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# Brilliant Integration Contest - Season 2 (Part 2)

Hi Brilliant! Due to many problems in part 1 it had become slow to load. So this is a sequel of Brilliant Integration - Season 2(Part 1)

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]** 

UPDATE:

• Tanishq Varshney has been banned from this contest indefinitely.

• Contour integration is allowed in the contest.

1 year ago

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Problem 44:

Evaluate $\displaystyle \int _{ 0 }^{ \infty }{ \left( \dfrac{ \ln { (1+x) } }{ { \ln }^{ 2 }(x)+{ \pi }^{ 2 } }\right) \frac { dx }{ {x}^{2} } } .$

###### Due to time constraint, the author decided to post their own solution.
· 11 months, 3 weeks ago

Lemma : $$\displaystyle I = \int _{ 0 }^{ \infty }{ \frac { { e }^{ -ky } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } =\frac{{ (-1) }^{ k }}{\pi} \int _{ k\pi }^{ \infty }{ \frac { sin(x) }{ x } dx }$$ , for $$k$$ is a whole number.

Proof : We being with noting that :

$$\displaystyle \frac { 1 }{ w } \int _{ 0 }^{ \infty }{ { e }^{ -st }sinwtdt } =\frac { 1 }{ { s }^{ 2 }+{ w }^{ 2 } }$$

So we can rewrite the integral on the left side as :

$$\displaystyle I = \frac { 1 }{ \pi } \int _{ 0 }^{ \infty }{ { e }^{ -ky }dy } \int _{ 0 }^{ \infty }{ { e }^{ -ty }sin\pi tdt } =\frac { 1 }{ \pi } \int _{ 0 }^{ \infty }{ sin\pi t\int _{ 0 }^{ \infty }{ { e }^{ -(k+t)y }dy } dt }$$

$$\displaystyle \Rightarrow I = \frac { 1 }{ \pi } \int _{ 0 }^{ \infty }{ \frac { sin\pi t }{ t+k } dt }$$

$$\displaystyle \Rightarrow I = \frac { 1 }{ \pi } \int _{ k }^{ \infty }{ \frac { sin(\pi t-\pi k) }{ t } dt }$$

$$\displaystyle \Rightarrow I = { (-1) }^{ k }\int _{ k }^{ \infty }{ \frac { sin(\pi t) }{ t } dt }$$

$$\displaystyle \Rightarrow I = { (-1) }^{ k }\int _{ k\pi }^{ \infty }{ \frac { sin(t) }{ t } dt }$$

Hence proved

Solution :

We start with $$x={e}^{-y}$$ to get out integral as :

$$\displaystyle I = \int _{ -\infty }^{ \infty }{ \frac { { e }^{ y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy }$$

Break in into two parts :

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

In the second part put $$y=-x$$ to get :

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

Manipulating it we have :

$$\displaystyle I = \int _{ 0 }^{ \infty }{ \frac { { e }^{ y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } +\int _{ 0 }^{ \infty }{ \frac { { e }^{ -y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } +\int _{ 0 }^{ \infty }{ \frac { { ye }^{ -y } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy }$$

$$\displaystyle \Rightarrow I= \int _{ 0 }^{ \infty }{ \frac { ({ e }^{ y }+{ e }^{ -y })\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } +\int _{ 0 }^{ \infty }{ \frac { { ye }^{ -y } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy }$$

We write $$I = J+K$$

$$\displaystyle J = \int _{ 0 }^{ \infty }{ \frac { ({ e }^{ x }+{ e }^{ -x })\ln { (1+{ e }^{ -x }) } }{ { x }^{ 2 }+{ \pi }^{ 2 } } } dx$$

Wrting $$\ln(1+{e}^{-x})$$ in it's taylor series we have :

$$\displaystyle J = \int _{ 0 }^{ \infty }{ \frac { ({ e }^{ x }+{ e }^{ -x }) }{ { x }^{ 2 }+{ \pi }^{ 2 } } \sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r-1 } }{ r } { e }^{ -rx } } } dx$$

We interchange the integral and sum :

$$\displaystyle J = \sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r-1 } }{ r } (\int _{ 0 }^{ \infty }{ \frac { { e }^{ -(r-1)x } }{ { x }^{ 2 }+{ \pi }^{ 2 } } } dx+\int _{ 0 }^{ \infty }{ \frac { { e }^{ -(r+1)x } }{ { x }^{ 2 }+{ \pi }^{ 2 } } } dx } )$$

Using the lemma we have :

$$\displaystyle J = \frac { 1 }{ \pi } \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ r } (\int _{ (r-1)\pi }^{ \infty }{ \frac { \sin { (x) } }{ x } dx } +\int _{ (r+1)\pi }^{ \infty }{ \frac { \sin { (x) } }{ x } } dx) }$$

Now it is worthy noting that :

$$\displaystyle \int _{ r\pi }^{ \infty }{ \frac { \sin { (x) } }{ x } dx } =\int _{ 1 }^{ \infty }{ \frac { sin(r\pi x) }{ x } dx }$$ , for all positive $$r$$, what if it was applied for $$r=0$$, then the left integral is $$\frac{\pi}{2}$$, while the write one is 0, so for $$r=0$$, we can write it as :

$$\displaystyle \int _{ r\pi }^{ \infty }{ \frac { \sin { (x) } }{ x } dx } =\int _{ 1 }^{ \infty }{ \frac { sin(r\pi x) }{ x } dx } +\frac { \pi }{ 2 }$$, for $$r=0$$, having said that we proceed further :

$$\displaystyle J = \frac { 1 }{ \pi } \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ r } (\int _{ \frac{1}{2} }^{ \infty }{ \frac { \sin { (2(r-1)\pi x) } +\sin { (2(r+1)\pi x) } }{ x } dx } ) } +\frac { 1 }{ 2 }$$

$$\displaystyle \Rightarrow J = \frac { 2 }{ \pi } \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ r } (\int _{ \frac{1}{2} }^{ \infty }{ \frac { cos(2\pi x)\sin { (2r\pi x) } }{ x } dx } ) } +\frac { 1 }{ 2 }$$

Again changing sum and integral we have :

$$\displaystyle \Rightarrow J = \frac { 2 }{ \pi } \int _{ \frac { 1 }{ 2 } }^{ \infty }{ \cos { (2\pi x) } \sum _{ r=1 }^{ \infty }{ \frac { \sin { (2r\pi x) } }{ rx } } dx } +\frac { 1 }{ 2 }$$

It is a well known result that :

$$\displaystyle \sum _{ r=1 }^{ \infty }{ \frac { \sin { (2r\pi x) } }{ r\pi } } =\left\lfloor x \right\rfloor +\frac { 1 }{ 2 }-x$$

Using this we have :

$$\displaystyle J = 2\int _{ \frac { 1 }{ 2 } }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } (-x+\left\lfloor x \right\rfloor +\frac { 1 }{ 2 } )dx } +\frac { 1 }{ 2 }$$

Leaving this let's evaluate $$K$$ first :

Using the same techniques as used in proving the lemma, we can show that :

$$\displaystyle K = \int _{ 0 }^{ \infty }{ \frac { x{ e }^{ -x } }{ { x }^{ 2 }+{ \pi }^{ 2 } } dx } =-\int _{ 1/2 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } dx }$$

Now $$\displaystyle I = J+K = 2\int _{ \frac { 1 }{ 2 } }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } (-x+\left\lfloor x \right\rfloor +\frac { 1 }{ 2 } )dx } +\frac { 1 }{ 2 } -\int _{ 1/2 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } dx } =\frac { 1 }{ 2 } +2\int _{ 1/2 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } \left\{ x \right\} dx }$$

$$\displaystyle I = \frac { 1 }{ 2 } +2\int _{ 1 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } \left\{ x \right\} dx } +2\int _{ \frac { 1 }{ 2 } }^{ 1 }{ \cos { (2\pi x) } dx } =\frac { 1 }{ 2 } +2\int _{ 1 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } \left\{ x \right\} dx }$$

Now we will be evaluating :

$$\displaystyle M = \int _{ 1 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } \left\{ x \right\} dx }$$

It can be written as :

$$\displaystyle M = \sum _{ n=1 }^{ \infty }{ \int _{ n }^{ n+1 }{ \frac { \cos { (2\pi x) } }{ x } (x-n)dx } }$$

$$\displaystyle \Rightarrow M = -\sum _{ n=1 }^{ \infty }{ n\int _{ n }^{ n+1 }{ \frac { \cos { (2\pi x) } }{ x } dx } }$$

$$M$$ can also be written as :

$$\displaystyle M =-\sum _{ n=1 }^{ \infty }{ \int _{ n }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } dx } }$$

This manipulation can easily be justified with the help of properties of double summations :

Using integration by parts we have :

$$\displaystyle M = \sum _{ n=1 }^{ \infty }{ \int _{ n }^{ \infty }{ \frac { \sin { (2\pi x) } }{ 2\pi { x }^{ 2 } } dx } } -\sum _{ n=1 }^{ \infty }{ \frac { \sin { (2\pi n) } }{ 2\pi n } }$$

$$\displaystyle \Rightarrow M = \sum _{ n=1 }^{ \infty }{ \int _{ n }^{ \infty }{ \frac { \sin { (2\pi x) } }{ 2\pi { x }^{ 2 } } dx } }$$

$$\displaystyle \Rightarrow M = \sum _{ n=1 }^{ \infty }{ \int _{ 1 }^{ \infty }{ \frac { \sin { (2\pi nx) } }{ 2\pi { nx }^{ 2 } } dx } }$$

Again changing sum and integral we have :

$$\displaystyle \Rightarrow M = \quad \int _{ 1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { \sin { (2\pi nx) } }{ 2\pi { nx }^{ 2 } } } dx }$$

Using the result we have used above we have :

$$\displaystyle M = \frac { 1 }{ 2 } \int _{ 1 }^{ \infty }{ \frac { 1 }{ 2{ x }^{ 2 } } -\frac { \left\{ x \right\} }{ { x }^{ 2 } } dx }$$

$$\displaystyle \Rightarrow M = \quad \frac { 1 }{ 2 } \int _{ 1 }^{ \infty }{ \frac { 1 }{ 2{ x }^{ 2 } } -\frac { \left\{ x \right\} }{ { x }^{ 2 } } dx }$$

For solving $$\displaystyle H = \int _{ 1 }^{ \infty }{ \frac { \left\{ x \right\} }{ { x }^{ 2 } } dx }$$

write it as a sum :

$$\displaystyle H = \sum _{ n=1 }^{ \infty }{ \int _{ n }^{ n+1 }{ \frac { 1 }{ x } -\frac { n }{ { x }^{ 2 } } dx } }$$

$$\displaystyle \Rightarrow H = \sum _{ n=1 }^{ \infty }{ \ln { (n+1) } -\ln { (n) } -\frac { 1 }{ n+1 } }$$

$$\displaystyle H = \lim _{ n\rightarrow \infty }{ 1+\ln { (n+1)-{ H }_{ n+1 } } } =1-\gamma$$

Hence $$\displaystyle M = \frac { \gamma }{ 2 } -\frac { 1 }{ 4 }$$

Finally we got :

$\Large \boxed { I=\gamma }$ · 11 months, 3 weeks ago

Problem 26:

Show that $\large\int _{ 0 }^{ \pi }{ { e }^{ 2\cos { x } }\sin ^{ 2 }{ \left( \sin { x } \right) } \, dx } =\frac { \pi }{ 2 } \left( J_{ 0 }\left( 2i \right) -1 \right).$

Where $$J_n\left(z\right)$$ denote the Bessel Function of First Kind.

###### This problem has been solved by Tanishq Varshney.
· 1 year ago

$$\text{Solution to Problem 26}$$

Firstly some common things

$$2 \sin ^{2} x= 1-\cos (2x)$$

$$\large{\pi I_{n} (z)=\displaystyle \int^{\pi}_{0} e^{z \cos \theta} \cos (n \theta) d \theta}$$

where $$I_{n} (z)$$ is modified Bessel function of first kind

Also $$\large{I_{n}(z) =e^{\frac{-in \pi}{2}} J_{n} (i z)}$$ where $$i^2=-1$$ and $$J_{n} (z)$$ is Bessel function of first kind.

In this question we have $$n=0$$

Now in the integral

$$\large{\displaystyle \int^{\pi}_{0} e^{2 \cos x} \left(\frac{1- \cos (2 \sin x)}{2}\right) dx}$$

On separating

$$\large{= \frac{\pi}{2} I_{0} (2) -\frac{1}{2}\displaystyle \int^{\pi}_{0} e^{2 \cos x} \cos (2 \sin x) dx}$$

Let $$\large{L=\displaystyle \int^{\pi}_{0} e^{2 \cos x} \cos (2 \sin x) dx}$$

$$\large{L=\Re \left(\displaystyle \int^{\pi}_{0} e^{2 \cos x}e^{2i \sin x} dx \right)}$$

$$\large{L=\Re \left(\displaystyle \int^{\pi}_{0} e^{2 (\cos x+i \sin x)} dx \right)}$$

$$\large{L=\Re \left(\displaystyle \int^{\pi}_{0} e^{2e^{ix}} dx \right)}$$

$$\large{L=\Re \left(\displaystyle \int^{\pi}_{0}\displaystyle \sum_{k=0}^{\infty} \frac{(2e^{ix})^{k}}{k!}dx \right)}$$

$$\large{L=\displaystyle \int^{\pi}_{0}\displaystyle \sum_{k=0}^{\infty} \frac{2^{k} \cos (kx)}{k!}dx}$$

Notice that $$\large{\displaystyle \int^{\pi}_{0} \cos (kx) dx=0 \quad \quad \forall k \geq 1 \\ \text{k is an integer}}$$

Thus only $$k=0$$ case is considered and hence $$L=\pi$$

On combining we have

$$\large{ \frac{\pi}{2} \left(I_{0}(2)-1 \right)}$$

$$\large{ \frac{\pi}{2} \left(J_{0}(2i)-1 \right)}$$ · 1 year ago

Problem 36:

Evaluate $\displaystyle \int _{ 0 }^{ \infty }{ \dfrac { \sin x }{ { e }^{ x }-1 } \, dx } .$

###### This problem has been solved by Deep Seth.
· 12 months ago

Comment deleted 12 months ago

Comment deleted 12 months ago

$$\large{\sum _{ r=1 }^{ \infty }{ \int _{ 0 }^{ \infty }{ { e }^{ -rx }\sin { \left( x \right) } dx } } \\ Using\quad laplace\quad tranform\quad of\quad \sin { \left( x \right) } \\ \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ { r }^{ 2 }+1 } } }$$

Lemma

$$\large{\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 }-{ a }^{ 2 } } =\frac { 1 }{ 2a } -\frac { \pi }{ 2a } } \cot { \left( \pi a \right) } }$$

Proof

$$\large{\frac { 1 }{ 2a } \left( \frac { 1 }{ n-a } -\frac { 1 }{ n+a } \right) \\ =\frac { 1 }{ 2a } \left( \psi \left( 1+a \right) -\psi \left( 1-a \right) \right) \\ =\frac { 1 }{ 2a } \left( \frac { 1 }{ a } +\psi \left( a \right) -\psi \left( 1-a \right) \right) \\ Using\quad reflection\quad formula\\ =\frac { 1 }{ 2{ a }^{ 2 } } -\frac { \pi }{ 2a } \cot { \left( \pi a \right) } }$$

here $$a\rightarrow i$$

we get $$\large{-\frac { 1 }{ 2 } -\frac { \pi }{ 2i } \cot { \left( i\pi \right) } \\ =-\frac { 1 }{ 2 } +\frac { \pi }{ 2 } \coth { \left( \pi \right) } }$$ · 12 months ago

Problem 40:

Prove that $\large \int_{0}^{{\pi} / {2}} x\left(\dfrac{\ln(\sin x)-2}{\sqrt{\sin x}}\right) \cot x \ \mathrm{d}x = - \sqrt{\frac{\pi}{8}} \left[\Gamma \left(\dfrac{1}{4}\right)\right]^{2}$

###### This problem has been solved by Surya Prakesh.
· 11 months, 4 weeks ago

$\int_{0}^{\pi /2} x \left(\dfrac{\ln(\sin x) - 2}{\sqrt{\sin x}} \right) \cot (x) dx \\ = x \left(\int\left(\dfrac{\ln(\sin x) - 2}{\sqrt{\sin x}} \right) \cot (x) dx \right) \Biggl|_{0}^{\pi /2} - \int_{0}^{\pi /2} \left(\int\left(\dfrac{\ln(\sin x) - 2}{\sqrt{\sin x}} \right) \cot (x) dx \right) dx$

\begin{align*} \int \left(\dfrac{\ln(\sin x) - 2}{\sqrt{\sin x}} \right) \cot x dx &= \int \left(\dfrac{\ln(t) - 2}{t \sqrt{t}} \right) dt \\ &=2 \int \dfrac{\ln t d(\sqrt{t} ) - \sqrt{t} d(\ln(t))}{(\sqrt{t}) ^2} \\ &= -2\dfrac{\ln(t)}{\sqrt{t}}\\ &= -2 \dfrac{\ln(\sin(x))}{\sqrt{\sin(x)}} \end{align*}

So,

\begin{align*} x \left(\int\left(\dfrac{\ln(\sin (x)) - 2}{\sqrt{\sin x}} \right) \cot x dx \right) \Biggl|_{0}^{\pi /2} &= \dfrac{-2x\ln(\sin(x))}{\sqrt{\sin(x)}} \Biggl|_{0}^{\infty}\\ &= 0 - \lim_{a \rightarrow 0} \dfrac{-2a \ln(\sin(a))}{\sqrt{\sin(a)}}\\ &= \lim_{a \rightarrow 0} \dfrac{2a \ln(\sin(a))}{\sqrt{\sin(a)}}\\ &= 0 \end{align*}

Above limit is evaluated using L'Hospital rule.

So,

$\int_{0}^{\pi /2} x \left(\dfrac{\ln(\sin x) - 2}{\sqrt{\sin x}} \right) \cot x dx = - \int_{0}^{\pi /2} \left(\int\left(\dfrac{\ln(\sin x) - 2}{\sqrt{\sin x}} \right) \cot x dx \right) dx = 2 \int_{0}^{\pi /2} \dfrac{\ln(\sin(x))}{\sqrt{\sin(x)}} dx$

Let $$I(a) = \int_{0}^{\pi /2} \sin^{2a-1} (x) dx$$. So, our required integral is $$I'(1/4)$$. By taking $$t = \sin ^2 (x)$$, the integral transforms to $I(a) = \dfrac{1}{2} \int_{0}^{1} t^{a-1} (1-t)^{-1/2} dt = \dfrac{1}{2} B(a, 1/2) = \dfrac{1}{2} \dfrac{\Gamma(a) \Gamma(1/2)}{\Gamma(a+1/2)}$

Differentiate it w.r.t $$a$$,

$I'(a) = \dfrac{1}{2} \dfrac{\Gamma(a) \Gamma(1/2)}{\Gamma(a+1/2)} \left( \psi(a) - \psi(a+1/2) \right)$

$I'(1/4) = \dfrac{1}{2} \dfrac{\Gamma(1/4) \Gamma(1/2)}{\Gamma(3/4)} \left( \psi(1/4) - \psi(3/4) \right)$

Using Euler's reflection formula. We get

$\Gamma(1/4) \Gamma(3/4) = \dfrac{\pi}{\sin(\pi/4)} = \sqrt{2} \pi \implies \Gamma(3/4) = \dfrac{\sqrt{2} \pi}{\Gamma(1/4)}$

Also, $$\psi(1/4) - \psi(3/4) = \pi \cot(3 \pi /4) = - \pi$$.

Therefore the required integral evaluates to

$-\sqrt{\dfrac{ \pi}{8}} \left( \Gamma \left(\dfrac{1}{4} \right) \right) ^2$ · 11 months, 4 weeks ago

Problem 34:

Prove That $\large \int_{0}^{\infty} \left(\dfrac{\sin(x^2)}{x^2}\right)\cdot \left(x^2+x-1\right) \ \mathrm{d}x = \dfrac{\pi}{4} - \sqrt{\dfrac{\pi}{8}}.$

###### This problem has been solved by Surya Prakesh.
· 12 months ago

Lemma:

Consider the Gamma Function $\Gamma(s) = \int_{0}^{\infty} e^{-t} t^{s-1} dt$

Take the substitution, $$t=pu^n$$.

$\Gamma(s) = n p^{s} \int_{0}^{\infty} e^{-pu^n} u^{ns-1} du$

Taking $$p= a+ ib$$ and comparing imaginary parts we get,

$\int_{0}^{\infty} e^{-au^n} u^{ns-1} \sin \left(bu^n \right) du = \dfrac{\Gamma(s)}{n |p|^s} \sin(\alpha s )$

where $$p=a+ib$$ and $$\alpha$$ is the argument of $$p$$.

1st Integral:

$\int_{0}^{\infty} \sin \left(x^2 \right) dx$

Comparing this with our lemma, we get $$a=0$$, $$b=1$$, $$n=2$$, $$s=1/2$$. Which gives us that $$|p| = 1$$ and $$\alpha = \pi /2$$.

On substitution we get the integral evaluated to be $$I_{1} = \dfrac{\sqrt{\pi}}{2 \sqrt{2}}$$.

2nd Integral:

$\int_{0}^{\infty} \dfrac{\sin (x^2)}{x} dx$

Take the substitution, $$x = t^{1/2}$$. We get,

$\dfrac{1}{2} \int_{0}^{\infty} \dfrac{\sin(t)}{t} dt$

This is well known Integral (Dirichlet integral): $$\int_{0}^{\infty} \dfrac{\sin(x)}{x} dx = \dfrac{\pi}{2}$$.

So, this integral evaluates to be $$I_{2} = \dfrac{\pi}{4}$$.

3rd Integral:

This can evaluated in similar way as we obtained for first integral. And this is evaluated to be $$I_{3} = \dfrac{\sqrt{\pi}}{\sqrt{2}}$$.

So, the overall integral $$I = I_{1} + I_{2} - I_{3} =\large{ \boxed{ \dfrac{\pi}{4} - \dfrac{\sqrt{\pi}}{2 \sqrt{2}}}}$$ · 12 months ago

Problem 29:

Show that $\large \int_0^1 \dfrac{\ln \left(\cos(\frac{\pi x}{2})\right)}{x(x+1)} \, dx = \dfrac{1}{2} (\ln2)^2 - (\ln \pi)( \ln 2 ) .$

###### This problem has been solved by Tanishq Varshney.
· 1 year ago

Solution to Problem 29

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

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

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

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

In the second integral $$x \rightarrow -x$$

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

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

Now in the second integral $$x+1 \rightarrow x$$

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

$$\large{= \displaystyle \lim_{a \to 0} \left( \displaystyle \int^{1}_{a} \frac{\ln \left(\frac{\sin (\pi x)}{2} \right)}{x}dx-\displaystyle \int^{2}_{a} \frac{\ln \left(\sin \left( \frac{\pi x}{2}\right) \right)}{x}dx \right)}$$

Again in the second integral $$\large{\frac{x}{2} \rightarrow x}$$

$$\large{= \displaystyle \lim_{a \to 0} \left( -\ln 2 \displaystyle \int^{1}_{a} \frac{1}{x} dx+\displaystyle \int^{1}_{a} \frac{\ln (\sin (\pi x))}{x} dx-\displaystyle \int^{1}_{\frac{a}{2}} \frac{\ln (\sin (\pi x))}{x} dx \right)}$$

$$\large{= \displaystyle \lim_{a \to 0} \left( -\ln 2 \displaystyle \int^{1}_{a} \frac{1}{x} dx -\displaystyle \int^{a}_{\frac{a}{2}} \frac{\ln (\sin (\pi x))}{x} dx \right)}$$

We know $$\sin \theta \approx \theta$$ when $$\theta$$ approaches zero.

$$\large{= \displaystyle \lim_{a \to 0} \left( -\ln 2 \displaystyle \int^{1}_{a} \frac{1}{x} dx -\displaystyle \int^{a}_{\frac{a}{2}} \frac{\ln (\pi x)}{x} dx \right)}$$

$$\large{= \displaystyle \lim_{a \to 0} \left( \ln 2 \ln a- \ln \pi \ln 2 -\frac{1}{2} \left((\ln a)^2-(\ln \frac{a}{2})^2 \right) \right)}$$

using $$a^2-b^2=(a-b)(a+b)$$ and on simplifying the terms containing $$a$$ get cancelled and we finally have

$$\Large{\boxed{\frac{1}{2} (\ln 2)^2-(\ln \pi) (\ln 2)}}$$ · 1 year ago

# CHEATER!!!

· 12 months ago

Problem 28:

Evaluate $\large{\displaystyle\int _{ 0 }^{ \infty }{ \frac { { x }^{ 5 }\sin { (x) } }{ { (1+{ x }^{ 2 }) }^{ 3 } } \, dx }}.$

###### This problem has been solved by Sudeep Salgia.
· 1 year ago

Solution to Problem 28:

The following result is directly used in the solution:$$\displaystyle \int_0^{\infty} \frac{ \cos (mx)}{(a^2+x^2)^2} dx = \frac{\pi}{4a^3} e^{-am} ( 1 + am)$$.
This can be obtained easily by differentiating the integral w.r.t. $$a$$ evaluated in solution of problem 15 of this contest. Consider,
$$\displaystyle I(b) = \int_0^{\infty} \frac{x \sin(bx) }{ (1+ x^2)^3} dx$$. The required integral is $$\dfrac{\partial^4}{\partial b^4} I(b)$$. Integrating by parts, we get,
$$\displaystyle I(b) = - \frac{\sin(bx)}{4(1+x^2)^2} \bigg|_0^{\infty} + \frac{b}{4} \int_0^{\infty} \frac{ \cos (bx)}{(1+x^2)^2} dx = \frac{\pi}{16} e^{-b} ( b+ b^2)$$

Differentiate four times w.r.t $$b$$ and put $$b = 1$$ to get the value of the integral as $$\dfrac{\pi}{8e}$$. · 1 year ago

Problem 27:

Evaluate

$\large{\displaystyle \int^{2}_{0} \frac{dx}{(2x^2-x^3)^{{1} / {3}}}}.$

###### This problem has been solved by Rajorshi Chaudhuri.
· 1 year ago

Substitute x=2t So the equation becomes $\large{\displaystyle \int^{1}_{0} \ { t }^{ -2/3 }{ (1-t) }^{ -1/3 } } dt$ This comes out as $\large{\displaystyle \Gamma (\frac { 1 }{ 3 } )\Gamma (\frac { 2 }{ 3 } )}$ which is equal to $\large{\displaystyle\frac { 2\pi }{ \sqrt { 3 }} }$ · 1 year ago

Can you guys join Slack? I intend to form a discussion group for the series+integrals enthusiast for us.

Do let me know if you do not know how to log in. · 11 months, 3 weeks ago

Hellolo sir, could I be part of this discussion? i'm also pretty interested in Integrals and series as well · 8 months, 3 weeks ago

Join Slack first. · 8 months, 3 weeks ago

Thanks! · 8 months, 3 weeks ago

How do you do you that? It isn't very clear · 8 months, 3 weeks ago

I'm in. · 11 months, 3 weeks ago

message me (3.14159han) · 11 months, 3 weeks ago

After entering my email Id and clicking on get my invite, nothing is happening. Please help · 6 months ago

Hi participants. Contour integration is allowed from now onwards. · 11 months, 3 weeks ago

Problem 38:

Find the Cauchy Principal value of: $\large\int_0^2\frac{\ln((1-0.25x^2)^3+x^3)-3\ln(x)}{(1+0.25x^2)\ln(1-0.25x^2)-(1+0.25x^2)\ln(x)}\, dx.$

###### This problem has been solved by Ronak Agarwal.
· 12 months ago

Begin with the substitution $$y=\dfrac{x}{2}$$ we have :

$$\displaystyle I = 2\int _{ 0 }^{ 1 }{ \frac { \ln { (\frac { 1 }{ 8 } { (\frac { 1 }{ y } -y) }^{ 3 }+1) } }{ (1+{ y }^{ 2 })\ln { (\frac { 1 }{ 2 } (\frac { 1 }{ y } -y)) } } } dy$$

Again we have the substitution $$\dfrac { 1 }{ 2 } (\dfrac { 1 }{ y } -y)=x$$ we have :

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

Honestly I don't believe this integral should converge as it blows up at x=1, but still I will proceed.

From problem 35 we have :

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

Now the integral can be shown to be equal to 0( although the function to be integrated blows up to infinity)

hence answer is $$I = \dfrac{3\pi}{4}$$. · 12 months ago

Problem 37:

Evaluate $\large{\displaystyle \int^{\frac{\pi}{2}}_{0} \cot \left(\frac{\theta}{2} \right) \sqrt{\cos \theta} \ln (\cos \theta) \, d \theta}.$

###### This problem has been solved by Surya Prakesh (first) and Kenny Lau (second) almost at the same time.
· 12 months ago

$\int_{0}^{\pi/2} \cot \left(\dfrac{\theta }{2} \right) \sqrt{\cos \theta} \ln \left( \cos \theta \right) d\theta = \int_{0}^{1} \dfrac{\sqrt{t} \ln(t) }{1-t} dt$

In the above integral I took $$\cos \theta = t$$ and I used the identity $$\cot \dfrac{\theta}{2} = \sqrt{\dfrac{1+\cos \theta}{1- \cos \theta}}$$.

Let $$I(a) = \int_{0}^{1} \dfrac{t^a}{1-t} dt$$. So, we have to find $$I'(1/2)$$.

\begin{align*} I(a) &= \int_{0}^{1} \dfrac{t^a}{1-t} dt \\ &= \int_{0}^{1} t^a \sum_{n=0}^{\infty} t^n dt \\ &= \sum_{n=0}^{\infty} \int_{0}^{1} t^{n+a} dt \\ &= \sum_{n=0}^{\infty} \dfrac{1}{n+a+1} \end{align*}

Differentiate it on both sides w.r.t $$a$$. which gives,

$\dfrac{d}{da} \sum_{n=0}^{\infty} \dfrac{1}{n+a+1} = - \sum_{n=1}^{\infty} \dfrac{1}{(n+a)^2}$

Take $$a=1/2$$ to evaluate the required integral.

$\sum_{n=1}^{\infty} \dfrac{1}{(n+1/2)^2} = 4\sum_{n=1}^{\infty} \dfrac{1}{(2n+1)^2}$

But we know that

\begin{align*} \dfrac{\pi^2}{6} &= 1 + \dfrac{1}{2^2 } + \dfrac{1}{3^2} + \ldots \\ &= 1 + \dfrac{1}{3^2} + \dfrac{1}{5^2} + \ldots + \dfrac{1}{4} \left( 1+ \dfrac{1}{2^2 } + \dfrac{1}{3^2} + \ldots \right) \end{align*} \\ \implies 1+ \dfrac{1}{3^2} + \dfrac{1}{5^2} + \ldots = \dfrac{\pi^2}{8} \\ \implies \sum_{1}^{\infty} \dfrac{1}{(2n+1)^2} = \dfrac{\pi^2}{8} - 1

So, $\int_{0}^{\pi /2} \cot \left(\dfrac{\theta }{2} \right) \sqrt{\cos \theta} \ln \left( \cos \theta \right) d\theta = -4\sum_{n=1}^{\infty} \dfrac{1}{(2n+1)^2} = \boxed{\dfrac{8-\pi^2}{2}}$ · 12 months ago

I first apply the fact that $$\cot(x/2) = \frac{1+\cos x}{\sin x}$$.

$$\quad\displaystyle\int^{\frac{\pi}{2}}_{0} \cot \left(\frac{\theta}{2} \right) \sqrt{\cos \theta} \ln (\cos \theta) d \theta =\displaystyle\int^{\frac{\pi}{2}}_{0} \frac{1+\cos\theta}{\sin\theta} \sqrt{\cos \theta} \ln (\cos \theta) d \theta =\displaystyle\int^{0}_{1} \frac{1+u}{\sqrt{1-u^2}} \sqrt{u} \ln (u) \frac{-d u}{\sqrt{1-u^2}}$$

$$=\displaystyle\int^{1}_{0} \frac{1+u}{1-u^2} \sqrt{u} \ln (u) d u$$

$$=\displaystyle\int^{1}_{0} \frac{\sqrt{u} \ln (u)}{1-u} d u$$

$$=\displaystyle\int^{1}_{0} \frac{\sqrt{t^2} \ln (t^2)}{1-t^2} 2tdt$$

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

$$=\displaystyle-\int^{1}_{0} \ln(t) d\left(4t+2\ln(1-t)-2\ln(1+t)\right)$$

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

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

$$= 4 + 2\displaystyle\int^{1}_{0} \frac1t \sum_{n=1}^{\infty}(-1)^{n+1}\frac{(-t)^n-t^n}n dt$$

$$= 4 - 2\displaystyle\int^{1}_{0} \frac1t \sum_{n=1}^{\infty}\frac{2t^{2n-1}}{2n-1} dt$$

$$= 4 - 2\displaystyle\int^{1}_{0} \sum_{n=1}^{\infty}\frac{2t^{2n-2}}{2n-1} dt$$

$$= 4 - 4\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}= 4 - 4\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2} + 4\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n)^2} = 4 - \frac{4\pi^2}6 + \frac{\pi^2}6 =\boxed{ 4 - \frac{\pi^2}2}.$$ · 12 months ago

Problem 32:

Let $$f(x) = \dfrac12 \ln^2 (1-x)$$ for $$0\leq x \leq \pi$$. Use Parseval's Theorem to show that $\large \int _{ 0 }^{ \pi }{ t^{ 2 }\ln { ^{ 2 } } \left( 2\cos \left( \frac { t }{ 2 } \right) \right) \, dt }= \frac{11}{180}\pi^5.$

###### Due to time constraint, the author decided to post their own solution.
· 1 year ago

(fixed)

Here's the synopsis of the proof:

• I'm going to use the fact that $$\displaystyle \sum_{n=1}^\infty \dfrac{H_n^2}{(n+1)^2} = \dfrac{11}{360}\pi^4$$. This value can be proven by seeing Cody's or robjohn's answer. (The proof of this is very long so I've omitted the proof)

• And use Parseval's Theorem to evaluate the integral above by first noting that the integrand is an even function.

Recall that:

$\displaystyle f(x)=\sum _{ n=1 }^{ \infty }{ { b }_{ n } } { x }^{ n } \quad \Rightarrow \quad f({ e }^{ ix })=\sum _{ n=1 }^{ \infty }{ { b }_{ n } } { e }^{ ixn }$

Now according to Fourier series,

$F(x)=\frac { 1 }{ 2 } { a }_{ 0 }+\sum _{ n=1 }^{ \infty }{ { b }_{ n } } { \sin { (nx) } }+\sum _{ n=1 }^{ \infty }{ { a }_{ n } } { \cos { (nx) } }$

Parseval's Theorem then states that

$\frac { 1 }{ \pi } \int _{ -\pi }^{ \pi }{ { F(x) }^{ 2 }dx } =\frac { 1 }{ 2 } { a }_{ 0 }^{ 2 }+\sum _{ n=1 }^{ \infty }{ { a }_{ n }^{ 2 }+{ b }_{ n }^{ 2 } }$

Substituting $$F(x)=\Im { (f({ e }^{ ix })) }$$, $${ a }_{ n }=0$$, $$b_n=\frac{H_n}{n+1}$$ gives

$\frac { 1 }{ \pi } \int _{ -\pi }^{ \pi }{ \Im({ f(e }^{ ix } } ))^2dx=\sum _{ n=1 }^{ \infty }{ { b }_{ n }^{ 2 } }$

Applying these facts above, we get

$\Im (f(e^{ ix }))=\frac { 1 }{ 2 } \left( x-\pi \right) \ln { \left( 2\sin \left( \frac { 1 }{ 2 } x \right) \right) } ,0\le x\le \pi$

$\frac { 1 }{ \pi } \int _{ -\pi }^{ \pi }{ \frac { 1 }{ 4 } { (x-\pi ) }^{ 2 }\ln { ^{ 2 } } \left( 2\sin \left( \frac { 1 }{ 2 } x \right) \right) dx } =\frac { 1 }{ 2\pi } \int _{ 0 }^{ \pi }{ { x }^{ 2 }\ln { ^{ 2 } } \left( 2\cos \left( \frac { 1 }{ 2 } x \right) \right) dx } =\sum _{ n=1 }^{ \infty }{ { b }_{ n }^{ 2 } } =\frac { 11 }{ 360 } { \pi }^{ 4 }$ · 12 months ago

Problem 31:

Prove that $\large \lim_{n \rightarrow \infty} \int_{0}^{1} \dfrac{x^n - x^{2n}}{1-x} \, dx = \ln (2).$

###### This problem has been solved by Julian Poon.
· 1 year ago

Using basic concepts in integration:

$$\quad\displaystyle\int_0^1\frac{x^n-x^{2n}}{1-x}\mathrm dx=\int_0^1x^n\frac{1-x^n}{1-x}\mathrm dx=\int_0^1x^n\sum_{k=0}^{n-1}x^k\ \mathrm dx=\sum_{k=0}^{n-1}\int_0^1x^{n+k}\ \mathrm dx=\sum_{k=0}^{n-1}\frac1{n+k+1}=\frac1n\sum_{k=1}^{n}\frac1{1+\frac kn}$$

Then:

$$\displaystyle\lim_{n\to\infty}\int_0^1\frac{x^n-x^{2n}}{1-x}\mathrm dx=\lim_{n\to\infty}\frac1n\sum_{k=1}^{n}\frac1{1+\frac kn}=\int_0^1\frac1{1+x}\mathrm dx=[\ln(1+x)]_0^1=\ln2$$ · 12 months ago

Substituting $$\displaystyle \sum _{ k=0 }^{ \infty }{ x^{ k } } =\frac { 1 }{ 1-x }$$

$\int _{ 0 }^{ 1 }{ \sum _{ k=0 }^{ \infty } x^{ k+n }-\sum _{ k=0 }^{ \infty } x^{ k+2n } } dx=\sum _{ k=1 }^{ \infty } \frac { 1 }{ k+n } -\sum _{ k=1 }^{ \infty } \frac { 1 }{ k+2n }$

$={ H }_{ 2n }-{ H }_{ n }=\ln(2)$

The last step is from the approximation $$H_n=\gamma+\ln(n)$$ as $$n\to \infty$$ · 1 year ago

Problem 30:

Prove that $\large{\displaystyle \int^{{\pi} / {8}}_{0} \ln (\tan 2x) \, dx=-\frac{G}{2}}.$

Where $${G}$$ denotes the Catalan's constant, $$\displaystyle G = \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^2} \approx 0.915 965 59$$.

###### This problem has been solved by Surya Prakesh (first) and Julian Poon (second) almost at the same time.
· 1 year ago

Lemma : $\int_{0}^{1} t^{a} \ln(t) dt = -\dfrac{1}{(a+1)^2}$

This lemma is already proved in many problems in this contest.

$\int_{0}^{\pi/8} \ln(\tan(2x))dx =1/2 \int_{0}^{1} \dfrac{\ln(t)}{1+t^2} dt$

I used the substitution $$t=\tan(2x)$$.

\begin{align*} \int_{0}^{1} \dfrac{\ln(t)}{1+t^2} dt &= \int_{0}^{1} \sum_{i=0}^{\infty}(-1)^{i} t^{2i} \ln(t)dt \\ &= \sum_{i=0}^{\infty}(-1)^{i} \int_{0}^{1} t^{2i} \ln(t) dt \\ &= -\sum_{i=0}^{\infty} (-1)^{i} \dfrac{1}{(2i+1)^2} \\ &= - \beta(2) \\ &= -G \end{align*}

Therefore,

$\int_{0}^{\pi/8} \ln(\tan(2x))dx = \dfrac{-G}{2}$ · 1 year ago

Not sure if this is considered but...

$y=\ln \left(\tan \left(2x\right)\right)$

$x=\frac{\tan ^{-1}\left(e^y\right)}{2}$

$\int _{ 0 }^{ \frac { \pi }{ 8 } }{ \ln { \left( \tan \left( 2x \right) \right) } } dx=\int _{ -\infty }^{ 0 }{ \frac { \tan ^{ -1 } \left( e^{ y } \right) }{ 2 } } dy=-\frac { 1 }{ 2 } \int _{ 0 }^{ \infty }{ \tan ^{ -1 } \left( e^{ -y } \right) } dy$

A known identity of the catalan constant is $G=\int _{ 0 }^{ \infty }{ \tan ^{ -1 } \left( e^{ -y } \right) } dy$

Hence the integral is $$-\frac{G}{2}$$

The identity above can be deduced through the Taylor Series of $$\tan^{-1}(x)$$ · 1 year ago

Problem 50:

Evaluate $\large \int_{0}^{\infty} \dfrac{\ln(1+x)}{\sqrt[4]{x^3}(1+x)}\, dx$ · 11 months, 3 weeks ago

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

$$\displaystyle I = -\int _{ 0 }^{ 1 }{ { y }^{ (-1/4) }{ (1-y) }^{ -3/4 }\ln { (y) } dy }$$

Use the definition of beta function we have :

$$\displaystyle \int _{ 0 }^{ 1 }{ { y }^{ m-1 }{ (1-y) }^{ n-1 }dy } = \frac { \Gamma (m)\Gamma (n) }{ \Gamma (m+n) }$$

Differentiating it with respect to $$m$$ we have :

$$\displaystyle \frac { \Gamma (m)\Gamma (n) }{ \Gamma (m+n) } (\psi (m)-\psi (m+n)) = \int _{ 0 }^{ 1 }{ { y }^{ m-1 }{ (1-y) }^{ n-1 }\ln { (y) } dy }$$

Put $$\displaystyle m=\dfrac{3}{4} , n=\dfrac{1}{4}$$ to get :

$$\displaystyle I = \frac { \Gamma (3/4)\Gamma (1/4) }{ \Gamma (1) } (\psi (1)-\psi (3/4))$$

Using reflection formula for the gamma function we have :

$$\displaystyle \Gamma(1/4)\Gamma(1-1/4) = \dfrac{\pi}{\sin(\dfrac{\pi}{4})}=\sqrt{2}\pi$$

We will use that $$\displaystyle \psi(x+1) + \gamma = \int _{ 0 }^{ 1 }{ \frac { 1-{ t }^{ x } }{ 1-t } dt }$$

$$\displaystyle \Rightarrow \psi(x+1)-\psi(y+1) = \int _{ 0 }^{ 1 }{ \frac { { t }^{ y }-{ t }^{ x } }{ 1-t } dt }$$

$$\displaystyle \Rightarrow \psi(1)-\psi(3/4) = \int _{ 0 }^{ 1 }{ \frac { { t }^{ -1/4 }-1 }{ 1-t } dt }=J$$

For evaluting $$J$$ we will put $$t={x}^{4}$$, to get :

$$\displaystyle J = 4\int _{ 0 }^{ 1 }{ \frac { 1-x }{ 1-{ x }^{ 4 } } { x }^{ 2 }dx }$$

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

$$\displaystyle J =2\int _{ 0 }^{ 1 }{ \frac { dx }{ 1+x } } +\int _{ 0 }^{ 1 }{ \frac { 2x }{ 1+{ x }^{ 2 } } dx } -2\int _{ 0 }^{ 1 }{ \frac { dx }{ 1+{ x }^{ 2 } } }$$

$$\displaystyle J = 3\ln { (2) }-\dfrac{\pi}{2}$$

$$\displaystyle \Rightarrow I = 3\sqrt { 2 } \pi \ln { (2) } -\frac { { \pi }^{ 2 } }{ \sqrt { 2 } }$$ · 11 months, 3 weeks ago

Alternative method:

$\dfrac{\ln(1+x)}{1+x} = - \sum_{n=1}^{\infty} H_{n}(-x)^{n}$

Now use RMT. That's it. $$\ddot \smile$$ · 11 months, 2 weeks ago

Problem 47:

Evaluate $\displaystyle \large \int _{ 0 }^{ \infty }{ { e }^{ -{ x }^{ 2 } }\ln { (x) } \, dx. }$

###### This problem has been solved by Julian Poon.
· 11 months, 3 weeks ago

Lemma $\int _{ 0 }^{ \infty }{ x^{ a }\cdot e^{ -x^{ 2 } }dx } =\frac { \Gamma \left( \frac { a+1 }{ 2 } \right) }{ 2 }$

Giving the substitution $$x^2=t$$, the integral becomes:

$\int _{ 0 }^{ \infty }{ x^{ a }\cdot e^{ -x^{ 2 } }dx } =\frac { 1 }{ 2 } \int _{ 0 }^{ \infty }{ t^{ \frac { a-1 }{ 2 } }\cdot e^{ -t }dt } =\frac { \Gamma \left( \frac { a+1 }{ 2 } \right) }{ 2 }$

So the required integral is

$=\frac { 1 }{ 4 } \left[ \Gamma \left( \frac { 1 }{ 2 } \right) \psi \left( \frac { 1 }{ 2 } \right) \right] =-\frac { 1 }{ 4 } \sqrt { \pi } (\gamma +\ln{(4)})$

To show that $$\psi(1/2)=-\gamma-\ln(4)$$, I'll use $$\psi(a)=H_{a-1}-\gamma$$

$\psi(1/2)=H_{-1/2} - \gamma$

From the definition of harmonic numbers:

$H_{-1/2}=H_{1/2} - 2$

$H_{1/2}=\int _{ 0 }^{ 1 }{ \frac { 1-\sqrt { x } }{ 1-x } dx } =2\int _{ 0 }^{ 1 }{ t\left( \frac { 1-t }{ 1-t^{ 2 } } \right) dt }$

I used the substitution $$t=\sqrt{x}$$ above

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

Hence

$H_{-1/2}=H_{1/2} - 2=2\left(1-\ln \left(2\right)\right)-2=-\ln(4)$

$\psi(1/2)=-\ln(4) - \gamma$ · 11 months, 3 weeks ago

Problem 45:

Prove that $\large \int_0^{\pi /2} \sin^6 (x) \tan (x ) \sin(\tan x) \, dx = \dfrac{5\pi}{96e} .$

###### This problem has been solved by Ishan Singh.
· 11 months, 3 weeks ago

Substitute $$\displaystyle \tan x \mapsto x$$

$$\displaystyle \implies \text{I} = \int_{0}^{\infty} \dfrac{x^{7}\sin x}{(1+x^2)^4} \mathrm{d}x$$

From solution of Problem 28, we have,

$$\displaystyle \int_{0}^{\infty} \dfrac{x\sin mx}{(a^2+x^2)^3} \mathrm{d}x = \dfrac{\pi b e^{-a}}{16a^3} (1+am) \quad \quad \quad \quad \tag{*}$$

Differentiating $$(*)$$ $$6$$ times w.r.t. $$m$$ , $$1$$ time w.r.t. $$a$$, at $$a=m=1$$ and simplifying, we get,

$$\displaystyle \text{I} = \boxed{\dfrac{5 \pi}{96e}}$$ · 11 months, 3 weeks ago

Problem 42:

Evaluate $\displaystyle \large \int _{ 0 }^{ \frac{\pi}{2} }{ \ln { \left( 2\cos \left( \frac { x }{ 2 } \right) \right) } \,dx }$

###### This problem has been solved by Surya Prakesh.
· 11 months, 3 weeks ago

$\ln \left( 2\cos \left(\dfrac{x}{2} \right) \right) = \ln \left(e^{ix/2} + e^{-ix/2} \right) = \ln \left( 1 + e^{ix} \right) - i\dfrac{x}{2}$

So,

\begin{align*} \int_{0}^{\pi / 2} \ln \left( 2\cos \left(\dfrac{x}{2} \right) \right) dx &= \int_{0}^{\pi / 2} \ln \left( 1 + e^{ix} \right) dx - i \int_{0}^{\pi / 2}\dfrac{x}{2} dx \\ &= \int_{0}^{\pi /2} \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} e^{irx}}{r} dx - i \dfrac{\pi^2}{16} \end{align*}

We shall evaluate the first integral

\begin{align*} \int_{0}^{\pi /2} \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} e^{irx}}{r} dx &= \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1}}{r} \int_{0}^{\pi /2} e^{irx} dx \\ &= \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1}}{r} \left[ \dfrac{e^{irx}}{ir} \right|_{0}^{\pi/2}\\ &= \dfrac{1}{i} \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} \left(e^{ir \pi /2} - 1\right)}{r^2} \end{align*}

Now,

$e^{ ir\pi /2 }-1=\quad \begin{cases} i-1 & \quad \quad r=4k+1 \\ -2 & \quad \quad r=4k+2 \\ -i-1 & \quad \quad r=4k+3 \\ 0 & \quad \quad r=4k+4 \end{cases}$

\begin{align*}\sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} \left(e^{ir \pi /2} - 1\right)}{r^2} &= \sum_{k=0}^{\infty} \dfrac{(-1)^{4k+2}(i-1)}{(4k+1)^{2}} + \sum_{k=0}^{\infty} \dfrac{(-1)^{4k+3}(-2)}{(4k+2)^{2}} + \sum_{k=0}^{\infty} \dfrac{(-1)^{4k+4}(-i-1)}{(4k+3)^{2}} + \sum_{k=0}^{\infty} \dfrac{(-1)^{4k+5}(0)}{(4k+4)^{2}} \\ &= \sum_{k=0}^{\infty} \dfrac{(i-1)}{(4k+1)^{2}} + \sum_{k=0}^{\infty} \dfrac{2}{(4k+2)^{2}} + \sum_{k=0}^{\infty} \dfrac{-i-1}{(4k+3)^{2}} + 0 \\ &= \sum_{k=0}^{\infty} \dfrac{(i-1)}{(4k+1)^{2}} + \sum_{k=0}^{\infty} \dfrac{2}{(4k+2)^{2}} + \sum_{k=0}^{\infty} \dfrac{-i-1}{(4k+3)^{2}} \end{align*}

$\sum_{k=0}^{\infty} \dfrac{1}{(4k+1)^2} = \dfrac{1}{16} \zeta \left(2 , \dfrac{1}{4} \right) \\ \sum_{k=0}^{\infty} \dfrac{1}{(4k+2)^2} = \dfrac{1}{16} \zeta \left(2 , \dfrac{1}{2} \right) \\ \sum_{k=0}^{\infty} \dfrac{1}{(4k+3)^2} = \dfrac{1}{16} \zeta \left(2 , \dfrac{3}{4} \right)$

So,

\begin{align*} \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} \left(e^{ir \pi /2} - 1\right)}{r^2} &= (i-1) \dfrac{1}{16} \zeta \left(2 , \dfrac{1}{4} \right) + 2 \dfrac{1}{16} \zeta \left(2 , \dfrac{1}{2} \right) + (-i-1) \dfrac{1}{16} \zeta \left(2 , \dfrac{3}{4} \right)\\ &= \dfrac{1}{16} \left( i \left( \zeta \left(2 , \dfrac{1}{4} \right) - \zeta \left(2 , \dfrac{3}{4} \right) \right) - \left( \zeta \left(2 , \dfrac{1}{4} \right) + \zeta \left(2 , \dfrac{3}{4} \right) \right) + 2 \zeta \left(2 , \dfrac{1}{2} \right) \right) \end{align*}

1:
$\zeta \left(2 , \dfrac{1}{4} \right) - \zeta \left(2 , \dfrac{3}{4} \right) = 16 \sum_{k=0}^{\infty} \left(\dfrac{1}{(4k+1)^2} - \dfrac{1}{(4k+3)^2}\right) = 16 \sum_{k=0}^{\infty} \dfrac{(-1)^{k}}{(2k+1)^2} = 16G$

2:

\begin{align*} \zeta \left(2 , \dfrac{1}{4} \right) + \zeta \left(2 , \dfrac{3}{4} \right) &= 16 \sum_{k=0}^{\infty} \left(\dfrac{1}{(4k+1)^2} + \dfrac{1}{(4k+3)^2}\right) \\ &= 16 \sum_{k=0}^{\infty} \dfrac{1}{(2k+1)^2}\\ &=16 \dfrac{\pi^2}{8}\\ &= 2 \pi^2 \end{align*}

3:

$\zeta \left(2 , \dfrac{1}{2} \right) = 4 \sum_{k=0}^{\infty} \dfrac{1}{(2k+1)^2} = 4 \dfrac{\pi^2}{8} = \dfrac{\pi^2}{2}$

So,

\begin{align*} \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} \left(e^{ir \pi /2} - 1\right)}{r^2} &= \dfrac{1}{16} \left( i \left( \zeta \left(2 , \dfrac{1}{4} \right) - \zeta \left(2 , \dfrac{3}{4} \right) \right) - \left( \zeta \left(2 , \dfrac{1}{4} \right) + \zeta \left(2 , \dfrac{3}{4} \right) \right) + 2 \zeta \left(2 , \dfrac{1}{2} \right) \right)\\ &= \dfrac{1}{16} \left( i (16G) - (2\pi^2) + 2 \dfrac{\pi^2}{2} \right)\\ &= iG - \dfrac{\pi^2}{16} \end{align*}

\begin{align*} \int_{0}^{\pi / 2} \ln \left( 2\cos \left(\dfrac{x}{2} \right) \right) dx &= \dfrac{1}{i} \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} \left(e^{ir \pi /2} - 1\right)}{r^2} - i \dfrac{\pi^2}{16}\\ &= \dfrac{1}{i} \left( iG - \dfrac{\pi^2}{16} \right) - i \dfrac{\pi^2}{16} \\ &= \boxed{G} \end{align*}

## where, $$G$$ is Catalan's constant.

If you have an easy way, please do post it. · 11 months, 3 weeks ago

Intended solution:

Lemma:

$\displaystyle \ln (2\cos \left(x/2\right))=-\sum _{n=1}^{\infty}\frac{\left(-1\right)^n\cdot \cos \left(nx\right)}{n}$

$\sum _{n=1}^{\infty}\frac{\cos \left(nx\right)}{n}=\frac{1}{2}\sum _{n=1}^{\infty}\frac{e^{inx}+e^{-inx}}{n}=-\frac{1}{2}\left(\ln \left(1-e^{-ix}\right)+\ln \left(1-e^{ix}\right)\right)$ $=-\frac{1}{2}\left(2-2\cos \left(x\right)\right)=-\ln \left(2\sin \left(\frac{x}{2}\right)\right)$ $\ln \left(2\cos \left(\frac{x}{2}\right)\right)=\ln \left(2\sin \left(\frac{x+\pi }{2}\right)\right)=\sum _{n=1}^{\infty}\frac{\cos \left(n\left(x+\pi \right)\right)}{n}=-\sum _{n=1}^{\infty}\frac{\left(-1\right)^n\cdot \cos \left(nx\right)}{n}$

So,

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( 2\cos \left( \frac { x }{ 2 } \right) \right) } dx } =-\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sum _{ n=1 }^{ \infty } \frac { \left( -1 \right) ^{ n }\cdot \cos \left( nx \right) }{ n } dx } =-\sum _{ n=1 }^{ \infty } \frac { \left( -1 \right) ^{ n }\cdot \sin \left( \frac { 1 }{ 2 } \pi n \right) }{ { n }^{ 2 } }$ $=\sum _{ n=0 }^{ \infty } \frac { \left( -1 \right) ^{ n } }{ \left( 2n+1 \right) ^{ 2 } } =G$ · 11 months, 3 weeks ago

Problem 48:

Evaluate $\int _{ 0 }^{ \pi /2 }{ \frac { \csc ( x ) \sqrt { \tan x } }{ \sin x+\cos x } \, dx . }$

###### This problem has been solved by Surya Prakesh.
· 11 months, 3 weeks ago

Take the substitution $$t= \tan x$$.

\begin{align*} \int_{0}^{\pi/2} \dfrac{ \csc x \sqrt{\tan x}}{\sin x + \cos x} dx &= 2\int_{0}^{\pi/2} \dfrac{\sqrt{\tan x}}{(1+\tan x ) (\sin 2x)} dx \\ &= 2\int_{0}^{\infty} \dfrac{\sqrt{t}}{(1+t)(2t)} dt \\ &= \int_{0}^{\infty} \dfrac{\sqrt{t}}{(1+t)(t)} dt \end{align*}

Take $$t = tan^2 (y)$$.

$\int_{0}^{\infty} \dfrac{\sqrt{t}}{(1+t)(t)} dt = \int_{0}^{\pi /2} \dfrac{\tan y}{\sec^2 (y) \tan^2 (y)} (2 \tan(y) \sec^2 (y) dy) = 2 \int_{0}^{\pi/2}dy = \pi.$ · 11 months, 3 weeks ago

Problem 46:

Evaluate

$\large \lim_{n \to \infty} \int_{0}^{{\pi} / {4}} n \left(\cos^{2n} x - \sin^{2n} x \right) \tan 2x \ \mathrm{d}x.$ · 11 months, 3 weeks ago

Take $$\displaystyle f(n) = \int _{ 0 }^{ \frac { \pi }{ 4 } }{ n(\cos ^{ 2n }{ \theta } -\sin ^{ 2n }{ \theta } )\tan { (2\theta ) } d\theta }$$

Take $$cos(2\theta)=x$$, to get :

$$\displaystyle f(n) = \frac { n }{ { 2 }^{ n+1 } } \int _{ 0 }^{ 1 }{ \frac { { (1+x) }^{ n }-{ (1-x) }^{ n } }{ x } dx }$$

We have to find $$\displaystyle \lim _{ n\rightarrow \infty }{ f(n) }$$

We will first find the functional equation :

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

$$\displaystyle f(n) = \frac { nf(n-1) }{ 2(n-1) } +\frac { 1 }{ 2 }$$

Let the limit be $$L$$

$$\displaystyle \lim _{ n\rightarrow \infty }{ f(n) } =\lim _{ n\rightarrow \infty }{ \frac { nf(n-1) }{ 2(n-1) } +\frac { 1 }{ 2 } }$$

Hence $$\displaystyle L = \dfrac{1}{2}L+\dfrac{1}{2}$$

$$\Rightarrow \boxed{L=1}$$ · 11 months, 3 weeks ago

Problem 41: Evaluate $\large \int \sqrt{x+\sqrt{x^2 + 2}} \, dx.$

###### This problem has been solved by Pi Han Goh.
· 11 months, 4 weeks ago

This is a simple case of Euler substitution. Let $$y = \sqrt{x + \sqrt{x^2 + 2}}$$, then $y^2 = x + \sqrt{x^2 + 2 } \Rightarrow 2y \cdot \dfrac{dy}{dx} = 1 + \dfrac x{\sqrt{x^2+2}}= \dfrac {y^2}{\sqrt{x^2+2}} \Rightarrow dx = \dfrac{2\sqrt{x^2+2}}y dy.$

The integral becomes $\int y \cdot \dfrac{2\sqrt{x^2+2}}y dy = 2\int \sqrt{x^2 + 2} \, dy = 2 \int (y^2 - x) \, dy$

Because $$y^2 = x + \sqrt{x^2 + 2}$$, then $(y^2-x)^2 = x^2 + 2 \Rightarrow y^4 - 2xy^2 = 2 \Rightarrow x = \dfrac{y^4-2}{2y^2} = \dfrac12 y^2 - \dfrac1{y^2} .$

Continue: $2 \int \left (y^2 - \left(\dfrac12 y^2 - \dfrac1{y^2}\right) \right) \, dy = 2 \int \left (\dfrac12 y^2 + \dfrac1{y^2} \right) \, dy = \int \left (y^2 + \dfrac2{y^2} \right) \, dy = \dfrac13 y^3 - \dfrac2y + C .$

Substituting back gives $\dfrac13 \left( \sqrt{x + \sqrt{x^2 + 2}}\right)^3 - \dfrac2{ \sqrt{x + \sqrt{x^2 + 2}}} + C.$ · 11 months, 4 weeks ago

I'm just going to post my solution despite Pi Han's solution coming first.

$\sqrt{x+\sqrt{x^2+2}}=y --(1)$

$x=\frac{y^4-2}{2y^2}--(2)$

So the area under the graph of equation (1) would be equal to the area of the rectangle, $$x\times\sqrt{x+\sqrt{x^2+2}}$$, minus the area under equation (2) (Or the blue and red area minus the blue area): Link

So, integrating equation (2) to get the blue area:

$\int { \frac { y^{ 4 }-2 }{ 2y^{ 2 } } dy } =\frac { y^{ 4 }+6 }{ 6y }$

Hence the area under equation (1) can be expressed as:

$\left(x\sqrt{x+\sqrt{x^2+2}}\right)-\frac{\left(x+\sqrt{x^2+2}\right)^2+6}{6\sqrt{x+\sqrt{x^2+2}}}+C$

To validate the steps I made, I have to show that $$f=\sqrt{x+\sqrt{x^2+2}}$$ is a one-to-one function, or that it never takes the same value twice.

This can be shown by noticing that $$f$$ is a monotonically increasing function from $$x>0$$, since $$\sqrt{x^2+2}$$ is monotonically increasing from $$x>0$$ and hence $$\sqrt{x+\sqrt{x^2+2}}$$ is too. · 11 months, 4 weeks ago

Problem 39:

Evaluate $\displaystyle \int _{ 0 }^{ \infty }{ \left(\dfrac { \sin { (x) } }{ x } -\dfrac { 1 }{ 1+x } \right) } \dfrac { dx }{ x }.$

###### This problem has been solved by Ishan Singh.
· 12 months ago

Using Integration By Parts, we have,

$$\displaystyle \text{I} = 1 + \int_{0}^{\infty} \left(\cos x - \dfrac{1}{x+1} \right) \dfrac{\mathrm{d}x}{x}$$

$$\displaystyle = 1 + \text{J}$$ (Let)

$$\displaystyle \text{J} = \int_{0}^{\infty} \int_{0}^{\infty} \cos x \cdot e^{-ax} - \dfrac{e^{-ax}}{x+1} \mathrm{d}x \ \mathrm{d}a$$

$$\displaystyle = \int_{0}^{\infty} \dfrac{a}{a^2+1} \mathrm{d}a -\int_{0}^{\infty}\int_{0}^{\infty} \dfrac{e^{-ax}}{x+1}\mathrm{d}x \ \mathrm{d}a$$

Substitute $$ax \mapsto x$$

$$\displaystyle \implies \text{J} = \int_{0}^{\infty} \dfrac{a}{a^2+1} \mathrm{d}a -\int_{0}^{\infty}\int_{0}^{\infty} \dfrac{e^{-x}}{x+a}\mathrm{d}x \ \mathrm{d}a$$

Since $$\displaystyle \int_{0}^{\infty} e^{-x}\mathrm{d}x = 1$$

$$\displaystyle \implies \text{J} = \int_{0}^{\infty} \int_{0}^{\infty} \dfrac{ae^{-x}}{a^2+1} - \dfrac{e^{-x}}{x+a}\mathrm{d}x \ \mathrm{d}a$$

$$\displaystyle = \int_{0}^{\infty} e^{-x} \int_{0}^{\infty} \dfrac{a}{a^2+1} - \dfrac{1}{x+a}\mathrm{d}a \ \mathrm{d}x$$

$$\displaystyle = \int_{0}^{\infty} e^{-x} \left[\ln\left(\dfrac{\sqrt{a^2+1}}{a+x}\right)\right]_{a=0}^{a \to \infty} \mathrm{d}x$$

$$\displaystyle =\int_{0}^{\infty} e^{-x} \ln x \ \mathrm{d}x$$

$$\displaystyle = -\gamma$$

$$\displaystyle \implies \text{I} = \boxed{1 - \gamma}$$ · 11 months, 4 weeks ago

Problem 33:

Show that $\large \int _{ 0 }^{ \infty }{ \sin ^{ -1 } ( e^{ -x } ) } \, dx =\frac { 1 }{ 2 } \pi \ln { ( 2 ) } .$

###### This problem has been solved by Ishan Singh.
· 12 months ago

Substitute $$e^{-x} \mapsto \sin x$$

$$\displaystyle \implies \text {I} = \int_{0}^{\frac {\pi}{2}} x \cot x \ \mathrm{d}x$$

Using Integration by parts,

$$\displaystyle \text {I} = - \int_{0}^{\frac {\pi}{2}} \ln (\sin x) \ \mathrm{d} x$$

$$\displaystyle= \dfrac{\pi}{2} \ln 2$$ · 12 months ago

Problem 49:

Evaluate $\large \displaystyle \int _{ 0 }^{ 1 }{ \ln { (\Gamma (x)) } dx. }$

###### This problem has been solved by Surya Prakesh.
· 11 months, 3 weeks ago

$I = \int_{0}^{1} \ln(\Gamma(x)) dx = \int_{0}^{1} \ln(\Gamma(1-t)) dt$

I used the substitution $$x+t=1$$. But from Euler's reflection formula for Gamma function i.e.

$\Gamma(x) \Gamma(1-x)= \pi \csc(\pi x) \implies \ln(\Gamma(1-x)) + \ln(\Gamma(x)) = \ln(\pi) + \ln(\csc(\pi x))$

\begin{align*} I &= \int_{0}^{1} \ln(\Gamma(1-x)) dx\\ I &= -\int_{0}^{1} \ln(\Gamma(x)) dx + \int_{0}^{1} \ln(\pi) dx+ \int_{0}^{1} \ln(\csc(\pi x))dx \\ I &= -I + \ln(\pi) - \int_{0}^{1} \ln(\sin(\pi x))dx \\ 2I &= \ln(\pi) - \dfrac{1}{\pi} \int_{0}^{\pi} \ln(\sin(x))dx \\ 2I &= \ln(\pi) - \dfrac{2}{\pi} \int_{0}^{\pi / 2} \ln(\sin(x))dx \\ 2I &= \ln(\pi) - \dfrac{2}{\pi} \left(-\dfrac{\pi}{2} \ln(2) \right) \\ 2I &= \ln(\pi) + \ln(2) \\ I &= \dfrac{1}{2} \ln(2 \pi) \end{align*}

In the above solution I used the result that $$\int_{0}^{\pi/2} \ln(\sin(x))dx = -\dfrac{\pi}{2} \ln(2)$$. This is proved here as follows:

\begin{align*} J &=\int_{0}^{\pi/2} \ln(\sin(x))dx = \int_{0}^{\pi/2} \ln(\sin(\pi/2 - x))dx = \int_{0}^{\pi/2} \ln(\cos(x))dx \\ 2J &= \int_{0}^{\pi/2} \ln(\sin(x))dx + \int_{0}^{\pi/2} \ln(\cos(x))dx\\ 2J &= \int_{0}^{\pi/2} \ln(\sin(x) \cos(x))dx \\ 2J &= \int_{0}^{\pi/2} \ln(\sin(2x))dx - \int_{0}^{\pi/2} \ln(2)dx \\ 2J &= \dfrac{1}{2} \int_{0}^{\pi} \ln(\sin(x))dx - \dfrac{\pi}{2} \ln(2) \\ 2J &= \dfrac{1}{2}\left( 2\int_{0}^{\pi/2} \ln(\sin(x))dx \right) - \dfrac{\pi}{2} \ln(2) \\ 2J &= J - \dfrac{\pi}{2} \ln(2) \\ J &=- \dfrac{\pi}{2} \ln(2) \end{align*} · 11 months, 3 weeks ago

Problem 35:

Let $$a$$ and $$b$$ be constants, find the closed form of $\Large \int _{ 0 }^{ \infty } \frac { \ln\left( \frac { 1+x^{ a } }{ 1+x^{ b } } \right) }{ \left( 1+x^{ 2 } \right) \ln x } \, dx$ in terms of $$a$$ and/or $$b$$.

###### This problem has been solved by Ronak Agarwal.
· 12 months ago

Define $$\displaystyle f(a)=\int _{ 0 }^{ \infty }{ \dfrac { ln(1+{ x }^{ a }) }{ (1+{ x }^{ 2 })ln(x) } dx }$$

Differentiating with respect to $$a$$ we have :

$$\displaystyle f'(a)= \int _{ 0 }^{ \infty }{ \dfrac { { x }^{ a } }{ (1+{ x }^{ 2 })(1+{ x }^{ a }) } dx }$$

Use the substitution $$x=tan(\theta)$$ we have :

$$\displaystyle f'(a) = \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \dfrac { \tan ^{ a }{ \theta } d\theta }{ (1+\tan ^{ a }{ \theta } ) } }$$

Using $$\displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ f(x)dx } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ f(\dfrac { \pi }{ 2 } -x)dx }$$ we have :

$$\displaystyle f'(a) = \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \dfrac { d\theta }{ (1+\tan ^{ a }{ \theta } ) } }$$

Using these two forms we have :

$$\displaystyle 2f'(a) = \int _{ 0 }^{ \frac { \pi }{ 2 } }{ d\theta } =\dfrac { \pi }{ 2 }$$

$$\Rightarrow f'(a)=\dfrac { \pi }{ 4 }$$

Integrating this with upper limit as $$a$$ and lower limit as $$b$$ we have :

$$\displaystyle \int _{ 0 }^{ \infty }{ \dfrac { ln\left(\dfrac { 1+{ x }^{ a } }{ 1+{ x }^{ b } } \right) }{ (1+{ x }^{ 2 })ln(x) } dx } = \dfrac{\pi}{4} (a-b)$$ · 12 months ago

Problem 43:

Evaluate $\large \int \dfrac{x^3 e^{x^2} }{(x^2 + 1)^2} \, dx .$

###### This problem has been solved by Ronak Agarwal.
· 11 months, 3 weeks ago

Put $${x}^{2}=y$$ to get the integral as :

$$\displaystyle I = \frac { 1 }{ 2 } \int { \frac { y{ e }^{ y } }{ { (y+1) }^{ 2 } } dy }$$

It can be also written as :

$$\displaystyle I = \frac { 1 }{ 2 } \int { \frac { { e }^{ y } }{ (y+1) } -\frac { { e }^{ y } }{ { (y+1) }^{ 2 } } dy } =\frac { 1 }{ 2 } \int { \frac { d(\frac { { e }^{ y } }{ y+1 } ) }{ dy } dy }$$

$$I = \dfrac { { e }^{ y } }{ 2(y+1) } +C=\dfrac { { e }^{ { x }^{ 2 } } }{ 2({ x }^{ 2 }+1) } +C$$ · 11 months, 3 weeks ago