# Brilliant Integration Contest - Season 1

I am interested in holding an Integration Contest here on Brilliant.org like any other online forums such as AoPS or Integrals and Series. 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

1. I will start by posting the first problem. If there is a user solves it, then (s)he must post a new one.
2. 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.
3. Please make a substantial comment.
4. Make sure you know how to solve your own problem before posting it in case there is no one can answer it within a week, then you must post the solution and you have a right to post another problem.
5. 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.
6. The scope of questions is only computation of integrals either definite or indefinite integrals.
7. You are NOT allowed to post a multiple integrals problem as well as a complex integral problem.
8. You are also NOT allowed to post a solution using a contour integration or residue method.
9. The final answer can ONLY contain the following special functions: gamma function, beta function, Riemann zeta function, Dirichlet eta function, dilogarithm, digamma function, and trigonometric integral.

Format your post is as follows:

SOLUTION OF PROBLEM xxx (number of problem) :

PROBLEM xxx (number of problem) :

Please share this note so that lots of users here know this contest and take part in it. (>‿◠)✌

Okay, let the contest begin! Here is the first problem:

PROBLEM 1 :

For $a>0$, show that

$\int_0^a \frac{\ln x}{\sqrt{ax-x^2}}\,dx=\pi\ln\left(\frac{a}{4}\right)$

P.S. You may also want to see Brilliant Integration Contest - Season 1 (Part 2) and Brilliant Integration Contest - Season 1 (Part 3).

Note by Anastasiya Romanova
5 years, 3 months ago

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Ans:ln4a/2a^2

- 5 years, 3 months ago

Solution of Problem 15 :

We have $\int_0^1 \left(\frac{1}{1-x} + \frac{1}{\ln x} \right)\,dx=\int_0^1 \frac{\ln x+1-x}{(1-x)\ln x}\,dx$

Proposition :

$I(s)=\int_0^1 \frac{s\ln x+1-x^s}{(1-x)\ln x}\,dx=\ln\Gamma(s+1) +\gamma s$

Proof :

Let

$I(s)=\int_0^1 \frac{s\ln x+1-x^s}{(1-x)\ln x}\,dx$

then

\begin{aligned} I'(s)&=\int_0^1 \frac{1-x^s}{1-x}\,dx\\ I''(s)&=-\int_0^1 \frac{x^s\ln x}{1-x} \,dx\\ &=-\int_0^1\sum_{n=0}^\infty x^{n+s}\ln x\,dx\\ &=-\sum_{n=0}^\infty\partial_s\int_0^1 x^{n+s}\,dx\\ &=-\sum_{n=0}^\infty\partial_s\left[\frac{1}{n+s+1}\right]\\ &=\sum_{n=0}^\infty\frac{1}{(n+s+1)^2}\\ &=\psi_1(s+1)\\ I'(s)&=\int\psi_1(s+1)\,ds\\ I'(s)&=\int\frac{\partial}{\partial s}\bigg[\psi(s+1)\bigg]\,ds\\ I'(s)&=\psi(s+1)+C\\ \end{aligned} For $s=0$, we have $I'(0)=0$. Implying $C=-\psi(1)=\gamma$, then \begin{aligned} I'(s)&=\psi(s+1)+\gamma\\ I(s)&=\int \psi(s+1)\,ds +\gamma s+C\\ &=\int \frac{\partial}{\partial s}\bigg[\ln\Gamma(s+1)\bigg]\,ds +\gamma s+C\\ &=\ln\Gamma(s+1) +\gamma s+C\\ \end{aligned}

where $\psi(z)$ is the digamma function. For $s=0$, we have $I(0)=0$. Implying $C=0$, then

$I(s)=\int_0^1 \frac{s\ln x+1-x^s}{(1-x)\ln x}\,dx=\ln\Gamma(s+1) +\gamma s\qquad\qquad\square$

For $s=1$, we have

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

where $\gamma$ is The Euler–Mascheroni constant.

Problem 16 :

Prove

$\int_0^{\Large\frac\pi2}\cos^{n-1}x\cos ax\ dx=\frac{\pi}{2^n n\ \operatorname{B}\left(\frac{n+a+1}{2},\frac{n-a+1}{2}\right)}$

where $\operatorname{B}\left(x,y\right)$ is the beta function.

## PS : POST YOUR SOLUTION BELOW EACH PROBLEM THREAD AND POST YOUR PROPOSED PROBLEM AS A NEW THREAD. PUT THEM IN SEPARATED THREAD. SO THAT THE POSTS LOOK MORE ORGANIZED. THANKS.

- 5 years, 3 months ago

Can you please post a all together new note of the integration contest it's getting messy and messier, Please move this contest to a new note @Anastasiya Romanova

- 5 years, 3 months ago

OK

## The Brilliant Integration Contest - Season 1 (Part 1) moves to The Brilliant Integration Contest - Season 1 (Part 2). Enjoy it!! (>‿◠)✌

- 5 years, 3 months ago

_ Solution of problem 14 _

$I = \displaystyle \int_{0}^{1} \ln \bigg(\dfrac{3+x}{3-x} \bigg) \dfrac{1}{\sqrt{x(1-x^2)}} {\mathrm dx}$

= $\displaystyle \int_{0}^{1} \ln(1 + x/3) \dfrac{1}{\sqrt{x(1-x)}} {\mathrm dx}$

= $\displaystyle \int_{0}^{\pi/2} 2 \ln\bigg(1 + \frac{1}{3} \sin^2 \theta\bigg) {\mathrm d} \theta - \displaystyle \int_{0}^{\pi/2} 2 \ln\bigg(1 - \frac{1}{3} \sin^2 \theta\bigg) {\mathrm d} \theta$

Lemma $\displaystyle \int_{0}^{\pi/2} \ln(1+a \sin^2 \theta) = \pi \ln \bigg(\dfrac{1+\sqrt{a+1}}{2}\bigg)$

Proof

Let $I(a) = \displaystyle \int_{0}^{\pi/2} \ln(1+a \sin^2 x) {\mathrm dx}$

$I'(a) = \displaystyle \int_{0}^{\pi/2} \dfrac{\sin^2 x}{1+ a \sin^2 x} {\mathrm dx}$

$I'(a) = \displaystyle \int_{0}^{\pi/2} \dfrac{1}{a}\bigg(1 - \dfrac{\sec^2 x}{1+(a+1) \tan^2x} {\mathrm dx}\bigg)$

= $\dfrac{\pi}{2a} - \dfrac{1}{a} \bigg(\displaystyle \int_{0}^{\infty} \dfrac{dz}{1+(a+1)z^2}\bigg)$

=$\dfrac{\pi}{2a} \bigg(1 - \dfrac{1}{\sqrt{a+1}}\bigg)$

Hence, $I(a) - I(0) = \displaystyle \int_{0}^{a} \dfrac{\pi}{2} \bigg(\dfrac{1}{x} - \dfrac{1}{x\sqrt{x+1}}\bigg) {\mathrm dx}$

= $\pi \ln(\sqrt{x+1} +1)\bigg|_{0}^{a} = \pi\ln\bigg(\dfrac{1+\sqrt{a+1}}{2}\bigg)$ (Note that $I(0) = 0$)

Put $a=1/3$ , and $-1/3$, then substitute back in original integral to get:

$I = 2 \pi \ln \bigg(\dfrac{1+\sqrt{4/3}}{1+\sqrt{2/3}} \bigg) = \boxed{\pi \ln \bigg(\dfrac{7 + 4\sqrt{3}}{5+2\sqrt{6}}\bigg)}$

- 5 years, 3 months ago

Nice solution Jatin, +1! But you're a bit late about 10 min than Ronak, so the right to propose a new problem goes to Ronak.

- 5 years, 3 months ago

@Anastasiya Romanova I think maybe we should change the format of the contest. Maybe each problem should start in a new comment and the solutions must be replies to that comment. It would keep the contest more organised and understandable. The present format is just messing things up with the problem and its solution in separate comments.

- 5 years, 3 months ago

$Solution\quad of\quad problem\quad 14$

Lemma

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

Proof :

$\displaystyle I(a)=\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln({ a }^{ 2 }{ cos }^{ 2 }x+{ b }^{ 2 }{ sin }^{ 2 }x)dx }$

Differentiating with respect to $a$ we get :

$\displaystyle {I}^{'}(a)=2a\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { { cos }^{ 2 }x }{ { a }^{ 2 }{ cos }^{ 2 }x+{ b }^{ 2 }{ sin }^{ 2 }x } dx }$

Put $tan(x)=t$ to get our integral as :

$\displaystyle {I}^{'}(a)=2a\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { dt }{ ({ a }^{ 2 }+{ b }^{ 2 }{ t }^{ 2 })(1+{ t }^{ 2 }) } }$

$\displaystyle =2a\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { dt }{ ({ a }^{ 2 }+{ b }^{ 2 }{ t }^{ 2 })(1+{ t }^{ 2 }) } } =\frac { 2a }{ { b }^{ 2 }-{ a }^{ 2 } } (\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { { b }^{ 2 }dt }{ { a }^{ 2 }+{ b }^{ 2 }{ t }^{ 2 } } } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { dt }{ 1+{ t }^{ 2 } } } )$

Solving it we get :

$\displaystyle {I}^{'}(a)=\frac { 2\pi }{ a+b }$

$\Rightarrow \displaystyle I(a)=2\pi ln(a+b)+C$

Put $a=b=1$ to get $C=-2\pi ln(2)$

Hence $\displaystyle I(a)=2\pi ln(\frac{a+b}{2})$

In our integral in the given question put $x=\frac{1-sin\theta }{2}$ to get the integral as :

$I= \displaystyle \int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(7-sin(\theta))dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(5+sin(\theta))dx }$

Using the property $\displaystyle \int _{ a }^{ b }{ f(x)dx } =\int _{ a }^{ b }{ f(a+b-x)dx }$ we get :

$\displaystyle I=\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(7+sin(\theta))dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(5-sin(\theta))dx }$

Adding these two we get :

$\displaystyle I=\frac { 1 }{ 2 } (\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(49-{ sin }^{ 2 }\theta)dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(25-{ sin }^{ 2 }(\theta))dx } )$

$\displaystyle I=\frac { 1 }{ 2 } (\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(49{ cos }^{ 2 }\theta+48{ sin }^{ 2 }\theta)dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(25{ cos }^{ 2 }\theta+24{ sin }^{ 2 }\theta)dx } )$

Using our given lemma we get :

$\displaystyle I=\frac { 1 }{ 2 } (2\pi ln(\frac{7+4\sqrt { 3 }}{2} )-2\pi ln(\frac{5+2\sqrt { 6 }}{2} ))=\pi ln(\frac { 7+4\sqrt { 3 } }{ 5+2\sqrt { 6 } } )$

$Problem\quad 15$

Evaluate $\displaystyle I=\int _{ 0 }^{ 1 }{ \frac { 1 }{ ln(x) } +\frac { 1 }{ 1-x } dx }$

- 5 years, 3 months ago

According to timeline, you're faster than @jatin yadav to post your solution, then the right to propose a new problem is yours.

- 5 years, 3 months ago

Solution of Problem 13 :

First we prove the following proposition:

Proposition :

$\int_0^1 x^a \ln^k x\ dx=\frac{(-1)^k\, k!}{(a+1)^{k+1}} \qquad\text{for }\ k=0,1,2,\ldots$

Proof :

Note that $\int_0^1 x^a\ dx=\frac1{a+1}\qquad\text{for }\ \alpha>-1.$ Differentiating equation above $k$ times w.r.t. $a$ we have $\int_0^1 \frac{\partial^k}{\partial a^k}\left(x^a\right)\ dx=\int_0^1 x^a \ln^k x\ dx=\frac{(-1)^k\, k!}{(a+1)^{k+1}}\qquad\text{Q.E.D.}$

Now, we will evaluate the general case of

$\int_0^\infty\frac{x^{a-1}}{e^{bx}-1}\,dx=\int_0^\infty\frac{x^{a-1}\,e^{-bx}}{1-e^{-bx}}\,dx\qquad\text{for }\ a,b>0$

Set $y=e^{-bx}$, then

$\int_0^\infty\frac{x^{a-1}}{e^{bx}-1}\,dx=\frac{(-1)^{a-1}}{b^a}\int_0^1\frac{\ln^{a-1}y}{1-y}\,dy$

Use a geometric series for $\dfrac{1}{1-y}$ then interchange the integral and summation sign which is justified by the Fubini–Tonelli theorem and apply the previous proposition, we have

\begin{aligned} \int_0^\infty\frac{x^{a-1}}{e^{bx}-1}\,dx&=\frac{(-1)^{a-1}}{b^a}\int_0^1\sum_{k=0}^\infty y^k\,\ln^{a-1}y\,dy\\ &=\frac{(-1)^{a-1}}{b^a}\sum_{k=0}^\infty\int_0^1 y^k\,\ln^{a-1}y\,dy\\ &=\frac{(-1)^{a-1}}{b^a}\sum_{k=0}^\infty \frac{(-1)^{a-1}\, (a-1)!}{(k+1)^{a}}\\ &=\frac{(a-1)!}{b^a}\sum_{k=1}^\infty \frac{1}{k^{a}}\\ &=\frac{\Gamma(a)\,\zeta(a)}{b^a} \end{aligned}

where $\Gamma(a)$ is the gamma function and $\zeta(a)$ is the Riemann zeta function.

Hence, by setting $a=n+1$ and $b=1$, we have

$\int_0^\infty\frac{x^{n}}{e^{x}-1}\,dx=\Gamma(n+1)\,\zeta(n+1)$

Problem 14 :

Prove that

\begin{aligned} \int_{0}^1 \ln\left(\frac{3+x}{3-x}\right)\,\frac{dx}{\sqrt{x(1-x)}}=\pi\ln\left(\dfrac{7+4\sqrt{3}}{5+2\sqrt{6}}\right)\\ \end{aligned}

- 5 years, 3 months ago

$Solution\quad of\quad Problem\quad 12$

It's a very good disguised integral.

First note that the function is an even function hence our integral can be written as :

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

Now we know that $\displaystyle cos(arctan(2x))=\frac{1}{\sqrt{1+4{x}^{2}}}$. Using this our integral becomes :

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

Splliting it by partial fractions we get :

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

$\displaystyle I=\frac { 2 }{ 3 } (2{ tan }^{ -1 }(2x)\overset { \infty }{ \underset { 0 }{ | } } -{ tan }^{ -1 }(x)\overset { \infty }{ \underset { 0 }{ | } } )$

$\displaystyle I=\frac { \pi }{ 3 }$

$Problem\quad 13$

Find closed form of $\displaystyle I=\int _{ 0 }^{ \infty }{ \frac { { x }^{ n } }{ { e }^{ x }-1 } dx }$

- 5 years, 3 months ago

Wait a sec!? Could you rectify your solution, because the original problem is written as $\arctan(2x)$. And one thing, what did you mean by "Now we know that $\cos(\arctan x)=\frac{1}{\sqrt{1+4x^2}}$"? Could you elaborate?

- 5 years, 3 months ago

That was my typing mistake, also Let$arctan(2x)=\theta$, hence $2x=tan(\theta)$, hence $cos(\theta)=\frac{1}{\sqrt{1+4{x}^{2}}}$

$\Rightarrow cos(arctan(2x))=\frac{1}{\sqrt{1+4{x}^{2}}}$

- 5 years, 3 months ago

Solution of Problem 11 :

Using substitution $u=\dfrac{\alpha^2}{x}\;\Rightarrow\;x=\dfrac{\alpha^2}{u}\;\Rightarrow\;dx=-\dfrac{\alpha^2}{u^2}\ du$, then

\begin{aligned} \int_0^\infty\frac{\ln x}{x^2+\alpha^2}\ dx&=\int_0^\infty\frac{\ln \left(\dfrac{\alpha^2}{u}\right)}{\left(\dfrac{\alpha^2}{u}\right)^2+\alpha^2}\cdot \dfrac{\alpha^2}{u^2}\ du\\ I(\alpha)&=\int_0^\infty\frac{2\ln \alpha-\ln u}{\alpha^2+u^2}\ du\\ &=2\ln \alpha\int_0^\infty\frac{1}{\alpha^2+u^2}\ du-\int_0^\infty\frac{\ln u}{u^2+\alpha^2}\ du\\ &=2\ln \alpha\int_0^\infty\frac{1}{\alpha^2+u^2}\ du-I(\alpha)\\ I(\alpha)&=\ln \alpha\int_0^\infty\frac{1}{\alpha^2+u^2}\ du. \end{aligned}

The last integral can easily be evaluated by using substitution $u=\alpha\tan\theta$, then

\begin{aligned} \int_0^\infty\frac{\ln x}{x^2+\alpha^2}\ dx&=\frac{\ln \alpha}{\alpha}\int_0^{\Large\frac\pi2} \ d\theta\\ &=\frac{\pi\ln \alpha}{2\alpha} \end{aligned}

Problem 12 :

Prove

$\int_{-\infty}^{\infty}\frac{\cos \left(\, \arctan 2x\right)}{(1+x^2)\sqrt{1+4x^2}}\,dx=\frac{\pi}{3}$

- 5 years, 3 months ago

The last substitution should be $u=\alpha \tan \theta$ :)

- 5 years, 3 months ago

This is a simple kind of NCERT problem. It can be easily done by substituting $t = tan^{-1}\dfrac{x}{\alpha}$ and then applying by parts

- 5 years, 1 month ago

Fixed it. Sorry, I'm too hasty. Thanks... :)

- 5 years, 3 months ago

Solution to problem 10

First integrate by parts -

$I= \int_{0}^{\infty} \dfrac{\sin^3{x}}{x^2} dx =\int_{0}^{\infty} \dfrac{3\sin^2{x} \cos{x}}{x} dx$

Then, use the property of the laplace transform that $\mathcal{L} \left\{\dfrac{f(t)}{t}\right\}(s)=\displaystyle\int_{s}^{\infty} F(p) dp$ (in our case, $s \rightarrow 0$).

$I=3\int_{0}^{\infty} \mathcal{L} \{\sin^2{t} \cos{t}\}(p) \ dp$

$I=3\int_{0}^{\infty} \dfrac{2s}{(s^2+1)(s^2+9)} ds=\boxed{\dfrac{3\log {3}}{4}}$ The last integral can be found by using partial fractions.

Note : $\mathcal{L}\{f(t)\}(s)=\displaystyle\int_{0}^{\infty} e^{-st} f(t) \mathrm{d}t$

Problem 11

Find

$\int_{0}^{\infty} \dfrac{\log {x}}{x^2+\alpha^2} \mathrm{d}x \ \ \ \ \ \text{for} \ \alpha>0$

- 5 years, 3 months ago

Solution 9: First, consider the following Lemma.

LEMMA : $\int_{0}^{1} \frac{\ln{x}}{1-x^2} dx=-\frac{\pi^2}{8}$

Proof: Since, $\frac{1}{1-x^2}=\sum\limits_{n=0}^{\infty} {x^{2n}}$, we have,

$\int_0^1 \frac{\ln{x}}{1-x^2}=\sum\limits_{n=0}^{\infty} \int_0^1 x^{2n}{\ln{x}} \: dx$

$=-\sum\limits_{n=0}^{\infty} {\frac{1}{(2n+1)^2}}$ (Using Integration By Parts)

$=\sum\limits_{n=0}^{\infty} {\frac{1}{(2n)^2}} - \sum\limits_{n=0}^{\infty} {\frac{1}{(n)^2}}$

$=\frac{-3}{4}\zeta(2)= -\frac{\pi^2}{8}$

Now,

$\text{I}= \int_{0}^{1} \ln({\frac{1+x}{1-x}}) \frac{dx}{x\sqrt{1-x^2}}$

Substituting $x=\frac{1-y^2}{1+y^2}$, we have,

$\text{I}=-4 \times \int_{0}^{1} \frac{\ln{y}}{1-y^2} dx$

$=\frac{\pi^2}{2}$ (Using the Lemma)

- 5 years, 3 months ago

Nice answer, +1! But I'm really sorry, the one who deserves to propose the next problem is Ronak since according to the timeline his answer comes first than the other answers. Maybe, you can try to answer his proposed problem. $\ddot\smile$

- 5 years, 3 months ago

@Anastasiya Romanova I think maybe we should change the format of the contest. Maybe each problem should start in a new comment and the solutions must be replies to that comment. It would keep the contest more organised and understandable. The present format is just messing things up with the problem and its solution in separate comments.

- 5 years, 3 months ago

Ya, I agree with you. I have edited the rule accordingly. Well, this is the first contest here and I hope you understand with that.

- 5 years, 3 months ago

$Solution\quad of\quad problem\quad 9$

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

Put $x=cos(\theta)$ to get our integral as :

$\displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln({ cot }^{ 2 }\frac { \theta }{ 2 } )\frac { d\theta }{ cos\theta } }$

$\Rightarrow \displaystyle I=(-2)\int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln(tan\frac { \theta }{ 2 } )\frac { d\theta }{ cos\theta } }$

Put $tan(\theta/2)=x$ to get our integral as :

$\displaystyle (-4)\int _{ 0 }^{ 1 }{ \frac { ln(t)dt }{ 1-{ t }^{ 2 } } }$

I proved in my solution to problem 4 that :

$\displaystyle \int _{ 0 }^{ 1 }{ \frac { ln(t)dt }{ 1-{ t }^{ 2 } } } =\frac { -{ \pi }^{ 2 } }{ 8 }$

Using this I get :

$I=\frac{{\pi}^{2}}{2}$

$Problem\quad 10$

Find $\displaystyle \int _{ 0 }^{ \infty }{ \frac { { sin }^{ 3 }x }{ { x }^{ 2 } } dx }$

- 5 years, 3 months ago

Since @Pranav Arora is unable to propose a problem (I hope it's only temporary) and to make this contest sustains, then according to rule 4, I, as the last solver, have a right to propose a new one. Here is the problem:

PROBLEM 9

Prove that

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

- 5 years, 3 months ago

Solution -Problem - 9 We substitute, $t = \frac{1+x}{1-x} \implies x = \frac{1-t}{1+t}$ Doing the substitution and , simplifying , gives, $I = - \int_{0}^{1} \frac{\ln(t)}{(\sqrt{t})(1-t)} dt$ Now we substitute, $t=\sin^2(x)$ $I = -4\int_{0}^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{\cos(x)} dx$ Using, Problem -$4$ $I = -4 \times \frac{-\pi^2}{8} = \frac{\pi^2}{2}$

- 5 years, 3 months ago

I had already posted the solution to the problem and a new problem also you should not reply instead post your solution as a seperate comment as I did @Shivang Jindal

- 5 years, 3 months ago

Lol, time difference of 5mins. I think i was typing solution when you posted the solution .

- 5 years, 3 months ago

You can try my posted question. @Shivang Jindal

- 5 years, 3 months ago

I am able to write the integral as a series - $I=\pi \sum_{n=0}^{\infty} \dfrac{(2n)!}{2^{2n}(n!)^2(2n+1)}$

- 5 years, 3 months ago

Solution of Problem 7:

Proposition :

\begin{aligned} \int_0^{\pi} \frac{dx}{p+\cos x}\,dx=\frac{\pi}{\sqrt{p^2-1}} \end{aligned}

Proof :

It can be proven by using Weierstrass substitution: $t=\tan\left(\dfrac{x}{2}\right)$, then

\begin{aligned} \\ \int_0^{\pi} \frac{dx}{p+\cos x}\,dx &=\int_0^{\infty} \frac{2}{p+1+(p-1)t^2}\,dt\qquad\Rightarrow\qquad t=\sqrt{\frac{p+1}{p-1}}\tan y\\ &=\left.\frac{2}{\sqrt{p^2-1}}\arctan t\;\right|_0^{\infty}\\ &=\frac{\pi}{\sqrt{p^2-1}}\qquad\qquad\text{Q.E.D.} \end{aligned}

Differentiating the proposition w.r.t. $p$ twice and setting $p=\sqrt{10}$, we have

\begin{aligned} \\ \frac{\partial^2}{\partial p^2}\int_0^\pi\frac{1}{p+\cos x}\, dx&=\frac{\partial^2}{\partial p^2}\left[\frac{\pi}{\sqrt{p^2-1}}\right]\\ \int_0^\pi\frac{2}{\left(p+\cos x\right)^3}\, dx&=\frac{\pi\left(2p^2+1\right)}{\sqrt{\left(p^2-1\right)^5}}\\ \int_0^\pi\frac{1}{\left(\sqrt{10}+\cos x\right)^3}\, dx&=\frac{7\pi}{162} \end{aligned}

Problem 8 :

Prove

$\int_0^{\Large\frac{\pi}{2}}\frac{dx}{1+8\sin^2(\tan x)}=\frac{\pi}{6}\left(\frac{2e^2+1}{2e^2-1}\right)$

- 5 years, 3 months ago

Sorry for very late response. Using elementary techniques,-

Solution - Problem 7

$\displaystyle \int_{0}^{\pi} \dfrac{dx}{p - cosx}$

$2I = \displaystyle \int_{0}^{\pi} \dfrac{2p~dx}{p^2 - cos^2x}$

$I = \displaystyle \int_{0}^{\pi} \dfrac{p~dx}{p^2(1 + tan^2x) - 1}$

$I = p \displaystyle \int_{0}^{\pi} \dfrac{sec^2x~dx}{p^2tan^2x + \sqrt{(p^2 - 1)^2}}$

tanx = t

$I = 2p \displaystyle \int_{0}^{\infty} \dfrac{dt}{p^2t^2 + \sqrt{(p^2 - 1)^2}}$

$I = \dfrac{2p}{ \sqrt{p^2 - 1}} \times \dfrac{1}{p} \Big[_{0}^{\infty} tan^{-1} \dfrac{pt}{ \sqrt{p^2 - 1}}$

$I = \dfrac{\pi}{ \sqrt{p^2 - 1}}$

- 5 years, 1 month ago

Solution of Problem 8:

Substitute $\tan x\mapsto x$, then the integral is:

$\int_0^{\infty} \frac{dx}{(1+x^2)(1+8\sin^2x)}=\int_0^{\infty} \frac{dx}{(1+x^2)(5-4\cos(2x))}$

$=\int_0^{\infty} \frac{1}{1+x^2}\left(\frac{1}{3}\left(1+2\sum_{k=1}^{\infty} \frac{\cos(2kx)}{2^k}\right)\right)=\frac{\pi}{6}+\frac{2}{3}\sum_{k=1}^{\infty} \frac{1}{2^k}\int_0^{\infty} \frac{\cos(2kx)}{1+x^2}\,dx$

$=\frac{\pi}{6}+\frac{\pi}{3}\sum_{k=1}^{\infty} \left(\frac{1}{2e^2}\right)^k=\frac{\pi}{6}+\frac{\pi}{3}\frac{1}{2e^2-1}=\boxed{\dfrac{\pi}{6}\left(\dfrac{2e^2+1}{2e^2-1}\right)}$

I have used the following result:

$\int_0^{\infty} \frac{\cos(mx)}{x^2+a^2}\,dx=\frac{\pi}{2a}e^{-am}$

I cannot think of a challenging problem at the moment, I request somebody else to post one. Thanks!

- 5 years, 3 months ago

Ingenious! But I have a doubt - In your fourth step, how is it guaranteed that the function $\dfrac{\cos (2kx)}{2^k}$ is positive on $(0,\infty)$? PS : you forgot to write the $\mathrm{d}x$ in the third step.

- 5 years, 3 months ago

I don't know what did you mean by positive/negative on $(0,\infty)$, but if you meant to swap between integral & summation sign, it is valid because $\frac{\cos(2kx)}{1+x^2}$ is continuous, finite & integrable, therefore swapping those two signs can be justified by Fubini's theorem.

- 5 years, 3 months ago

Right. I confused it with Tonelli's theorem.

- 5 years, 3 months ago

If pranav refuses to put up a question then who put will put it up @Anastasiya Romanova .

- 5 years, 3 months ago

According to the rule 4, the one is the last solver and it turns out that one is me. See PROBLEM 9 below, I've just proposed it.

- 5 years, 3 months ago

Solution of Problem 5 :

I will prove the general form of

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

It 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. Hence, by setting $m=1$, we have

$\int_0^\infty\frac{1}{1+x^n}\,dx=\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}=\frac{\pi}{n}\csc\left(\frac{\pi}{n}\right)$