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

- I will start by posting the first problem. If there is a user solves it, then (s)he 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. - Please make a
**substantial comment**. - 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. - 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 as well as a complex integral problem.
- 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, and trigonometric integral.

Please post your solution and your proposed problem in **a single new thread**.

Format your post is as follows:

SOLUTION OF PROBLEM xxx (number of problem) :[Post your solution here]

PROBLEM xxx (number of problem) :[Post your problem here]

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

\[\begin{equation} \int_0^a \frac{\ln x}{\sqrt{ax-x^2}}\,dx=\pi\ln\left(\frac{a}{4}\right) \end{equation}\]

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

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## Comments

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TopNewestHere is the solution: \(\displaystyle I = \int \limits_0^a \frac{ \text{ln } x }{\sqrt{ax - x^2}} \text{d}x \)

\(\displaystyle I = \int \limits_0^a \frac{ \text{ln } (a-x) }{\sqrt{ax - x^2}} \text{d}x \)

\(\displaystyle \Rightarrow 2I = \int \limits_0^a \frac{ \text{ln } (ax - x^2) }{\sqrt{ax - x^2}} \text{d}x \)

Put \(\displaystyle x = \frac{a}{2} (1 - \sin t ) \Rightarrow ax - x^2 = \frac{a^2}{4} \cos ^2 t \)

Therefore, the integral becomes, \(\displaystyle 2I = \int \limits_{\frac{\pi}{2}}^{\frac{-\pi}{2}} \frac{ \text{ln } (\frac{a^2}{4} \cos ^2 t ) }{\sqrt{\frac{a^2}{4} \cos ^2 t }} (\frac{a}{2} \cos t )(-\text{d}t) \)

Rearranging, we obtain,

\(\displaystyle 2I = 4\int \limits_0^{\frac{\pi}{2}} \text{ln } (\frac{a}{2} \cos t ) \text{d}t \)

The value of \(\displaystyle \int \limits_0^{\frac{\pi}{2}} \text{ln } ( \cos \theta ) \text{d}\theta \) is \(\displaystyle \frac{ -\pi \text{ ln } 2 }{2} \) (which I have calculated separately and I can post if it is required).

Thus, we obtain, \(\displaystyle I = \pi \text{ ln } \bigg( \frac{a}{2} \bigg) - 2\frac{ \pi \text{ ln } 2 }{2} = \pi \text{ ln } \bigg(\frac{a}{4} \bigg)\)

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Find a closed form expression for the integral : \(\displaystyle I = \int \sin (2015 x) \sin ^{2013} x \text{ d}x \)

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\( Solution \quad of \quad problem \quad 2\)

Split \(sin(2015x)\) as \(sin(2014x+x)\) and our integral becomes :

\(I=\displaystyle \int { ({ sin }^{ 2014 }x.cos(2014x)+sin(2014x)cos(x){ sin }^{ 2013 }(x))\quad dx } \)

Multiply and divide it by \(2014\) to get :

\(\frac { 1 }{ 2014 } \displaystyle \int { ({ sin }^{ 2014 }x.(2014cos(2014x))+sin(2014x)(2014{ sin }^{ 2013 }(x)cos(x)))\quad dx } \)

\(\Rightarrow \displaystyle I=\frac { 1 }{ 2014 } \int { { sin }^{ 2014 }(x)dsin(2014x)+sin(2014x)(d{ sin }^{ 2014 }x) } \)

\(\Rightarrow \displaystyle I=\frac { 1 }{ 2014 } \int { d({ sin }^{ 2014 }(x)sin(2014x)) } \)

\(\large \displaystyle I=\frac { { sin }^{ 2014 }(x)sin(2014x) }{ 2014 } +C\)

\(\large Problem \quad 3\)

Evaluate \(I=\displaystyle \int _{ 0 }^{ \infty }{ { (\frac { sin(x) }{ x } })^{ 2 }dx } \)

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@Sudeep Salgia. See the rules & the format post. Thank you. :)

Nicely done, +1. Now you must post a new problem, but before that. I think it would be nice if you post this as a new post, no need to replyLog in to reply

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@Anastasiya Romanova I have posted my problem sorry for holding the contest.

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Nicely done, +1! But you forget to post your problem anyway. You have a privilege for that so that this contest keeps going on. Post

PROBLEM 2in your thread solution. Thanks :)Log in to reply

Well, there must be a rule for people like Ronak! Solving a problem doesn't give you the right to withhold the contest. In such a case, someone should be allowed to interfere, right?

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within a day. The one who has a right to post a problem is the last solver before him/her. Is this fair?Log in to reply

Solution of Problem 3 :Use integration by parts by taking \(u=\sin^2x\) and \(dv=\dfrac{dx}{x^2}\), we get

\[\begin{align} \int_0^\infty\frac{\sin^2x}{x^2}\,dx&=-\left.\frac{\sin^2x}{x}\right|_0^\infty+\int_0^\infty\frac{2\sin x\cos x}{x}\,dx\\ &=0+\int_0^\infty\frac{\sin 2x}{x}\,dx\\ &=\int_0^\infty\frac{\sin t}{t}\,dt\qquad\Rightarrow\qquad t=2x\\ \end{align}\]

Now consider

\[\begin{equation} I(a)=\int_0^\infty\frac{e^{-at}\sin t}{t}\,dt\qquad a\ge0 \end{equation}\]

so that \(I(\infty)=0\) and our considered integral is \(I(0)\). Differentiating w.r.t. \(a\) and then integrating back, we get

\[\begin{align} I'(a)&=-\int_0^\infty e^{-at}\sin t\,dt\\ &=-\frac{1}{1+a^2}\qquad\Rightarrow\qquad\text{integration by parts twice}\\ I(a)&=-\int \frac{1}{1+a^2}\,da\\ &=-\arctan(a)+C \end{align}\]

For \(I(\infty)=0\), implying \(C=\dfrac{\pi}{2}\). Hence

\[\begin{equation} I(a)=\int_0^\infty\frac{e^{-at}\sin t}{t}\,dt=\frac{\pi}{2}-\arctan(a) \end{equation}\]

and

\[\begin{equation} I(0)=\int_0^\infty\frac{\sin t}{t}\,dt=\frac{\pi}{2} \end{equation}\]

Thus

Problem 4 :Prove

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\(\large Solution\quad of\quad Problem\quad 4\)

Firstly we will prove a general result

Result

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

Proof

\(I=\displaystyle \int _{ 0 }^{ 1 }{ { x }^{ a }ln(x)dx }\)

Integrating by parts \(u=ln(x),dv={x}^{a}dx\)

\(I=\displaystyle ln(x)\frac { { x }^{ a+1 } }{ a+1 } \overset { 1 }{ \underset { 0 }{ | } } -\frac { 1 }{ a+1 } \int _{ 0 }^{ 1 }{ { x }^{ a }dx }\)

\(I=\frac { -1 }{ { (1+a) }^{ 2 } }\)

Hence proved

Now we have :

\(I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { ln(cos(x)) }{ sin(x) } dx }\)

Take \(cos(x)=y\) to get our integral as :

\(\displaystyle \int _{ 0 }^{ 1 }{ \frac { ln(y)dy }{ 1-{ y }^{ 2 } } } \)

Now \(\frac { 1 }{ 1-{ y }^{ 2 } } =\displaystyle \sum _{ n=0 }^{ \infty }{ { y }^{ 2n } } \)

Our integral becomes :

\(I=\displaystyle \sum _{ n=0 }^{ \infty }{ \int _{ 0 }^{ 1 }{ { y }^{ 2n }ln(y)dy } } \)

Using our proved result we get :

\(I=-\displaystyle \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { (2n+1) }^{ 2 } } } \)

Also the summation can we written as :

\(I=-(\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 } } } -\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { (2n) }^{ 2 } } } )\)

\(I=-(\zeta (2)-\frac { 1 }{ 4 } \zeta (2))=\frac { -3 }{ 4 } \zeta (2)=\large \frac{-{\pi}^{2}}{8}\)

Where \(\zeta(x)\) is the zeta function

\(\large Problem \quad 5\)

Find closed form of \(\displaystyle I=\int _{ 0 }^{ \infty }{ \frac { dx }{ 1+{ x }^{ n } } }\)

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Keep going fella, +1! But please post your solution & proposed problem in a new thread like I did. No need to reply other threads to post a solution & a problem. Anyway, I've posted the solution of your problem & also proposed a new problem.

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Solution of Problem 7:Proposition :Proof :It can be proven by using Weierstrass substitution: \(t=\tan\left(\dfrac{x}{2}\right)\), then

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

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

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

Problem 8 :Prove

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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!

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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.

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@Anastasiya Romanova .

If pranav refuses to put up a question then who put will put it upLog in to reply

PROBLEM 9below, I've just proposed it.Log in to reply

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

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

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

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Solution to problem 10First 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 11Find

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Solution of Problem 13 :First we prove the following proposition:

Proposition :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

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

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

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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 9Prove that

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Solution -Problem - 9We 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 -\(4 \) \[ I = -4 \times \frac{-\pi^2}{8} = \frac{\pi^2}{2} \]Log in to reply

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

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@Shivang Jindal

You can try my posted question.Log in to reply

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

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\(Solution\quad of\quad problem\quad 9\)

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

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 \(\displaystyle \int _{ 0 }^{ \infty }{ \frac { { sin }^{ 3 }x }{ { x }^{ 2 } } dx } \)

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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

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

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

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

Problem 12 :Prove

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

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

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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

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\(Solution\quad of\quad Problem\quad 12\)

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 \(\displaystyle I=\int _{ 0 }^{ \infty }{ \frac { { x }^{ n } }{ { e }^{ x }-1 } dx } \)

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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?

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

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

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\(Solution\quad of\quad problem\quad 14\)

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

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

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_ 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\(\displaystyle \int_{0}^{\pi/2} \ln(1+a \sin^2 \theta) = \pi \ln \bigg(\dfrac{1+\sqrt{a+1}}{2}\bigg)\)ProofLet \(I(a) = \displaystyle \int_{0}^{\pi/2} \ln(1+a \sin^2 x) {\mathrm dx}\)

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

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

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

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

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:

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

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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.

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@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.

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Solution of Problem 15 :We have \[\begin{equation} \int_0^1 \left(\frac{1}{1-x} + \frac{1}{\ln x} \right)\,dx=\int_0^1 \frac{\ln x+1-x}{(1-x)\ln x}\,dx \end{equation}\]

Proposition :Proof :Let

\[\begin{equation} I(s)=\int_0^1 \frac{s\ln x+1-x^s}{(1-x)\ln x}\,dx \end{equation}\]

then

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

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

\[\begin{equation} I(s)=\int_0^1 \frac{s\ln x+1-x^s}{(1-x)\ln x}\,dx=\ln\Gamma(s+1) +\gamma s\qquad\qquad\square \end{equation}\]

For \(s=1\), we have

\[\begin{equation} I(1)=\int_0^1 \left(\frac{1}{1-x} + \frac{1}{\ln x} \right)\,dx=\int_0^1 \frac{\ln x+1-x}{(1-x)\ln x}\,dx=\gamma \end{equation}\]

where \(\gamma\) is The Euler–Mascheroni constant.

Problem 16 :Prove

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.

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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

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OK

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

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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

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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. \(\ddot\smile\)

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@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.

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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

Problem 6 :Show that

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

\(I = \displaystyle \int_{0}^{\pi/2} \dfrac{\cos^2 x}{\sin x + \cos x} {\mathrm dx}\) (As \(\displaystyle \int_{0}^{a} f(x) {\mathrm dx} = \int_{0}^{a} f(a-x) {\mathrm dx} \) )

Hence, \(2I = \displaystyle \int_{0}^{\pi/2} \dfrac{1}{\sin x + \cos x} {\mathrm dx}\)

Hence, \(I = \dfrac{1}{2} \displaystyle \int_{0}^{\pi/2} \dfrac{1 + \tan^2 (x/2)}{ 2 \tan(x/2) + 1 -\tan^2(x/2)} {\mathrm dx}\)

Putting \(\tan \dfrac{x}{2} = t\),

\(I = \displaystyle \int_{0}^{1} \dfrac{dt}{2 - (1-t)^2}\)

=\(\displaystyle \dfrac{1}{2 \sqrt{2}} \ln \bigg(\dfrac{\sqrt{2} + t - 1}{\sqrt{2} - t + 1} \bigg) \bigg|_{0}^{1} \)

= \(\displaystyle \dfrac{\ln(3 + 2 \sqrt{2})}{2 \sqrt{2}}\)

Problem 7Prove that \(\displaystyle \int_{0}^{\pi} \dfrac{1}{(\sqrt{10}+ \cos x)^3} = \dfrac{7 \pi}{162}\)

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\[ I(a) = \int_{0}^{\pi} \frac{1}{(a-\cos(x))}, a>1 \] Use Weierstrass substitution, its not hard to show that , this integral \( =\frac{\pi}{\sqrt{a^2-1}} \) Now , we differentiate, both sides with respect to \( a \). \[ I'(a) = \int_{0}^{\pi} -\frac{1}{(a-\cos(x))^2} = \frac{d}{da} \frac{\pi}{\sqrt{a^2-1}} \] Again differentiating both sides, \[ I''(a) = \int_{0}^{\pi} \frac{2}{(a-\cos(x))^3} = \frac{d}{d^2a} \frac{\pi}{\sqrt{a^2-1}} \] Taking, 2 to RHS gives, \[ I''(a) = \int_{0}^{\pi} \frac{1}{(a-\cos(x))^3} = \frac{d}{d^2a} \frac{\pi}{2\sqrt{a^2-1}} \] Note that, we need \( I''(\sqrt{10}) \) . Plugging, \( a=\sqrt{10} \) , and evaluating the expression in RHS gives, \( \frac{7\pi}{162} \)

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PROBLEM 9) that I've just posted below.Log in to reply

It looks like there's someone here who is interesting in knowing the closed-form of your previous problem 7 and he posted it on Math S.E.. I am wondering, does that really have a closed-form? I tried to solve it for hours but no success & you really gave me lots of trouble cracking that integral.

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this problem and got struck at it.

Well ,actually I had to change the problem as I didn't know its solution. Actually, few days back, at the same website I tried answeringLog in to reply

his question but only gets 3 upvotes. It's not worth enough. Luckily, he accepts my answer. I also answer one of his question on Integrals and Series, but of course not for free. LOL

OMG! Why did you propose then here on this contest? Anyway, just ignore that user. He always ask two or three hard integrals in a single question on Math S.E. I can answer one ofLog in to reply

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I hate him!), and lots of people on internet since I was 7. You might be interested in seeing this. You're very good at all math subjects & physics too. That's truly impressive!!Anyway, I've just remembered that the same user (Samurai, the one who posts your previous problem 7 on Math S.E.) also posted one of my problem on Brilliant at Math S.E. too. What a cheater!

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Because I wanna beat my big brother.Seriously, he always says to me that calculus is the highest level mathematics, so that's why I'm really interested in calculus specially integrals & series and almost every weekend we have a competition of solving integrals & series problems although he always beats me easily. Of course I love math, it means I also love the other fields of math but I'm just not too interested in them. Besides, my first crush on math because of calculus. I don't wanna tell the story because it's too personal. Special for you, I have answered two problems here that don't relate to Calculus to show my capability other than Calculus (It's also because I'm bored). Here they are Algebra problem: Advanced System Of Equations and Geometry problem: All Three Concurrency. Answering Algebra problem makes me level up, so I must down my level again. I just want the other know me because I'm sorta good at Calculus, not the other subjects. Knowing Calculus in such young age also gives me lots of advantages on the other forums where I join in. Okay, I think it's enough. I hope I am not making any immodest statements or showing narcissism on my part. \(\ddot\smile\)Log in to reply

farfrom the hardest discipline of mathematics. Learn abstract algebra, or complex analysis, or dynamical systems. Then you'll be able to appreciate fields of mathematics where there is still a lot of work left to be done, even today, and you'll probably end up knowing more math than your big brother. ;)Log in to reply

Ans:ln4a/2a^2

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