{2} -\left( \frac{\pi}{2}-2 \ln (2)\right)\right)}$$

$$\large{I=\frac{\pi}{2} \ln(2)}$$

- 2 years, 5 months ago

Problem 13: (Originally posted by Tanishq)

Evaluate $\int_0^1 x \ln \left( \dfrac{1+x}{1-x} \right) \, dx.$

###### This problem has been solved by Surya Prakesh.

- 2 years, 5 months ago

By using integration by parts,

$\int x \ln \left(\dfrac{1+x}{1-x} \right) dx = \dfrac{x^2}{2} \ln \left(\dfrac{1+x}{1-x} \right) + x - \dfrac{1}{2} \ln \left(1+x \right) + \dfrac{1}{2} \ln \left(1-x \right)$

So,

\begin{align*} \int_{0}^{1} x \ln \left(\dfrac{1+x}{1-x} \right) dx &= \lim_{a \rightarrow 0} \int_{0}^{1-a} x \ln \left(\dfrac{1+x}{1-x} \right) dx \\ &= \lim_{a \rightarrow 0} \dfrac{(1-a)^2}{2} \ln \left(\dfrac{2-a}{a} \right) + 1 - a - \dfrac{1}{2} \ln \left(2-a \right) + \dfrac{1}{2} \ln a \\ &= 1\end{align*}

- 2 years, 5 months ago

Problem 9:

Evaluate the indefinite integral: $\large{\displaystyle \int \frac{x^2+20}{(x \sin x+5 \cos x)^2} \, dx}.$

###### This problem has been solved by Surya Prakesh.

- 2 years, 5 months ago

\begin{align*} x^2 + 20 &= (x \sin(x) + 5 \cos(x))^2 + (5 \sin (x) - x \cos(x))^2 - 5 \\ &=(x \sin(x) + 5 \cos(x))^2 + (5 \sin (x) - x \cos(x))^2 - 5 \cos^2(x) - 5\sin^2 (x) -x \cos(x)\sin(x) + x\cos(x) \sin(x) \\ &= (x \sin(x) + 5 \cos(x)) ( 5 \cos(x) + x \sin(x) - \cos(x)) + (5 \sin(x) - x \cos(x))(5 \sin(x) - x \cos(x) - \sin(x)) \\ &= (x \sin(x) + 5 \cos(x)) ( 5 \cos(x) + x \sin(x) - \cos(x)) - (5 \sin(x) - x \cos(x))( x \cos(x) + \sin(x) - 5 \sin (x)) \\ &= (x \sin(x) + 5 \cos(x)) \dfrac{d(5 \sin(x) - x \cos(x))}{dx} - (5 \sin(x) - x \cos(x)) \dfrac{d(x \sin(x) + 5 \cos(x))}{dx} \end{align*}

Let $$u=x \sin(x) + 5 \cos(x)$$ and $$v=5 \sin(x) - x \cos(x)$$. So, $$x^2 + 20 = u\dfrac{dv}{dx} - v\dfrac{du}{dx} = u^2 \dfrac{d\left(\dfrac{v}{u}\right)}{dx}$$.

So, $\int \dfrac{x^2 + 20}{(x \sin(x) + 5 \cos(x))^2} dx = \int \dfrac{ u^2 \dfrac{d\left(\dfrac{v}{u}\right)}{dx}}{u^2} dx= \dfrac{v}{u} = \dfrac{5 \sin(x) - x \cos(x)}{x \sin(x) + 5 \cos(x)}$

- 2 years, 5 months ago

$$\text{Problem 3}$$

Evaluate:

$\large {\displaystyle \int^{\frac{\pi}{2}}_{0} \tan^{\frac{3}{5}} x \, dx}$

###### This problem has been solved by Aditya Kumar.

- 2 years, 5 months ago

Solution to Problem 3: $$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \left( tanx \right) }^{ \frac { 3 }{ 5 } } } dx=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \left( sinx \right) }^{ \frac { 3 }{ 5 } }{ \left( cosx \right) }^{ \frac { -3 }{ 5 } } } dx\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { B\left( \frac { 4 }{ 5 } ,\frac { 1 }{ 5 } \right) }{ 2 } ...........\left\{ B(x,y)\quad is\quad beta\quad function \right\} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { \Gamma \left( \frac { 4 }{ 5 } \right) \Gamma \left( \frac { 1 }{ 5 } \right) }{ 2\Gamma \left( 1 \right) } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { \pi }{ 2sin\left( \frac { \pi }{ 5 } \right) } \quad ...............\{ by\quad euler's\quad reflection\quad formula\} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { 2\pi }{ \sqrt { 10-2\sqrt { 5 } } }$$

- 2 years, 5 months ago

Problem 15:

Find the closed form for $$I(x)$$, where $I(x) = \large \int_{-\pi /2}^{\pi /2} \cos(x\tan(\theta)) \, d\theta .$

###### This problem has been solved by Tanishq Varshney via contour integration. And Surya Prakesh posted his intended solution.

- 2 years, 5 months ago

Another interesting elementary solution :

Lemma :

$\large{\displaystyle \int_{0}^{\infty} \frac{\cos mx}{x^2+a^2} \mathrm{d} x=\frac{\pi}{2a} e^{-am}}$

Proof :

$$\text{I}(m) = \displaystyle \int_{0}^{\infty} \dfrac{\cos mx}{x^2+a^2} \mathrm{d} x \tag{1}$$

Using Integration by parts,

$$\text{I}(m) = \displaystyle \int_{0}^{\infty} \dfrac{2x\sin mx}{m(x^2+a^2)^2} \mathrm{d} x$$

$$\implies m \cdot \text{I}(m) = \displaystyle \int_{0}^{\infty} \dfrac{2x \sin mx}{(x^2+a^2)^2} \mathrm{d} x \tag {2}$$

Partially Differentiating $$(2)$$ w.r.t. $$m$$, we have,

$$\text{I} + m\dfrac{\partial}{\partial m} \text{I} = \displaystyle \int_{0}^{\infty} \dfrac {2x^2\cos(mx)}{(x^2+a^2)^2} \mathrm{d}x = \displaystyle \int_{0}^{\infty} \dfrac {2\cos(mx)}{(x^2+a^2)} \mathrm{d}x - 2a^2\displaystyle \int_{0}^{\infty}\dfrac{\cos mx}{(x^2+a^2)^2} \mathrm{d}x$$

$$\implies a\dfrac{\partial}{\partial m} \text{I} = \text{I} - 2b^2\displaystyle \int_{0}^{\infty}\dfrac{\cos mx}{(x^2+a^2)^2} \mathrm{d}x \tag{3}$$

Again, partially differentiating $$(3)$$ w.r.t. $$m$$, we have,

$$m\dfrac{\partial^2}{\partial m^2} \text{I} = a^2 \displaystyle \int_{0}^{\infty} \dfrac{2x \sin mx}{(x^2+a^2)^2} \mathrm{d} x$$

$$\implies \dfrac{\partial^2}{\partial m^2} \text{I} = a^2 \text{I} \tag{*}$$ (from $$(2)$$)

Note that $$I(0) = \dfrac{\pi}{2a}$$ and $$I(\infty) = 0$$

Now, solving $$(*)$$, we have,

$$I = \dfrac{\pi}{2a} e^{-am}$$

- 2 years, 5 months ago

Oh, I suddenly noticed I reproduced your method independently after several days and just didn't noticed that it is already posted here.

- 2 years, 4 months ago

Solution to $$Problem ~15$$

Lemma

$$\large{\displaystyle \int^{\infty}_{0} \frac{\cos mx}{x^2+a^2} dx=\frac{\pi}{2a} e^{-am}}$$

$$\text{Proof}$$

Consider the integral $$\large{\displaystyle \int _{ C } f\left( z \right) dz}$$ where $$\large{f(z)=\frac{e^{imz}}{z^2+a^2}}$$ taken round the closed contour $$C$$ consisting of the upper half of a large circle $$|z|=R$$ and the real axis from $$-R$$ to $$R$$.

Poles of $$f(z)$$ are given by

$$z^2+a^2=0 \rightarrow z=\pm ai$$

The only pole which lies within the contour is at $$z=ai$$. The residue of $$f(z)$$ at $$z=ai$$

$$\large{= \displaystyle \lim_{z \to ai} \frac{(z-ai)e^{imz}}{z^2+a^2}= \displaystyle \lim_{z \to ai} \frac{e^{imz}}{z+ai}=\frac{e^{-am}}{2ai}}$$

Hence by cauchy residue theorem

$$\large{\displaystyle \int _{ C } f\left( z \right) dz=2 \pi i \times \text{sum of residues}}$$

$$\large{\displaystyle \int _{ C } f\left( z \right) dz=2 \pi i \times \frac{e^{-am}}{2ai} }$$

$$\large{\displaystyle \int _{ C }\frac{e^{imz}}{z^2+a^2} dz \rightarrow \displaystyle \int^{R} _{ -R }\frac{e^{imx}}{x^2+a^2} dx= \frac{ \pi e^{-am}}{a}}$$

Equating Real parts

$$\large{\displaystyle \int^{\infty} _{ -\infty }\frac{\cos mx}{x^2+a^2} dx= \frac{ \pi e^{-am}}{a}}$$

$$\large{\Longrightarrow \displaystyle \int^{\infty} _{ 0 }\frac{\cos mx}{x^2+a^2} dx= \frac{ \pi e^{-am}}{2a}}$$

Now in the integral put $$\tan \theta =t$$

$$\large{\displaystyle \int^{\infty} _{ - \infty }\frac{\cos xt}{t^2+1} dt=\pi e^{-x}}$$

I would like to see the solution which doesn't involve contour!!

- 2 years, 5 months ago

Lemma: $\int_{-\infty}^{\infty} \dfrac{\cos (xt)}{1+t^2} dt = \int_{-\infty}^{\infty} \dfrac{e^{-ixt}}{1+t^2} dt$

Try to prove this yourselves.

Consider,

$f(t) = \int_{-\infty}^{\infty} e^{-|x|} e^{ixt}dx = \dfrac{2}{1+t^2}$

Above integral is easy to evaluate.

So,

$f(t)=\dfrac{2}{1+t^2}$

Now by Fourier inversion formula,

$e^{-|x|} = \dfrac{1}{2\pi} \int_{-\infty}^{\infty} f(t) e^{-ixt} dt$

So we get that

$\int_{-\infty}^{\infty} \dfrac{\cos (xt)}{1+t^2} dt = \int_{-\infty}^{\infty} \dfrac{e^{-ixt}}{1+t^2} dt = \pi e^{-|x|}$

- 2 years, 5 months ago

Problem 14:

Evaluate $\Large {{\displaystyle \int^{1}_{0} \frac{1-x}{\ln x} \left(x+x^2+x^{2^2}+x^{2^3}+\cdots \right) \, dx}}.$

###### This problem has been solved by Surya Prakash.

- 2 years, 5 months ago

$\int_{0}^{1} \dfrac{(1-x)}{\ln (x)} ( x + x^2 + x^{2^2} + x^{2^3} + \ldots ) = \sum_{n=0}^{\infty} \int_{0}^{1} \dfrac{x^{2^n} - x^{2^n + 1}}{\ln (x)} dx$

Let $$f(a) = \int_{0}^{1} \dfrac{x^a}{\ln x} dx$$. Differentiate it w.r.t $$a$$. We get

$f'(a) = \int_{0}^{1} x^a dx = \dfrac{1}{a+1}$

So,

$\int_{0}^{1} \dfrac{x^{2^n} - x^{2^n + 1}}{\ln(x)} dx = \int_{2^n+1}^{2^n} f'(a) da = \ln \left(\dfrac{2^n + 1}{2^n + 2} \right)$

$\int_{0}^{1} \dfrac{x^{2^n} - x^{2^n + 1}}{\ln(x)} dx = \ln (2^n +1) - \ln (2^n +2)$

$\sum_{n=0}^{\infty} \int_{0}^{1} \dfrac{x^{2^n} - x^{2^n + 1}}{\ln(x)} dx = \lim_{N \rightarrow \infty} \sum_{n=0}^{N} \ln (2^n +1) - \ln (2^n +2) = \lim_{N \rightarrow \infty} \sum_{n=0}^{N} \ln (2^n +1) -\ln (2) - \ln (2^{n-1} +1)$

So, we have got a telescoping series which sums up to

$= \lim_{N \rightarrow \infty} -(N+1)\ln(2) - \ln(3/2) + \ln (2^N + 1) = -\ln(3)$

- 2 years, 5 months ago

Problem 12:

Prove or disprove:

If $$a$$ is not a natural number, then

$\large \int_0^{\pi /2} (\sin \theta)^{a-1} (\cos \theta)^{2t-a}\sin(2t\theta) \, d\theta = \dfrac{\pi \Gamma(2t-a)}{2\sin\left( \frac{\pi a}2\right) \Gamma(2t) \Gamma(-a) }.$

###### This problem has been solved by Tanishq Varshney.

- 2 years, 5 months ago

$$\large{Solution ~to~Problem ~12:}$$

$$\Large{f(x)=\frac{\sin (2 t \arctan \sqrt{x})}{\sqrt{x} (1+x)^t}=\displaystyle \sum^{\infty}_{k=0} \frac{\Gamma(2t+2k+1) \Gamma(k+1)}{\Gamma(2t) \Gamma(2k+2)}.\frac {(-x)^{k}}{k!}}$$

Using Ramanujan master theorem

$$\Large{\displaystyle \int^{\infty}_{0} x^{s-1} f(x)=\Gamma(s) \phi(-s)}$$

where, $$\phi(k)=\frac{\Gamma(2t+2k+1) \Gamma(k+1)}{\Gamma(2t) \Gamma(2k+2)}$$

$$x \rightarrow \tan^2 \theta$$

$$\Large{\displaystyle \int^{\frac{\pi}{2}}_{0} sin^{a-1} \theta (\cos \theta )^{2t-a} \cos (2t \theta) d \theta =\frac{\pi \Gamma(2t-a)}{2 \sin \left( \frac{\pi a}{2} \right) \Gamma(2t) \Gamma(-a)}}$$

- 2 years, 5 months ago

Problem 11:

Prove that

$\large \int_0^1 \dfrac{\ln(x+1)}{(x+1)(x^2 + 1)} \, dx = \dfrac1{192} (36 (\ln2)^2 + 12\pi (\ln2) - \pi^2 ).$

###### This problem has been solved by Aditya Kumar.

- 2 years, 5 months ago

Solution to Problem 11:

Pre-requisites:

$$I=\int \frac { -a\ln { \left( a+1 \right) } }{ \left( a-1 \right) \left( a^{ 2 }+1 \right) } da\\ { Apply\: Integral\: Substitution }:\quad \\ \int f\left( g\left( x \right) \right) \cdot g^{ ' }\left( x \right) dx=\int f\left( u \right) du,\: \quad u=g\left( x \right) \\ u=a\ln { \left( a+1 \right) } :\\ u=\frac { a }{ a+1 } +\ln { \left( a+1 \right) } da,\\ da=\frac { 1 }{ \frac { a }{ a+1 } +\ln { \left( a+1 \right) } } du\\ I=-\int \frac { u }{ \left( a-1 \right) \left( a^{ 2 }+1 \right) } \frac { 1 }{ \frac { a }{ a+1 } +\ln { \left( a+1 \right) } } du\\ I=-\frac { 1 }{ \left( a-1 \right) \left( a^{ 2 }+1 \right) \left( \frac { a }{ a+1 } +\ln { \left( a+1 \right) } \right) } \frac { u^{ 2 } }{ 2 } \\ On\quad substitution\quad \quad \\ I=-\frac { a^{ 2 }\ln { ^{ 2 } } \left( a+1 \right) }{ 2\left( a-1 \right) \left( a^{ 2 }+1 \right) \left( \frac { a }{ a+1 } +\ln { \left( a+1 \right) } \right) }$$

$$I\left( a \right) =\int _{ 0 }^{ 1 }{ \frac { ln\left( 1+ax \right) }{ \left( 1+x \right) \left( 1+{ x }^{ 2 } \right) } dx } \\ { I }^{ ' }\left( a \right) =\int _{ 0 }^{ 1 }{ \frac { x }{ \left( 1+x \right) \left( 1+ax \right) \left( 1+{ x }^{ 2 } \right) } dx } \\ On\quad splitting\quad as\quad partial\quad fractions\quad and\quad solving,\\ { I }^{ ' }\left( a \right) =\frac { -aln(a) }{ (a-1)({ a }^{ 2 }+1) } +\frac { ln2 }{ 2(a+1) } +\frac { ln2(1-a) }{ 4({ a }^{ 2 }+1) } \\ I\left( a \right) =\int _{ 0 }^{ 1 }{ \left( \frac { -aln(a) }{ (a-1)({ a }^{ 2 }+1) } +\frac { ln2 }{ 2(a+1) } +\frac { ln2(1-a) }{ 4({ a }^{ 2 }+1) } \right) da } \\ Now\quad on\quad solving\quad and\quad substituting\quad a=1,\quad we\quad get\\ I=\frac { 1 }{ 192 } \left( 36{ \left( ln2 \right) }^{ 2 }+12\pi \left( ln2 \right) -{ \pi }^{ 2 } \right)$$

- 2 years, 5 months ago

Problem 10:

Evaluate

$\large \int_{0}^{\pi/2} \dfrac{1}{(9 \tan^2 (x) + 16)^3} \, dx$

###### This problem has been solved by Pi Han Goh.

- 2 years, 5 months ago

Let $$I$$ denote the value of the integral. And let $$3\tan (x) = 4\tan(y)$$, differentiate with respect to $$x$$: $3\sec^2(x) = 4\sec^2(y) \left( \frac{dy}{dx} \right) \Rightarrow dx = \dfrac{4\sec^2(y)}{3\sec^2(x)} dy= \dfrac{12\sec^2(y)}{9\sec^2(x)} dy= \dfrac{12\sec^2(y)}{9\tan^2(x) + 9} dy= \dfrac{12\sec^2(y)}{16\tan^2(y) + 9} dy$

And of course, $$9\tan^2(x) + 16 = (3\tan(x))^2 + 16 = 16(\tan^2 (y) + 1) = 16\sec^2(y)$$.

When $$x = 0$$, $$y = 0$$; when $$x\to \frac\pi2^-$$, $$y \to \frac\pi2 ^-$$ as well. The integral becomes

$\begin{eqnarray} I &=& \int_0^{\pi/2} \dfrac{1}{(16\sec^2(y))^3} \cdot \dfrac{12\sec^2(y)}{16\tan^2(y) + 9} \, dy \\ &=& \dfrac{12}{16^3} \int_0^{\pi/2} \dfrac1{\sec^4(y)} \cdot \dfrac1{16\tan^2 (y) + 9} \, dy \\ &=& \dfrac{3}{1024} \int_0^{\pi/2} \dfrac{\cos^6(y)}{16\sin^2 (y) + 9\cos^2(y)} \, dy \\ &=& \dfrac{3}{1024} \int_0^{\pi/2} \dfrac{\cos^6(y)}{16(1-\cos^2(y)) + 9\cos^2(y)} \, dy \\ &=& \dfrac{3}{1024} \int_0^{\pi/2} \dfrac{\cos^6(y)}{16 - 7\cos^2(y)} \, dy. \\ \end{eqnarray}$

By long division, $$\dfrac{X^3}{16-7X} = -\dfrac17 X^2 - \dfrac{16}{49} X - \dfrac{4096}{343}\cdot \dfrac{1}{7X-16} - \dfrac{256}{343}$$. We contine from above,

$\begin{eqnarray} \dfrac{1024}3 I &=& \int_0^{\pi/2} \bigg [ -\dfrac17 \cos^4(y) - \dfrac{16}{49} \cos^2 (y) - \dfrac{4096}{343} \cdot \dfrac1{7\cos^2(y) - 16} - \dfrac{256}{343} \bigg ] \, dy \\ \dfrac{1024}3 I &=& -\dfrac17\int_0^{\pi/2} \cos^4(y) \, dy - \dfrac{16}{49} \int_0^{\pi/2} \cos^2 (y) \, dy - \dfrac{8192}{343} \cdot \int_0^{\pi/2} \dfrac1{7(2\cos^2(y)) - 32} \, dy - \dfrac{128\pi}{343} \\ &=& -\left(\dfrac17 \cdot \dfrac{3!!}{4!!} \cdot \dfrac\pi2\right) - \left(\dfrac{16}{49}\cdot \dfrac{1!!}{2!!} \cdot \dfrac{\pi}2\right) - \dfrac{8192}{343} \int_0^{\pi/2} \dfrac1{7\cos(2y) - 25} \, dy -\left(\dfrac{128\pi}{343} \right) \\ &=& -\dfrac{2643\pi}{5488} + \dfrac{8192}{343} \int_0^{\pi/2}\dfrac1{25 - 7\cos(2y)} \, dy. \\ \end{eqnarray}$

For the final integral, let $$t = \tan(y) \Rightarrow dy = \dfrac{dt}{t^2 + 1}$$, and the integral becomes

$\begin{eqnarray} \int_0^{\pi/2}\dfrac1{25 - 7\cos(2y)} \, dy &=& \int_0^\infty \dfrac{1}{25 - 7 \left( \frac{1-t^2}{1+t^2}\right) } \cdot \dfrac{dt}{t^2 + 1} \\ &=& \int_0^\infty \dfrac{t^2 + 1} {25 + 2t^2 - 7t + 7t^2} \cdot \dfrac{dt}{t^2 + 1} \\ &=& \int_0^\infty \dfrac{1}{32t^2 + 18} \, dt \\ &=& \dfrac1{32}\int_0^\infty \dfrac{1}{t^2 + (3/4)^2} \, dt \\ &=& \dfrac1{32} \cdot \left. \dfrac43 \tan^{-1} \left( \dfrac43 t \right) \right |_0^{t\to\infty} = \dfrac\pi{48}. \\ \end{eqnarray}$

Simplify everything and we get the answer of $$I = \boxed{\dfrac{263 \pi}{5619712}} \approx 0.00014702512653568645897079$$.

As Surya Prakesh has pointed out, there's an easier method.

It can be easily proved that $$\int_{0}^{\pi /2} \dfrac{1}{\tan^2 (x ) + p} dx = \dfrac{\pi}{2(p+\sqrt{p})}$$. Then differentiate it twice w.r.t $$p$$. Then take $$p= \dfrac{4}{3}$$.

- 2 years, 5 months ago

Problem 8:

Evaluate

$\large f(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-a^2t^2}e^{i\omega t}dt$

###### This problem has been solved by both Tanishq Varshney (first) and Surya Prakash (second) almost at the same time.

- 2 years, 5 months ago

$$\large{\text{Solution to problem 8}}$$

$$t \rightarrow -t$$

adding this with the original equation

$$\Large{2f(w)=\frac{2}{\sqrt{2 \pi}} \displaystyle \int^{\infty}_{-\infty}e^{-a^2 t^2} \cos wt dt}$$

as $$e^{iwt}+e^{-iwt}=2 \cos wt=2 \Re(e^{iwt})$$

$$\Large{f(w)=\frac{1}{\sqrt{2 \pi}} \Re \left(\displaystyle \int^{\infty}_{-\infty}e^{-a^2 t^2}e^{iwt} dt \right)}$$

Let $$at=z$$

$$\Large{f(w)=\frac{1}{a \sqrt{2 \pi}} \Re \left(\displaystyle \int^{\infty}_{-\infty}e^{-z^2}e^{i\frac{w}{a}z} dz \right)}$$

$$\Large{f(w)=\frac{1}{a \sqrt{2 \pi}} \Re \left(\displaystyle \int^{\infty}_{-\infty}e^{-(z-\frac{iw}{2a})^2}e^{-\frac{w^2}{4a^2}} dz \right)}$$

Using Gaussian integral

$$\Large{f(w)=\frac{1}{a \sqrt{2}} e^{-\frac{w^2}{4a^2}}}$$

- 2 years, 5 months ago

Comment deleted Nov 30, 2015

Alternative solution to Problem 8:

$f(\omega)=\frac 1{\sqrt{2\pi}} \int_{-\infty}^{\infty}e^{-a^2t^2+i \omega t}dt=\frac 1{\sqrt{2\pi}} e^{-\frac{\omega^2}{4a^2}}\int_{-\infty}^{\infty}e^{-a^2\left(t-\frac{i\omega}{2a^2}\right)^2}dt=\frac 1{\sqrt{2\pi}} e^{-\frac{\omega^2}{4a^2}}\int_{-\infty}^{\infty}e^{-a^2t^2}dt$ We get this upon shifting $$t \to t+\frac{i\omega}{2a^2}$$ which is justified by applying Cauchy’s theorem to the rectangle with vertices $$-T, T, T+\frac{i\omega}{2a^2}, -T+\frac{i\omega}{2a^2}$$ for $$T \to \infty$$ noting that the integrand has no singularities in this region. Therefore $\frac 1{\sqrt{2\pi}} e^{-\frac{\omega^2}{4a^2}}\int_{-\infty}^{\infty}e^{-a^2t^2}dt =\frac 1{\sqrt{2\pi}} e^{-\frac{\omega^2}{4a^2}}\frac{\sqrt{\pi}}a=\frac{1}{a\sqrt2}e^{\frac{-\omega^2}{4a^2}}$

- 2 years, 5 months ago

$f(\omega) = \dfrac{1}{\sqrt{2 \pi}} \int_{- \infty} ^{\infty} e^{-a^2 t^2} e^{i\omega t}dt$

It implies that $$f(\omega) = \big[\mathcal{F}(F) \big](\omega)$$, where $$F(t) = e^{-a^2 t^2}$$.

Now, differentiate $$F(t)$$, we get $$F'(t) = -2a^2 t F(t)$$.

Applying fourier transformation on both sides gives $$i \omega f(\omega) = -2i a^2 f'(\omega)$$.

$\implies \omega f(\omega) = -2a^2 \dfrac{d f(\omega)}{d \omega} \\ \implies f(\omega) = c e^{-\dfrac{\omega^{2}}{4a^{2}}}$

Since $$c = f(0) =\dfrac{1}{\sqrt{2 \pi}} \int_{- \infty} ^{\infty} e^{-a^2 t^2} dt = 2 \dfrac{1}{\sqrt{2 \pi}} \int_{0} ^{\infty} e^{-a^2 t^2} dt =\dfrac{1}{a \sqrt{2 \pi}} \Gamma(1/2) = \dfrac{1}{a\sqrt{2}}$$.

Therefore, $$f(\omega) =\dfrac{1}{a\sqrt{2}} e^{-\dfrac{\omega^{2}}{4a^{2}}}$$.

- 2 years, 5 months ago

Problem 6:
Show that for positive reals, $$a , b$$ where $$a<b$$, $\int_a^b \arccos \left( \frac{x}{\sqrt{(a+b)x - ab}} \right) \, dx = \frac{\pi}{4} \frac{ (b-a)^2}{a+b}$

Statutory warning: The substitutions need not be very intuitive.

###### This problem has been solved Aditya Kumar.

- 2 years, 5 months ago

Solution to Problem 6:

$$I=\int _{ a }^{ b }{ \text{arccos}\left( \frac { x }{ \sqrt { \left( a+b \right) x-ab } } \right) dx } \\ I=\int _{ a }^{ b }{ \text{arccos}\left( \frac { \frac { x }{ \sqrt { ab } } }{ \sqrt { \left( \frac { \left( a+b \right) }{ ab } \right) x-1 } } \right) dx } \\ \sqrt { \left( \frac { \left( a+b \right) }{ ab } \right) x-1 } =t\\ I=\int _{ \sqrt { \frac { a }{ b } } }^{ \sqrt { \frac { b }{ a } } }{ \left( \frac { 2t }{ a+b } arccos\left( \frac { \left( { t }^{ 2 }+1 \right) \sqrt { ab } }{ t } \right) \right) dt } \\ I=\int _{ \sqrt { \frac { a }{ b } } }^{ \sqrt { \frac { b }{ a } } }{ \left( \frac { 2t }{ a+b } arccos\left( \left( { t }+\frac { 1 }{ t } \right) \sqrt { ab } \right) \right) dt }$$

Now using Integration By Parts,

$$I=\int _{ \sqrt { \frac { a }{ b } } }^{ \sqrt { \frac { b }{ a } } }{ \left( \frac { { t }^{ 2 }+1 }{ { \left( \sqrt { \frac { b }{ a } } -\sqrt { \frac { a }{ b } } \right) }^{ 2 }-{ \left( { t }-\frac { 1 }{ t } \right) }^{ 2 } } \right) dt } \\ t=p+\sqrt { 1+{ p }^{ 2 } } \\ I=\int _{ \frac { 1 }{ 2 } \left( -\sqrt { \frac { b }{ a } } +\sqrt { \frac { a }{ b } } \right) }^{ \frac { 1 }{ 2 } \left( \sqrt { \frac { b }{ a } } -\sqrt { \frac { a }{ b } } \right) }{ \left( \frac { { 2p }^{ 2 } }{ \sqrt { \frac { \frac { a }{ b } +\frac { b }{ a } +2 }{ 4 } -{ p }^{ 2 } } } \right) dp }$$

Now,

$${ p }^{ 2 }=\left( \frac { \frac { a }{ b } +\frac { b }{ a } +2 }{ 4 } \right) sin\theta$$

This can be evaluated easily.

$$\therefore \quad I=\frac { \pi }{ 4 } \frac { { \left( b-a \right) }^{ 2 } }{ \left( a+b \right) }$$

Moral of the question:
To whatever extent we learn anything, we must never forget our basics.
Note that this question is solved using very basic concepts.

Any elegant modification to the solution will be accepted.

- 2 years, 5 months ago

Problem 25

Prove that

$\large{\displaystyle \int^{\frac{\pi}{2}}_{0}\frac{\ln(\sin x) \ln(\cos x)}{\tan x} dx=\frac{1}{8} \zeta (3)}.$

###### This problem has been solved by Julian Poon.

- 2 years, 5 months ago

Substitute $$t \to \sin(x)$$. The integral becomes:

$\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \ln { t } \ln { \left( 1-t^{ 2 } \right) } \left( \frac { 1 }{ t } \right) } dt$

Substitute in the Taylor series for $$\ln (1-x)$$:

$-\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \ln { t } \sum _{ n=1 }^{ \infty } \frac { t^{ 2n-1 } }{ n } } dt=-\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty } \frac { 1 }{ n } \int _{ 0 }^{ 1 }{ \ln { (t) } t^{ 2n-1 } } dt$

Finally, integrate by parts to obtain

$-\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty } \frac { 1 }{ n } \frac { -1 }{ 4{ n }^{ 2 } } =\frac { 1 }{ 8 } \sum _{ n=1 }^{ \infty } \frac { 1 }{ { n }^{ 3 } }$

- 2 years, 5 months ago

Alternative approach could be (i know its too late :P) could be considering Beta function $$(\dfrac{\Gamma(a) \Gamma(b)}{ \Gamma(a+b)})$$ in trignometric form and differentiating it with respect to a and b and then putting a = 0 and b = 1.

- 2 years, 4 months ago

Great! That's the standard approach! Can you post the full solution?

- 2 years, 4 months ago

$$\displaystyle \frac { \Gamma (a)\Gamma (b) }{ \Gamma (a+b) } = 2\int _{ 0 }^{ \pi /2 }{ \sin ^{ 2a-1 }{ x } \cos ^{ 2b-1 }{ x } dx }$$

Differentiating both sides first wrt a and then wrt b, we get

$$\displaystyle \frac { \Gamma (a)\Gamma (b) }{ \Gamma (a+b) } (((\psi (a)-\psi (a+b))(\psi (b)-\psi (a+b))-\psi '(a+b)) \displaystyle =8\int _{ 0 }^{ \pi /2 }{ log(sin(x))log(cos(x))\sin ^{ 2a-1 }{ x } \cos ^{ 2b-1 }{ x } dx }$$

Putting a = 0 and b = 1,

$$-\psi ' (1) = 8\int^{\frac{\pi}{2}}_{0}\frac{\ln(\sin x) \ln(\cos x)}{\tan x} dx$$

- 2 years, 4 months ago

Even I got the same. Now your job is to prove it equal to 1/8 zeta(3).

- 2 years, 4 months ago

I don't have any idea how to proceed from here.

- 2 years, 4 months ago

Hint: What is the definition of a digamma function?

- 2 years, 4 months ago

@Harsh Shrivastava Want to join our discussion group where we talk bout integrals and series?

- 2 years, 4 months ago

Well I first need to clear my basics of series, so maybe after March or so, I'll join the group, BTW if you guys discuss from basics, then I'll surely join right now.

- 2 years, 4 months ago

Well, you can ask simple questions as well, we're happy to explain to you... Still want to join?

- 2 years, 4 months ago

Yeah sure! I'll join right after my school , I have to go to school now, Thanks!

- 2 years, 4 months ago

Go here

- 2 years, 4 months ago

When i enter my email address, it says already invited, what to do now? I am not able to access further.

- 2 years, 4 months ago

How's the registration coming along?

- 2 years, 4 months ago

- 2 years, 4 months ago

Step 1: GO here and type your email.
Step 2: GO check your email which you have typed in.
Step 3: Click that email link (for verification...). Done!

If none of these seem to work, register another email address and repeat steps 1 to 3.

- 2 years, 4 months ago

- 2 years, 4 months ago

Problem 19:

Prove that

$\large{\displaystyle \int^{\frac{\pi}{3}}_{0} \ln^{2} \left( \frac{\sin x}{\sin \left(\frac{\pi}{3}+x \right)} \right) \, dx=\frac{5}{81} \pi^{3}}.$

###### This problem has been solved by Surya Prakesh.

- 2 years, 5 months ago

Alternate Solution to problem 19 :

Lemma: $$\displaystyle \int _{ 0 }^{ \infty }{ \frac { { x }^{ m-1 }{ ln }^{ 2 }(x) }{ 1+{ x }^{ n } } dx } ={ \left(\frac { \pi }{ n } \right) }^{ 3 }\left(\csc { \left(\frac { m\pi }{ n } \right) } \cot ^{ 2 }{ \left(\frac { m\pi }{ n } \right) } +\csc ^{ 3 }{ \left(\frac { m\pi }{ n } \right) } \right)\quad$$

Proof : We begin with identity $$\displaystyle I = \int _{ 0 }^{ \infty }{ \frac { dx }{ 1+{ x }^{ n } } } =\frac { \pi }{ n } \csc { (\frac { \pi }{ n } ) }$$

To prove this put $$\displaystyle y = \dfrac{1}{1+{x}^{n}}$$ to get :

$$\displaystyle I =\dfrac{1}{n} \int _{ 0 }^{ 1 }{ { y }^{ -1/n }{ (1-y) }^{ 1/n-1 }dy }$$

Use definition of beta function and euler reflection formula to get :

$$\displaystyle I = \dfrac{1}{n}\int _{ 0 }^{ 1 }{ { y }^{ -1/n }{ (1-y) }^{ 1/n-1 }dy } =\dfrac{1}{n}\frac { \Gamma (1/n)\Gamma (1-1/n) }{ \Gamma (1) } =\frac { \pi }{ n } \csc { (\frac { \pi }{ n } ) }$$

Now consider the integral $$\displaystyle J(m) = \int _{ 0 }^{ \infty }{ \frac { { x }^{ m-1 }dx }{ 1+{ x }^{ n } } }$$

Put $${x}^{m}=y$$ to get :

$$\displaystyle J(m) = \frac { 1 }{ m } \int _{ 0 }^{ \infty }{ \frac { dy }{ 1+{ y }^{ n/m } } }$$

Use the identity to get :

$$\displaystyle J(m) = \frac { \pi }{ n } \csc { (\frac { m\pi }{ n } ) }$$

Differentiate it two times with respect to $$m$$ to prove the lemma :

My initial steps are the same as surya prakash and I directly get till :

$$\displaystyle I = \frac { \sqrt { 3 } }{ 2 } \int _{ 0 }^{ 1 }{ \frac { (1+y){ ln }^{ 2 }(y) }{ 1+{ y }^{ 3 } } dy }$$

Put $$\displaystyle y=\dfrac{1}{x}$$ to get :

$$\displaystyle I = \frac { \sqrt { 3 } }{ 2 } \int _{ 1 }^{ \infty }{ \frac { (1+x){ ln }^{ 2 }(x) }{ 1+{ x }^{ 3 } } dx }$$

Adding these two forms we have :

$$\displaystyle I = \frac { \sqrt { 3 } }{ 4 } \int _{ 0 }^{ \infty }{ \frac { (1+x){ ln }^{ 2 }(x) }{ 1+{ x }^{ 3 } } dx }$$

$$\displaystyle I = \frac { \sqrt { 3 } }{ 4 } (\int _{ 0 }^{ \infty }{ \frac { { ln }^{ 2 }(x) }{ 1+{ x }^{ 3 } } dx } +\int _{ 0 }^{ \infty }{ \frac { x{ ln }^{ 2 }(x) }{ 1+{ x }^{ 3 } } dx } )$$

Use the lemma to get :

$$\displaystyle I = \frac { \sqrt { 3 } }{ 4 } ({ (\frac { \pi }{ 3 } ) }^{ 3 }(\frac { 2 }{ \sqrt { 3 } } .\frac { 1 }{ 3 } +\frac { 8 }{ 3\sqrt { 3 } } )+{ (\frac { \pi }{ 3 } ) }^{ 3 }(\frac { 2 }{ \sqrt { 3 } } .\frac { 1 }{ 3 } +\frac { 8 }{ 3\sqrt { 3 } } ))=\frac { 5{ \pi }^{ 3 } }{ 81 }$$

- 2 years, 4 months ago

Lemma: $\int_{0}^{1} x^{m} \ln ^{n} (x) dx = \dfrac{(-1)^n n!}{(m+1)^{n+1}}$

Let $$y=\dfrac{\sin x}{\sin \left(x+\pi/3 \right)} \implies \tan x = \dfrac{\sqrt{3}y}{2-y} \implies dx = \dfrac{\sqrt{3}}{2} \dfrac{1}{y^2 - y +1}dy$$.

So,

\begin{align*} \int_{0}^{\pi/3} \ln ^2 \left(\dfrac{\sin x}{\sin \left(\dfrac{\pi}{3} + x \right) } \right) dx &= \dfrac{\sqrt{3}}{2} \int_{0}^{1} \dfrac{\ln ^2 (y)}{y^2 - y +1} dy \\ &= \dfrac{\sqrt{3}}{2} \int_{0}^{1} \dfrac{(1+y)\ln ^2 (y)}{1+y^3} dy \end{align*}

Since, $$0 < y < 1$$, we can use the expansion $$\dfrac{1}{1+y^3} = 1-y^3 + y^6 + \ldots = \sum_{k=0}^{\infty} (-1)^k y^{3k}$$.

So, the integral becomes,

\begin{align*} \int_{0}^{1} \dfrac{(1+y)\ln ^2 (y)}{1+y^3} dy &= \int_{0}^{1} (1+y)\ln ^2 (y) \sum_{k=0}^{\infty} (-1)^k y^{3k} dy \\ &= \sum_{k=0}^{\infty} (-1)^k \Bigr[ \int_{0}^{1} y^{3k} \ln^2 (y) dy + \int_{0}^{1} y^{3k+1} \ln^2 (y) dy \Bigr] \\ &= 2 \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+1)^3} + 2 \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+2)^3} \end{align*}

So,

$\int_{0}^{\pi/3} \ln ^2 \left(\dfrac{\sin x}{\sin \left(\dfrac{\pi}{3} + x \right) } \right) dx = \sqrt{3} \Bigr[ \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+1)^3} + \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+2)^3} \Bigr]$

Here comes the tricky part,

Consider the series $$(0< q< 1)$$

$\dfrac{1}{q^3} - \dfrac{1}{(1+q)^3} + \dfrac{1}{(2+q)^3} - \ldots$

It is easy to prove that above series is equal to $$\dfrac{1}{8} \left( \zeta \left(3, \dfrac{q}{2} \right) - \zeta \left(3, \dfrac{q+1}{2} \right) \right)$$, where $$\zeta(s,q)$$ is Hurwitz Zeta Function.

$\dfrac{1}{q^3} - \dfrac{1}{(1+q)^3} + \dfrac{1}{(2+q)^3} - \ldots = \dfrac{1}{16} \left(\psi_{(2)} \left(\dfrac{q+1}{2} \right) - \psi_{(2)} \left(\dfrac{q}{2} \right) \right)$

So, finally

\begin{align*} \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+1)^3} + \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+2)^3} &= \dfrac{1}{27} \left( \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(k+1/3)^3} + \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(k+2/3)^3} \right) \\ &= \dfrac{1}{27} \dfrac{1}{16} \left(\psi_{(2)} \left(\dfrac{1}{6}\right)-\psi_{(2)}\left(\dfrac{2}{3}\right)+\psi_{(2)}\left(\dfrac{1}{3}\right)-\psi_{(2)}\left(\dfrac{5}{6}\right) \right) = \dfrac{5 \sqrt{3} \pi^3}{243} \end{align*}

Above final calculation is done using Euler's Reflection Formula for polygamma function

So, finally

$\int_{0}^{\pi/3} \ln ^2 \left(\dfrac{\sin x}{\sin \left(\dfrac{\pi}{3} + x \right) } \right) dx = \sqrt{3} \Bigr[ \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+1)^3} + \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+2)^3} \Bigr] = \sqrt{3} \dfrac{5 \sqrt{3}\pi^3}{243} = \dfrac{5 \pi^3}{81}$

Sorry for missing the calculation part, As it became too lengthy to type here.

Notify me if there are any typing mistakes.

- 2 years, 5 months ago

Thanks for everyone. Because of this competition and you guys I learnt a lot of integrations skills and develop my integration skills so fast. I thank everyone. @Pi Han Goh @Aditya Kumar @Tanishq Varshney @Sudeep Salgia @Abhishek Bakshi

- 2 years, 5 months ago

Problem 7:

For $$0 < v < 1$$ independent of $$x$$, prove that: $\large \int _{ 0 }^{ \infty }{ \left[ { x }^{ v-3 }\left( \gamma x+\log\Gamma \left( 1+x \right) \right) \right] \, dx } =\frac { \pi }{ \sin\left( \pi \left( v \right) \right) }\cdot \frac { \zeta \left( 2-v \right) }{ 2-v }.$

###### Due to time constraint, the author decided to post the solution.

- 2 years, 5 months ago

Solution to Problem 7:

Lemma:

$$\Gamma \left( x \right) =\frac { { e }^{ -\gamma x } }{ x } \prod _{ n=1 }^{ \infty }{ { \left( 1+\frac { x }{ n } \right) }^{ -1 }{ e }^{ \frac { x }{ n } } }$$

Proof:

$$\Gamma \left( x \right) =\lim _{ n\rightarrow \infty }{ \left( \frac { { n }^{ x } }{ x } \prod _{ i=1 }^{ n }{ \left( \frac { i }{ x+i } \right) } \right) } \\ \Gamma \left( x \right) =\lim _{ n\rightarrow \infty }{ \left( \frac { { n }^{ x } }{ x } \prod _{ i=1 }^{ n }{ { \left( 1+\frac { x }{ i } \right) }^{ -1 } } \right) } \\ \Gamma \left( x \right) =\lim _{ n\rightarrow \infty }{ \left( \frac { { e }^{ xlog\left( n \right) } }{ x } \prod _{ i=1 }^{ n }{ { \left( 1+\frac { x }{ i } \right) }^{ -1 } } \right) } \\ \Gamma \left( x \right) =\lim _{ n\rightarrow \infty }{ \left( \frac { { e }^{ \sum _{ i=1 }^{ \infty }{ \left( \frac { x }{ n } \right) -\sum _{ i=1 }^{ \infty }{ \left( \frac { x }{ n } \right) } + } xlog\left( n \right) } }{ x } \prod _{ i=1 }^{ n }{ { \left( 1+\frac { x }{ i } \right) }^{ -1 } } \right) } \\ But,\quad \gamma =\lim _{ n\rightarrow \infty }{ \left( \sum _{ i=1 }^{ \infty }{ \left( \frac { 1 }{ n } \right) } -log\left( n \right) \right) } \\ \Gamma \left( x \right) =\lim _{ n\rightarrow \infty }{ \left( \frac { { e }^{ \sum _{ i=1 }^{ \infty }{ \left( \frac { x }{ n } \right) -\gamma x } } }{ x } \prod _{ i=1 }^{ n }{ { \left( 1+\frac { x }{ i } \right) }^{ -1 } } \right) } \\ \Gamma \left( x \right) =\lim _{ n\rightarrow \infty }{ \left( \frac { { e }^{ -\gamma x } }{ x } \prod _{ i=1 }^{ n }{ { { e }^{ \left( \frac { x }{ n } \right) }\left( 1+\frac { x }{ i } \right) }^{ -1 } } \right) }$$

Now taking log on both sides and replacing x by x+1, we get

$$\log { \left( \Gamma \left( 1+x \right) \right) } =-\gamma x+\sum _{ k=2 }^{ \infty }{ \left( \frac { \zeta \left( k \right) }{ k } { \left( -x \right) }^{ k } \right) }$$

Now plugging these values into the question and by using Ramanujan's master theorem, we get

$\int _{ 0 }^{ \infty }{ \left[ \frac { { x }^{ v-1 }\left( \gamma x+\log \Gamma \left( 1+x \right) \right) }{ { x }^{ 2 } } \right] \, dx } =\frac { \pi }{ \sin \left( \pi \left( v \right) \right) } \cdot \frac { \zeta \left( 2-v \right) }{ 2-v }$

- 2 years, 5 months ago

Alternate Solution to Problem 7

Lemma : $$\displaystyle \int _{ 0 }^{ \infty }{ \frac { { x }^{ n } }{ { e }^{ x }-1 } dx } =\Gamma (n+1)\zeta (n+1)$$

Proof :$$\displaystyle I = \int _{ 0 }^{ \infty }{ \frac { { x }^{ n } }{ { e }^{ x }-1 } dx } =\int _{ 0 }^{ \infty }{ \frac { { x }^{ n }{ e }^{ -x } }{ 1-{ e }^{ -x } } dx } =\int _{ 0 }^{ \infty }{ { x }^{ n }\sum _{ r=1 }^{ \infty }{ { e }^{ -rx } } dx } =\sum _{ r=1 }^{ \infty }{ \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -rx }dx } }$$

Note that I have used the series expansion of $$\dfrac{y}{1-y}$$ and interchanged summation and integral.

$$\displaystyle I = \sum _{ r=1 }^{ \infty }{ \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -rx }dx } } =\quad \sum _{ r=1 }^{ \infty }{ \frac { \Gamma (n+1) }{ { r }^{ n+1 } } } =\Gamma (n+1)\zeta (n+1)$$

I have here used the definitions of gamma function and the zeta function.

Now back to our problem.

Integrate it by parts to get :

$$\displaystyle I = \frac { { x }^{ v-2 } }{ v-2 } (\gamma x+\log {\Gamma (1+x) } ){ | }_{ 0 }^{ \infty }+\frac { 1 }{ 2-v } \int _{ 0 }^{ \infty }{ { x }^{ v-2 }(\gamma +\psi (1+x))dx } = \frac { 1 }{ 2-v } \int _{ 0 }^{ \infty }{ { x }^{ v-2 }(\gamma +\psi (1+x))dx }$$

Using the definition that $$\displaystyle \gamma +\psi (1+x)=\int _{ 0 }^{ 1 }{ \frac { 1-{ y }^{ x } }{ 1-y } dy }$$ , we get :

$$\displaystyle I = \frac { 1 }{ 2-v } \int _{ 0 }^{ \infty }{ \int _{ 0 }^{ 1 }{ { x }^{ v-2 }(\frac { 1-{ y }^{ x } }{ 1-y } )dxdy } }$$

Changing the order of integration we have :

$$\displaystyle I = \frac { 1 }{ 2-v } \int _{ 0 }^{ 1 }{ \frac { 1 }{ 1-y } \int _{ 0 }^{ \infty }{ { x }^{ v-2 }(1-{ y }^{ x })dxdy } }$$

Integrating it by parts we have :

$$\displaystyle I = \frac { 1 }{ 2-v } (\int _{ 0 }^{ 1 }{ \frac { dy }{ 1-y } (\frac { { x }^{ v-1 }(1-{ y }^{ x }) }{ v-1 } { | }_{ 0 }^{ \infty }+\frac { \ln(y) }{ v-1 } \int _{ 0 }^{ \infty }{ { x }^{ v-1 }{ y }^{ x }dx } ) } )$$

$$\displaystyle I = \frac { 1 }{ (v-1)(2-v) } (\int _{ 0 }^{ 1 }{ \frac { \ln(y) dy }{ 1-y } (\int _{ 0 }^{ \infty }{ { x }^{ v-1 }{ y }^{ x }dx } ) } )$$

$$\displaystyle I = \frac { 1 }{ (v-1)(2-v) } (\int _{ 0 }^{ 1 }{ \frac { \ln(y) dy }{ (1-y) } (\int _{ 0 }^{ \infty }{ { x }^{ v-1 }{ e }^{ -x(-\ln { (y)) } }dx } ) } )$$

$$\displaystyle \Rightarrow I = \frac { 1 }{ (v-1)(2-v) } (\int _{ 0 }^{ 1 }{ \frac { \ln { (y) } dy }{ (1-y){ (-\ln { (y)) } }^{ v } } (\int _{ 0 }^{ \infty }{ { x }^{ v-1 }{ e }^{ -x }dx } ) } )$$

$$\displaystyle \Rightarrow I = \frac { \Gamma (v) }{ (v-1)(2-v) } (\int _{ 0 }^{ 1 }{ \frac { \ln { (y) } dy }{ (1-y){ (-\ln { (y)) } }^{ v } } } )$$

Put $$y={e}^{-x}$$

$$\displaystyle \Rightarrow I = \frac { \Gamma (v) }{ (1-v)(2-v) } (\int _{ 0 }^{ \infty }{ \frac { { x }^{ 1-v }dx }{ { e }^{ x }-1 } } )$$

Using Lemma we have :

$$\displaystyle I = \frac { \Gamma (v)\Gamma (2-v)\zeta (2-v) }{ (1-v)(2-v) } =\frac { \Gamma (v)\Gamma (1-v)\zeta (2-v) }{ 2-v }$$

Using gamma reflection formula it becomes :

$$\displaystyle I = \frac { \pi \zeta (2-v) }{ \sin { (\pi v) } (2-v) }$$

Hence Proved.

- 2 years, 4 months ago

Problem 22:

Evaluate the integral

$\large \int_{0}^{\pi} \dfrac{x \sin (x)}{(\cos^2 (x) + 3)^2} dx .$

###### This problem has been solved by Aditya Kumar.

- 2 years, 5 months ago

$I=\int _{ 0 }^{ \pi }{ \frac { xsinx }{ { \left( { cos }^{ 2 }x+3 \right) }^{ 2 } } dx } \\ I=\frac { \pi }{ 2 } \int _{ 0 }^{ \pi }{ \frac { sinx }{ { \left( { cos }^{ 2 }x+3 \right) }^{ 2 } } dx } \\ cosx=t\quad \\ I=\frac { \pi }{ 2 } \int _{ -1 }^{ 1 }{ \frac { 1 }{ { \left( { t }^{ 2 }+3 \right) }^{ 2 } } dt } \\ t=\sqrt { 3 } tanx\\ I=\frac { \pi \sqrt { 3 } }{ 36 } \int _{ \frac { -\pi }{ 6 } }^{ \frac { \pi }{ 6 } }{ \left( cos2x+1 \right) dx } \\ \therefore I=\frac { \pi \sqrt { 3 } }{ 18 } \left\{ \frac { \sqrt { 3 } }{ 4 } +\frac { \pi }{ 6 } \right\}$

In the first step, I used: $$\int _{ a }^{ b }{ f(x) } =\int _{ a }^{ b }{ f(a+b-x) }$$

- 2 years, 5 months ago

Problem 21:

For $$|b|<1$$, find the value of the integral below in terms of $$b$$.

$\int_{-\pi /2}^{\pi /2} \dfrac{ \ln(1+ b \sin x)}{\sin x} \, dx$

###### This problem has been solved by Surya Prakesh.

- 2 years, 5 months ago

$f(b) = \int_{-\pi/2}^{\pi/2} \dfrac{\ln ( 1+ b \sin (x))}{\sin(x)} dx \\ f'(b) = \int_{-\pi/2}^{\pi/2}\dfrac{1}{1+ b\sin(x)} dx$

Using Weierstrass Substitution i.e. $$t = \tan (x/2) \implies \sin(x) = \dfrac{2t}{1+t^2}$$.

\begin{align*} f'(b) &= 2\int_{-1}^{1} \dfrac{dt}{t^2 + 2bt + 1}\\ &= 2\int_{-1}^{1} \dfrac{dt}{(t+b)^2 + (\sqrt{1-b^2})^2}\\ &= \dfrac{2}{\sqrt{1-b^2}} \Bigr[\arctan\left( \dfrac{t+b}{\sqrt{1-b^2}} \right) \Bigr|_{-1}^{1}\\ &=\dfrac{2}{\sqrt{1-b^2}} \Bigr[\arctan\left( \dfrac{1+b}{\sqrt{1-b^2}} \right) - \arctan\left( \dfrac{-1+b}{\sqrt{1-b^2}} \right)\Bigr] \\ &=\dfrac{2}{\sqrt{1-b^2}} \Bigr[\arctan\left( \dfrac{1+b}{\sqrt{1-b^2}} \right) + \arctan\left( \dfrac{1-b}{\sqrt{1-b^2}} \right)\Bigr] \\&= \dfrac{2}{\sqrt{1-b^2}} \Bigr[\arctan\left(\sqrt{\dfrac{1+b}{1-b}} \right) + \arctan \left(\sqrt{\dfrac{1-b}{1+b}} \right)\Bigr] \\&= \dfrac{\pi}{\sqrt{1-b^2}} \end{align*}

I used the fact that $$\arctan(x) + \arctan(1/x) = \pi/2$$.

So,

$f'(b) = \dfrac{\pi}{\sqrt{1-b^2}} \\ f(b) = \int \dfrac{\pi}{\sqrt{1-b^2}} db \\ f(b) = \pi \arcsin(b) +c$

But $f(b) = \int_{-\pi/2}^{\pi/2} \dfrac{\ln ( 1+ b \sin (x))}{\sin(x)} dx \implies f(0) = \int_{-\pi/2}^{\pi/2} \dfrac{\ln ( 1)}{\sin(x)} dx =0$

So, $$c=0$$. Therefore, $$f(b) = \pi \arcsin(b)$$.

- 2 years, 5 months ago

Problem 18:

Evaluate

$\large \int_{0}^{\infty} \dfrac{\log^2 (t)}{1+t^2} \, dt$

###### This problem has been solved by Tanishq Varshney (first) and Sudeep Salgia (second) almost at the same time.

- 2 years, 5 months ago

$$\text{Solution to problem 18}$$

$$\large{\displaystyle \int^{1}_{0} \frac{\ln^{2} (t)}{1+t^2} dt+\displaystyle \int^{\infty}_{1} \frac{\ln^{2} (t)}{1+t^2} dt}$$

In the latter integral $$t\rightarrow \frac{1}{t}$$

The whole expression becomes

$$\large{2 \displaystyle \int^{1}_{0} \frac{\ln^{2} (t)}{1+t^2} dt}$$

Now

$$\large{\displaystyle \int _{ 0 }^{ 1 }{ { x }^{ a } } dx=\frac { 1 }{ a+1 } \\ \displaystyle \int _{ 0 }^{ 1 }{ \frac { { \partial }^{ s } }{ { \partial x }^{ s } } } { x }^{ a }\quad dx=\displaystyle \int _{ 0 }^{ 1 }{ { x }^{ a } } { \ln { ^{ s } } { x } }dx=\frac { { \left( -1 \right) }^{ s }s! }{ { \left( a+1 \right) }^{ s+1 } } }$$

we have

$$\large{\frac{1}{1+t^2}=\displaystyle \sum^{\infty}_{r=0} (-1)^{r} t^{2r}}$$

$$\large{\longrightarrow 2 \displaystyle \int^{1}_{0} \displaystyle \sum^{\infty}_{r=0} \ln^{2} (t) (-1)^{r} t^{2r} dt}$$

$$\large{\longrightarrow 2 \displaystyle \sum^{\infty}_{r=0} (-1)^{r} \displaystyle \int^{1}_{0} t^{2r} \ln^{2} (t) dt}$$

$$\large{4 \displaystyle \sum^{\infty}_{r=0} \frac{(-1)^{r}}{(2r+1)^{3}}}$$

Use of Dirichlet Beta function

which gives $$\large{4 \beta(3)=\frac{\pi^{3}}{8}}$$

- 2 years, 5 months ago

Comment deleted Dec 02, 2015

Here's one without it: I will use the general form,

$\int_0^\infty\frac{x^{m-1}}{1+x^n}\,dx=\frac{\pi}{n\sin\frac{m\pi}{n}}$

Thus,
$$\displaystyle I(m) = \int_0^\infty\frac{t^{m-1}}{1+t^2}\,dt=\frac{\pi}{2}\csc\left(\frac{m\pi}{2}\right)$$.
The required integral is $$I''(1)$$ which when evaluated turns to be $$\dfrac{\pi ^3}{8}$$.

The general form used above can easily be proven by taking substitution $$\displaystyle y=\frac{1}{1+x^n}$$ and the integral becomes Beta function

$\frac{1}{n}\int_0^1 y^{\large 1-\frac{m}{n}-1}\ (1-y)^{\large \frac{m}{n}-1}\,dy=\frac{\Gamma\left(1-\frac{m}{n}\right)\Gamma\left(\frac{m}{n}\right)}{n}=\frac{\pi}{n\sin\frac{m\pi}{n}}$

where the last part comes from Euler's reflection formula for the gamma function.

The solution is extremely concise. :P Nice solution @Tanishq Varshney . Waiting for next problem.

- 2 years, 5 months ago

Problem 17:

Prove that

$\large \left(\int_{-\infty}^\infty \dfrac{x^2}{x^4 - x^3+x^2-x+1} \, dx \right) \div \left(\int_{-\infty}^\infty \dfrac{x}{x^4 - x^3+x^2-x+1} \, dx \right) = \dfrac{3+\sqrt5}2.$

###### This problem has been solved by Surya Prakesh.

- 2 years, 5 months ago

\begin{align*} \int_{-\infty}^{\infty}\dfrac{x^2}{x^4 - x^3 + x^2 - x+1} dx &=\int_{-\infty}^{\infty} \dfrac{x^3 + x^2}{x^5+1}dx \\ &= \int_{-\infty}^{0} \dfrac{x^3 + x^2}{x^5+1}dx + \int_{0}^{\infty} \dfrac{x^3 + x^2}{x^5+1} dx \\ &= \int_{0}^{\infty} \dfrac{x^2 - x^3}{1-x^5}dx + \int_{0}^{\infty} \dfrac{x^3 + x^2}{x^5+1} dx \\ &= 2 \int_{0}^{\infty} \dfrac{x^2 - x^8}{1-x^{10}} dx \end{align*}

\begin{align*} \int_{0}^{\infty} \dfrac{x^2 - x^8}{1-x^{10}} dx &= \int_{0}^{1} \dfrac{x^2 - x^8}{1-x^{10}} dx + \int_{1}^{\infty} \dfrac{x^2 - x^8}{1-x^{10}} dx \\ &= \int_{0}^{1} \dfrac{1+ x^2 - x^6 - x^8}{1-x^{10}}dx \end{align*}

I just took the substitution $$t=\dfrac{1}{x}$$ in the above steps.

So,

\begin{align*} \int_{-\infty}^{\infty}\dfrac{x^2}{x^4 - x^3 + x^2 - x+1} dx &=2 \int_{0}^{1} \dfrac{1+ x^2 - x^6 - x^8}{1-x^{10}}dx \\ &= \dfrac{1}{5} \int_{0}^{1} \dfrac{t^{-9/10} + t^{-7/10} - t^{-3/10} - t^{-1/10}}{1-t} dt \\ &= \dfrac{1}{5} \left( H_{-1/10} + H_{-3/10} - H_{-7/10} - H_{-9/10} \right) \\&=\dfrac{1}{5} ((\psi(9/10) - \psi(1/10)) + (\psi(7/10) - \psi(3/10))) \end{align*}

By euler reflection formula for Digamma function i.e. $$\psi(1-x) - \psi(x) = \pi \cot(\pi x)$$, we get

$(\psi(9/10) - \psi(1/10)) + (\psi(7/10) - \psi(3/10)) = \pi \cot(\pi/10) + \pi \cot(3\pi/10)$

So,

$\int_{-\infty}^{\infty}\dfrac{x^2}{x^4 - x^3 + x^2 - x+1} dx = \dfrac{1}{5} ((\psi(9/10) - \psi(1/10)) + (\psi(7/10) - \psi(3/10))) = \dfrac{\pi}{5}\cot(\pi/10) + \dfrac{\pi}{5} \cot(3\pi/10)$

Similarly, evaluating the integral in the denominator, we get

$\int_{-\infty}^{\infty}\dfrac{x}{x^4 - x^3 + x^2 - x+1} dx = \dfrac{2\pi}{5} \cot (3\pi /10)$

So finally

$\left(\int_{-\infty}^{\infty}\dfrac{x^2}{x^4 - x^3 + x^2 - x+1} dx \right) \div \left(\int_{-\infty}^{\infty}\dfrac{x}{x^4 - x^3 + x^2 - x+1} dx \right) \\ = \dfrac{\dfrac{\pi}{5}\cot(\pi/10) + \dfrac{\pi}{5} \cot(3\pi/10)}{\dfrac{2\pi}{5} \cot (3\pi /10)} = \dfrac{\cot(\pi/10)+\cot(3\pi/10)}{2 \cot (3\pi /10)}= \dfrac{3+\sqrt{5}}{2}$

Note: $$\cot (\pi/10) = \sqrt{5+2\sqrt{5}}$$ and $$\cot (3 \pi/10) = \sqrt{5-2\sqrt{5}}$$.

- 2 years, 5 months ago

Problem 16:

Evaluate

$\large{\displaystyle \int^{1}_{0} \frac{t \arccos (t)}{ 1+t^4} dt}.$

###### This problem has been solved by Pi Han Goh.

- 2 years, 5 months ago

This is very similar to Problem 5. Integrate by Parts: with $$dv = \dfrac x{1+x^4} \, dx, u = \cos^{-1}(x)$$, then $$du =- \dfrac1{\sqrt{1-x^2}} dx, v = \dfrac12 \tan^{-1}(x^2)$$.

$\begin{eqnarray} \int u \, dv &= &uv - \int v \, du \\ &=& \require{cancel} \cancelto{0}{ \left . \cos^{-1}(x) \cdot \dfrac12 \tan^{-1} (x^2) \right |_{x=0}^{x=1}}+ \dfrac12 \int_0^1 \dfrac {\tan^{-1}(x^2)}{\sqrt{1-x^2}} \, dx \quad\quad && \text{ let } x = \sin(y) \Rightarrow dy = \dfrac{dx}{\sqrt{1-x^2}} \\ &=& \dfrac12 \int_0^{\pi /2} \tan^{-1} (\sin^2 (y)) \, dy \quad\quad && \text{ let } z = \dfrac \pi2 - y \\ &=& \dfrac12 \int_0^{\pi /2} \tan^{-1} (1 \cdot \cos^2 (z)) \, dz \quad\quad && \text{ refer to solution to problem 5} \\ &=& \dfrac12 \dfrac \pi2 \tan^{-1} \left ( \sqrt{\dfrac{\sqrt{1^2+1}+1}{2} }\right) \\ &=& \dfrac\pi4 \tan^{-1} \left( \sqrt{\dfrac{\sqrt2-1}{2}}\right) \approx 0.335426737379 \\ \end{eqnarray}$

- 2 years, 5 months ago

Problem 20:

Evaluate

$\large \int_{0}^{\infty} \dfrac{dx}{(1+x)^3 +1}$

###### This problem has been solved by Gautam Sharma.

- 2 years, 5 months ago

Solution to problem 20

Let $(1+x)=t \\ I=\large \int \dfrac{dt}{t^3 +1}$

By partial fractions

$I=\int \frac { 2-t }{ 3({ t }^{ 2 }-t+1) } dt+\frac { 1 }{ 3(t+1) } dt \\ I=\int \frac { 1 }{ 2({ t }^{ 2 }-t+1) } dt-\frac { 2t-1 }{ 6({ t }^{ 2 }-t+1) } dt+\frac { 1 }{ 3(t+1) } dt \\ I=\int \frac { 1 }{ 2({ (t-\frac { 1 }{ 2 } ) }^{ 2 }+\frac { 3 }{ 4 } ) } dt-\frac { 2t-1 }{ 6({ t }^{ 2 }-t+1) } dt+\frac { 1 }{ 3(t+1) } dt \\ I=\int \frac { 1 }{ \sqrt { 3 } } \tan ^{ -1 }{ \frac { 2t-1 }{ \sqrt { 3 } } } -\frac { 1 }{ 6 } \ln({ t }^{ 2 }-t+1)+\frac { 1 }{ 3 } \ln(t+1)+c \\ I=\int \frac { 1 }{ \sqrt { 3 } } \tan ^{ -1 }{ \frac { 2x+1 }{ \sqrt { 3 } } } +\frac { 1 }{ 6 } \ln\left(\frac { { (x+2) }^{ 2 } }{ { x }^{ 2 }+x+1 } \right)+c$

Putting limits and evaluating we get $$\frac{\pi}{3\sqrt3}-\frac{\ln2}{3}$$

- 2 years, 5 months ago