# 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.

5 years, 7 months 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.

- 5 years, 7 months 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 :

$\Large \boxed { I=\gamma }$

- 5 years, 7 months 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.

- 5 years, 7 months 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

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

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

- 5 years, 7 months ago

Problem 36:

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

###### This problem has been solved by Deep Seth.

- 5 years, 7 months 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.

- 5 years, 7 months 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 $\large{\displaystyle\frac { 2\pi }{ \sqrt { 3 }} }$

- 5 years, 7 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.

- 5 years, 7 months 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 $\dfrac{\pi}{8e}$.

- 5 years, 7 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.

- 5 years, 7 months 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

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

- 5 years, 7 months ago

# CHEATER!!!

- 5 years, 7 months 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.

- 5 years, 7 months 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{aligned} \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{aligned}

Therefore,

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

- 5 years, 7 months 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 $\tan^{-1}(x)$

- 5 years, 7 months 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.

- 5 years, 7 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 $I = I_{1} + I_{2} - I_{3} =\large{ \boxed{ \dfrac{\pi}{4} - \dfrac{\sqrt{\pi}}{2 \sqrt{2}}}}$

- 5 years, 7 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.

- 5 years, 7 months 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{aligned} \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{aligned}

So,

\begin{aligned} 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{aligned}

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

$-\sqrt{\dfrac{ \pi}{8}} \left( \Gamma \left(\dfrac{1}{4} \right) \right) ^2$

- 5 years, 7 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.

- 5 years, 7 months 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:

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

- 5 years, 7 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 $n\to \infty$