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).

## 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, \(\displaystyle I = \pi \text{ ln } \bigg( \frac{a}{2} \bigg) - 2\frac{ \pi \text{ ln } 2 }{2} = \pi \text{ ln } \bigg(\frac{a}{4} \bigg)\) – Sudeep Salgia · 1 year, 11 months ago

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– Sudeep Salgia · 1 year, 11 months ago

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

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 \(I=\displaystyle \int _{ 0 }^{ \infty }{ { (\frac { sin(x) }{ x } })^{ 2 }dx } \) – Ronak Agarwal · 1 year, 11 months ago

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– Sudeep Salgia · 1 year, 11 months ago

Perfect. Nice work Ronak. Expecting a problem soon.Log in to reply

– Anastasiya Romanova · 1 year, 11 months ago

Where is your proposed problem? We are waiting it. Please don't let us wait too long like this. You should post your solution and your proposed problem at once.Log in to reply

@Anastasiya Romanova I have posted my problem sorry for holding the contest. – Ronak Agarwal · 1 year, 11 months ago

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– Anastasiya Romanova · 1 year, 11 months ago

It's okay. I have posted the solution of your problem and also proposed a new problem. :)Log in to reply

@Sudeep Salgia. See the rules & the format post. Thank you. :) – Anastasiya Romanova · 1 year, 11 months ago

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

PROBLEM 2in your thread solution. Thanks :) – Anastasiya Romanova · 1 year, 11 months agoLog in to reply

– Danny Kills · 1 year, 11 months ago

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?Log in to reply

within a day. The one who has a right to post a problem is the last solver before him/her. Is this fair? – Anastasiya Romanova · 1 year, 11 months agoLog in to reply

– Ronak Agarwal · 1 year, 11 months ago

Sorry If that troubled all because actually I have to go to coaching centre, and my internet connection got down and I went to my coaching centre, I have returned just now and has posted the question.Log in to reply

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

– Pratik Shastri · 1 year, 11 months agoLog 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

– Anastasiya Romanova · 1 year, 11 months agoLog in to reply

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 \(\displaystyle I=\int _{ 0 }^{ \infty }{ \frac { dx }{ 1+{ x }^{ n } } }\) – Ronak Agarwal · 1 year, 11 months ago

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– Anastasiya Romanova · 1 year, 11 months ago

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.Log in to reply

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

– Anastasiya Romanova · 1 year, 11 months agoLog in to reply

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! – Pranav Arora · 1 year, 11 months ago

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– Pratik Shastri · 1 year, 11 months ago

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

– Anastasiya Romanova · 1 year, 11 months ago

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

– Pratik Shastri · 1 year, 11 months ago

Right. I confused it with Tonelli's theorem.Log in to reply

@Anastasiya Romanova . – Ronak Agarwal · 1 year, 11 months ago

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. – Anastasiya Romanova · 1 year, 11 months agoLog in to reply

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

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

\( I = \dfrac{\pi}{ \sqrt{p^2 - 1}}\) – Megh Choksi · 1 year, 9 months ago

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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

– Anastasiya Romanova · 1 year, 11 months agoLog in to reply

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

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

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 \(\displaystyle \int _{ 0 }^{ \infty }{ \frac { { sin }^{ 3 }x }{ { x }^{ 2 } } dx } \) – Ronak Agarwal · 1 year, 11 months ago

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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.

– Anastasiya Romanova · 1 year, 11 months agoLog in to reply

@Anastasiya Romanova – Ronak Agarwal · 1 year, 11 months ago

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

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

– Anastasiya Romanova · 1 year, 11 months agoLog in to reply

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

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 \(\displaystyle I=\int _{ 0 }^{ 1 }{ \frac { 1 }{ ln(x) } +\frac { 1 }{ 1-x } dx } \) – Ronak Agarwal · 1 year, 11 months ago

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@jatin yadav to post your solution, then the right to propose a new problem is yours. – Anastasiya Romanova · 1 year, 11 months ago

According to timeline, you're faster thanLog in to reply

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

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 \(\displaystyle I=\int _{ 0 }^{ \infty }{ \frac { { x }^{ n } }{ { e }^{ x }-1 } dx } \) – Ronak Agarwal · 1 year, 11 months ago

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– Anastasiya Romanova · 1 year, 11 months ago

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

\(\Rightarrow cos(arctan(2x))=\frac{1}{\sqrt{1+4{x}^{2}}}\) – Ronak Agarwal · 1 year, 11 months ago

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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

– Anastasiya Romanova · 1 year, 11 months agoLog in to reply

– Pratik Shastri · 1 year, 11 months ago

The last substitution should be \(u=\alpha \tan \theta\) :)Log in to reply

– Megh Choksi · 1 year, 9 months ago

This is a simple kind of NCERT problem. It can be easily done by substituting \( t = tan^{-1}\dfrac{x}{\alpha}\) and then applying by partsLog in to reply

– Anastasiya Romanova · 1 year, 11 months ago

Fixed it. Sorry, I'm too hasty. Thanks... :)Log in to reply

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

– Anastasiya Romanova · 1 year, 11 months agoLog in to reply

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 -\(4 \) \[ I = -4 \times \frac{-\pi^2}{8} = \frac{\pi^2}{2} \] – Shivang Jindal · 1 year, 11 months agoLog in to reply

@Shivang Jindal – Ronak Agarwal · 1 year, 11 months ago

I had already posted the solution to the problem and a new problem also you should not reply instead post your solution as a seperate comment as I didLog in to reply

– Shivang Jindal · 1 year, 11 months ago

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

@Shivang Jindal – Ronak Agarwal · 1 year, 11 months ago

You can try my posted question.Log in to reply

– Pratik Shastri · 1 year, 11 months ago

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

_ 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:

\(I = 2 \pi \ln \bigg(\dfrac{1+\sqrt{4/3}}{1+\sqrt{2/3}} \bigg) = \boxed{\pi \ln \bigg(\dfrac{7 + 4\sqrt{3}}{5+2\sqrt{6}}\bigg)}\) – Jatin Yadav · 1 year, 11 months ago

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– Anastasiya Romanova · 1 year, 11 months ago

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

@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. – Sudeep Salgia · 1 year, 11 months ago

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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

Lemma) – Ishan Singh · 1 year, 11 months agoLog in to reply

– Anastasiya Romanova · 1 year, 11 months ago

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

@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. – Sudeep Salgia · 1 year, 11 months ago

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– Anastasiya Romanova · 1 year, 11 months ago

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

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

– Anastasiya Romanova · 1 year, 11 months agoLog in to reply

\(I = \displaystyle \int_{0}^{\pi/2} \dfrac{\cos^2 x}{\sin x + \cos x} {\mathrm dx}\) (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 \(\displaystyle \int_{0}^{\pi} \dfrac{1}{(\sqrt{10}+ \cos x)^3} = \dfrac{7 \pi}{162}\) – Jatin Yadav · 1 year, 11 months ago

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– Shivang Jindal · 1 year, 11 months ago

\[ I(a) = \int_{0}^{\pi} \frac{1}{(a-\cos(x))}, a>1 \] 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, \( \frac{7\pi}{162} \)Log in to reply

– Shivang Jindal · 1 year, 11 months ago

-_- . Solution was already there :||||. Please, make this thread little structured. I didn't saw any solution after problem problem -7th, so wrote solution here. After posting here , i realized that, there is exactly same solution posted before.Log in to reply

PROBLEM 9) that I've just posted below. – Anastasiya Romanova · 1 year, 11 months agoLog in to reply

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. – Anastasiya Romanova · 1 year, 11 months ago

It looks like there's someone here who is interesting in knowing the closed-form of your previous problem 7 and he posted it onLog in to reply

this problem and got struck at it. – Jatin Yadav · 1 year, 11 months ago

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 – Anastasiya Romanova · 1 year, 11 months ago

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

– Jatin Yadav · 1 year, 11 months ago

My bad! Didn't read the rules carefully. By the way , seeing a 14 year old student having so much knowledge of calculus is really impressive! I knew only basic integration properly when I was 14.Log in to reply

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! – Anastasiya Romanova · 1 year, 11 months ago

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– Christopher Boo · 1 year, 11 months ago

If I'm not mistaken, you only solve / post problems in the calculus section. Why are you specifically interested in Calculus? How about other fields of Math?Log in to reply

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. \(\ddot\smile\) – Anastasiya Romanova · 1 year, 11 months agoLog 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. ;) – Michael Lee · 1 year, 11 months agoLog in to reply

Ans:ln4a/2a^2 – Shreyas Dighe · 1 year, 11 months ago

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