# Brilliant Integration Contest - Season 1 (Part 3)

This is Brilliant Integration Contest - Season 1 (Part 3) as a continuation of the previous contest Part 1 and Part 2. There is a major change in the rules of contest, so please read all of them carefully before take part in this contest.

I am interested in holding an Integration Contest here on Brilliant.org like any other online forums such as AoPS or Integrals and Series. The aims of the Integration Contest are to improve skills in the computation of integrals, to learn from each other as much as possible, and of course to have fun. Anyone here may participate in this contest.

The rules are as follows

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

Format your post is as follows:

SOLUTION OF PROBLEM xxx (number of problem) :

PROBLEM xxx (number of problem) :

Remember, put them separately.

## 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.

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

Okay, let the contest part 3 begin!

Note by Anastasiya Romanova
5 years ago

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PROBLEM 45

Evaluate:

$\displaystyle \huge \sum_{k=1}^\infty \frac{\left(2\mathcal{H}_{2k} - \mathcal{H}_k \right)^2}{k^2}$

- 2 years, 7 months ago

PROBLEM 44:

$\int_0^{\infty} \frac{\sinh(ax)\sin(bx)}{\left(\cosh(ax)+\cos(bx)\right)^2}\,dx=\frac{b}{a^2+b^2}\,\,\,(a,b>0)$

- 4 years, 11 months ago

I have not found a solution, so please post your solution. I have succeeded in writing the integrand as an absolutely convergent sum of some rational function summed from minus to plus infinity. But it seems that switching summation and integration is not allowed here, because I get the result zero.

- 4 years, 11 months ago

Long time I don't visit here and I thought this thread had already arrived at problem 60. Anyway, you may try this to evaluate the integral:

First, you evaluate $I(a)=-\int_0^\infty\frac{\sin bx}{x\,(\cosh ax + \cos bx)} dx$

and the original integral is $I'(a)$. To evaluate $I(a)$, you may use the following identity:

$2\;\sum_{n=1}^\infty\;(-1)^{n-1}e^{-anx}\sin(nbx)=\frac{\sin bx}{\cosh ax+\cos bx}\qquad,\qquad\text{for \\;a,b>0\}$

where it can be proven by noticing $\sum_{n=1}^\infty\;(-1)^{n-1}e^{-anx}\sin(nbx)= \Im\left(\sum_{n=1}^\infty(-1)^{n-1} e^{(ib-a)nx}\right)$

the rest can be done by using an infinite geometric progression.

P.S. I haven't try it yet, hehe... But I'm sure this works. I leave the rest for you. $\quad\ddot\smile$

- 4 years, 11 months ago

Please @Tunk-Fey Ariawan post a new problem enjoying to learn new things

According to rules , @Ruben Doornenbal you can post a new problem

- 4 years, 10 months ago

@Tunk-Fey Ariawan Ok that was easy. If you want, you can post the next problem.

- 4 years, 11 months ago

Problem 43

Find $\displaystyle \int_0^{\pi/2} \sqrt{1 + \sin^2 x} dx$

Elliptic integrals may be useful.

- 4 years, 12 months ago

I hope you all don't mind me participating a little. I won't anymore if you don't want me to. I realize this is for you math-gifted younger folks rather than the ancients like me :):). Someone provided a link to this contest and I noticed some fun integrals.

Anyway, with regards to problem 43 it is, by definition, an Elliptic Integral of the form $K(k)=\int_{0}^{\frac{\pi}{2}}\sqrt{1-k^{2}\sin^{2}(x)}dx$, with $k=\sqrt{-1}$

The obvious sub $t=\sin(x)$ gives:

$\int_{0}^{1}\frac{\sqrt{1+t^{2}}}{\sqrt{1-t^{2}}}dt$

Multiply top and bottom by $\sqrt{1+t^{2}}$:

$\int_{0}^{1}\frac{1+t^{2}}{\sqrt{1-t^{4}}}dt=\int_{0}^{1}\frac{1}{\sqrt{1-t^{4}}}+\int_{0}^{1}\frac{t^{2}}{\sqrt{1-t^{4}}}dt$

Now, it is ready to be hammered into a Beta function/Gamma function.

I am sure you all can take it from here. One should arrive at something like:

$\frac{\sqrt{\pi}\Gamma(1/4)}{4\Gamma(3/4)}+\frac{\Gamma^{2}(3/4)}{\sqrt{2\pi}}$

or some other equivalent form depending on how you would like to write it.

- 4 years, 11 months ago

Everyone is welcome to solve integrals here! Thank you for your contribution.

- 4 years, 11 months ago

Thanks. I was under the impression this site was for high school and under age only.

- 4 years, 11 months ago

Using elliptic integral of second kind, the answer is: $\boxed{E\left(\dfrac{\pi}{2},i\right)}$. Or you can write $\sqrt{1+\sin^2 x}=\sqrt{2-\cos^2 x}$ to get $\sqrt{2}\,E\left(\frac{\pi}{2},\frac{1}{\sqrt{2}}\right)$

- 4 years, 11 months ago

You are not allowed to use elliptic integrals in the final answer. The challenge is to express this in terms of the gamma function.

- 4 years, 11 months ago

PROBLEM 42 :

Show that $\int_{0}^{1}\,\left\{\frac{1}{x}\right\}\frac{\ln x}{\sqrt{x}}\,\mathrm{d}x = \left(4-\gamma-\ln8\pi-\frac{\pi}{2}\right)\zeta\left(\frac{1}{2}\right)-4$

where $\left\{\frac{1}{x}\right\}$ denotes the fractional part of $\frac{1}{x}$.

- 5 years ago

Let

$\displaystyle J(s) = \int_0^1 \left\{ \frac 1 x \right\} x^{s-1} dx.$

We want to calculate

$\displaystyle J'(1/2) = \int_0^1 \left\{ \frac 1 x \right\} \frac{ \ln x}{\sqrt x} dx.$

We have

\displaystyle \begin{aligned} J(s) &= \int_0^1 \left\{ \frac 1 x \right\} x^{s-1} dx \\&= \int_0^\infty \left\{ x \right\} x^{-s-1} dx \\&= \sum\limits_{n \geq 0} \int_n^{n+1} (x - n) x^{-s-1} dx \\&= \sum\limits_{n \geq 0} \left\{\frac n s \left[(n+1)^{-s} - n^{-s} \right] + \frac{1}{1-s}\left[(n+1)^{1-s} - n^{1-s}\right] \right\} \\&= \frac 1 s\sum\limits_{n \geq 0} \left[(n+1)^{1-s} - (n+1)^{-s} - n^{1-s} \right] - \frac{1}{1-s} \\&= \frac 1 s\sum\limits_{n \geq 1} -n^{-s}- \frac{1}{1-s} \\&= -\frac{\zeta(s)}{s}- \frac{1}{1-s}. \end{aligned}

Here we used some telescoping series.

Therefore

$\displaystyle J'(1/2) = 4 \zeta(1/2) - 2 \zeta'(1/2) -4.$

Using the value

$\zeta'(1/2) = \frac 1 4 \zeta(1/2) \left(\pi + 2 \gamma + 2 \log(8 \pi) \right)$

gives the result.

- 4 years, 12 months ago

PROBLEM 41

Evaluate, $\int_{0}^{\pi/12} \ln(\tan(x)) dx$

- 5 years ago

Why these types of ques can't be done with elementary techniques?

- 4 years, 11 months ago

There exist a elementary solution to this problem. See, (First solution) http://math.stackexchange.com/questions/983044/integral-int-0-pi-12-ln-tan-x-dx

- 4 years, 11 months ago

Using Fourier series representations of $\ln \sin x$ and $\ln \cos x$, $\ln \sin x=-\ln2-\sum_{k=1}^\infty \frac{\cos2kx}{k}$ and $\ln \cos x=-\ln2+\sum_{k=1}^\infty (-1)^{k+1}\frac{\cos2kx}{k}$ we then have $\ln \tan x=-2\sum_{k=0}^\infty \frac{\cos2(2k+1)x}{2k+1}$ Therefore \begin{aligned} \int_0^{\pi/12}\ln \tan x\,dx&=-2\sum_{k=0}^\infty \int_0^{\pi/12}\frac{\cos2(2k+1)x}{2k+1}\,dx\\ &=-\sum_{k=0}^\infty \frac{\sin\left(\frac{2k+1}{6}\right)\pi}{(2k+1)^2}\\ \end{aligned} The term $\sin\left(\frac{2k+1}{6}\right)\pi$ has a periodicity every six steps, namely $\frac{1}{2},1,\frac{1}{2},-\frac{1}{2},-1,-\frac{1}{2}$, then \begin{aligned} \int_0^{\pi/12}\ln \tan x\,dx &=-\left[\frac{\left(\frac{1}{2}\right)}{1^2}+\frac{\left(1\right)}{3^2}+\frac{\left(\frac{1}{2}\right)}{5^2}+\frac{\left(-\frac{1}{2}\right)}{7^2}+\frac{\left(-1\right)}{9^2}+\frac{\left(-\frac{1}{2}\right)}{11^2}+\cdots\right]\\ &=-\left[\frac{\left(\frac{1}{2}\right)}{1^2}+\frac{\left(\frac{3}{2}-\frac{1}{2}\right)}{3^2}+\frac{\left(\frac{1}{2}\right)}{5^2}-\frac{\left(\frac{1}{2}\right)}{7^2}-\frac{\left(\frac{3}{2}-\frac{1}{2}\right)}{9^2}-\frac{\left(\frac{1}{2}\right)}{11^2}+\cdots\right]\\ &=-\left[\frac{1}{2}\sum_{k=0}^\infty \frac{(-1)^{k}}{(2k+1)^2}+\frac{3}{2}\sum_{k=0}^\infty \frac{(-1)^{k}}{(6k+3)^2}\right]\\ &=-\frac{2}{3}\sum_{k=0}^\infty \frac{(-1)^{k}}{(2k+1)^2}\\ &=-\frac{2}{3}\text{G} \end{aligned} where $\text{G}$ is Catalan's constant.

- 5 years ago

The result is $\displaystyle -\frac 2 3 G$.

- 5 years ago

- 5 years ago

PROBLEM 40

For $a\ge0$, show that $\int_{-\infty}^{\infty}\frac{\cos\left(ax^2\right)-\sin\left(ax^2\right)}{1+x^4}\mathrm dx=\frac{\pi e^{-a}}{\sqrt{2}}$

- 5 years ago

Let $f(a) = \int_{-\infty}^{\infty} \frac{\cos(ax^2)-\sin(ax^2)}{1+x^4}$. It is easy to calculate, $f(0)$ Then, $f'(a) = \int_{-\infty}^{\infty} \frac{x^2(-\sin(ax^2)-\cos(ax^2))}{1+x^4}$ Again, differentiating gives, $-f''(a) = \int_{-\infty}^{\infty} \frac{x^4(\cos(ax^2)-\sin(ax^2))}{1+x^4}$ Thus, $f(a)-f''(a) = \int_{-\infty}^{\infty} \cos(ax^2) -\sin(ax^2)= 2\int_{0}^{\infty} \cos(ax^2)-\sin(ax^2) = 0$ (Fresnel integral) .

Solving this differential easily give the desired result. We get , $f(a) = e^{-a}{n}$ . $f(0) = \frac{\pi}{\sqrt{2}}$ So, $f(a) = \frac{\pi e^{-a}}{\sqrt{2}}$

- 5 years ago

brilliantly done +1

- 4 years, 11 months ago

PROBLEM 39

Prove

$\int_0^\infty\frac{\arctan(x)\arctan(2x)}{x^2}\,dx=\frac{\pi}{2}\ln\left(\frac{27}{4}\right)$

- 5 years ago

Actually it took me so much time to figure out the equivalent of  here, anyways, Here we go

$I(a,b)=\int_0^{\infty}\frac{\arctan(x)\arctan(2x)}{x^2}\,\mathrm dx=?$ Consider $I(a,b)=\int_0^{\infty}\frac{\arctan(ax)\arctan(bx)}{x^2}\,\mathrm dx$

$\frac{\partial }{\partial a}I(a,b)=\int_0^{\infty}\frac{\arctan(bx)}{x(1+a^2x^2)}dx$

$\frac{\partial }{\partial b}I(a,b)=\int_0^{\infty}\frac{\arctan(ax)}{x(1+b^2x^2)}dx$

$\frac{\partial ^2}{\partial a\partial b}I(a,b)=\frac{\partial ^2}{\partial b\partial a}I(a,b)=\int_0^{\infty}\frac{1}{(1+a^2x^2)(1+b^2x^2)}dx=\frac{\pi}{2(a+b)}$

$\frac{\partial }{\partial a}I(a,b)=\frac\pi 2\Big[\ln(a+b)-\log(a)\Big]\\ \frac{\partial }{\partial b}I(a,b)=\frac\pi 2\Big[ \ln(a+b)-\ln(b)\Big]$

$I(a,b)=\frac{\pi}{2}\Big[a \log (a+b)+b \ln (a+b)-b \ln (b)-a\ln(a)\Big]$

$I(a,b)=\frac{\pi}{2}\ln \left[\frac{(a+b)^{a+b}}{a^ab^b}\right]$

$I(2,1)=I(1,2)=\frac{\pi}{2}\ln \left(\frac{3^{3}}{2^2}\right)=\frac{\pi}{2}\ln \left(\frac{27}{4}\right)$

$\displaystyle\large\int_0^{\infty}\frac{\arctan(x)\arctan(2x)}{x^2}\,\mathrm dx=\frac{\pi}{2}\ln \left(\frac{27}{4}\right)$

I did'nt understood the fifth step , can you please elaborate more ,I don't know how to differentiate 2 variables simultaneously.

- 4 years, 11 months ago

Nice solution, +1. Welcome to this contest Integrator. I hope you have fun around here. $\LaTeX$ code in Math S.E. also works here except for $...$ and $$...$$ change to a backslash a left parenthesis math expression a backslash a right parenthesis and $\backslash[\ldots\backslash]$ .

- 5 years ago

Nice solution ^^

- 5 years ago

Let's first prove this integral for any value of a> 0. I will make use of this integral(If you want me to prove I'll do it): $\int_{0}^{\infty}\frac{ln(a^{2}x^{2} + 1)}{b^{2}x^{2} + 1}dx = \frac{\pi}{b}ln(\frac{a}{b}+1)$

Using parts we get: $I(a) = \int_{0}^{\infty} \frac{arctan(ax)}{x(1+x^{2})}dx + a\int_{0}^{\infty} \frac{arctan(x)}{x(1+a^{2}x^{2})}dx$ $I(a) = -a\int_{0}^{\infty} \frac{ln(x) - ln(x^{2} + 1)/2}{1+a^{2}x^{2}}dx - \frac{\pi aln(a))}{2}-a\int_{0}^{\infty} \frac{ln(x) - ln(a^{2}x^{2} + 1)/2}{1+x^{2}}dx$ $I(a) = -a\int_{0}^{\infty} \frac{ln(x)}{1+a^{2}x^{2}}dx + \frac{a}{2}\int_{0}^{\infty}\frac{ln(x^{2}+1)}{1+a^{2}x^{2}}dx -\frac{\pi aln(a))}{2} + \frac{a}{2}\int_{0}^{\infty}\frac{ln(a^{2}x^{2}+1)}{1+x^{2}}dx$

And using the above result we evaluate the following integrals: $I(a) = \frac{\pi}{2}ln(a) + \frac{\pi}{2}[ln(1+a)-aln(a)+aln(a+1)]$

And simplifying we get: $I(a) = \frac{\pi}{2}ln(\frac{(a+1)^{a+1}}{a^{a}})$

And plugging a = 2, we get: $I(2) = \frac{\pi}{2}ln(\frac{27}{4})$ I can't think of a challenging integral at the moment .So Anastasiya can post a new problem.

- 5 years ago

Nice solution, +1. Please next time you post a new problem. You can post any integral problems you want as long as it doesn't break the rules. for now, I'll post a new one for you.

- 5 years ago

PROBLEM 38 Compute , $\int_{0}^{\pi} \frac{2+2\cos(x)-\cos((2^{8}-1)x)-2\cos(2^{8}x)-\cos((2^8+1)x)}{1-\cos(2x)}$

- 5 years ago

I did, it by checking few values of $n$. I guessed the relation , $I = n\pi$, and then i proved it easily by proving that sequence is an AP.

- 5 years ago

SOLUTION OF PROBLEM 38 :

I'm affraid that no one will answer this question so I decide to answer it. So here is an answer.

Rewrite the integrand as

$\frac{2-2\cos(256 x)+\cos(x)-\cos(255x)+\cos(x)-\cos(257x)}{1-\cos(2x)}$

Using my post and my answers on Math S.E. (see 1, 2, and 3), it is clearly the term $\cos(x)-\cos(255x)+\cos(x)-\cos(257x)$ is a red herring since $2n\neq1,255,257$ for $n$ integer. The integral of that term cancels each other. Hence, our integrand reduces to

$2\int_0^{\pi}\frac{1-\cos(256x)}{1-\cos(2x)}\,dx=2\int_0^{\pi}\frac{\sin^2(128x)}{\sin^2(x)}dx$

$\int_0^{\pi}\frac{\sin^2(nx)}{\sin^2(x)}\,dx=n\pi$

Thus

$\int_0^{\pi}\frac{2-2\cos(256 x)+\cos(x)-\cos(255x)+\cos(x)-\cos(257x)}{1-\cos(2x)}\,dx=2(128)\pi=256\pi$

and the result agrees numerically.

- 5 years ago

Hint: Replace $2^{8}$ by $n$ and then calculate it for $0,1,2,3..$

- 5 years ago

If You replace 2^8 with 2^n as a general you will get the answer as $2^{n}\pi$ but I m not sure how to prove it I tried induction but it didn't work. So the answer will be $2^{8}\pi$

- 5 years ago

$2^{8}$ is just to confuse :D. This results is true for any n., try to prove it now, you are close..

- 5 years ago

This Is the problem I cant prove it, but here's the simplified form of the integral: $I=\displaystyle \int_{0}^{\pi} \frac{sin(2^{7}x)^{2}}{sin(\frac{x}{2})^{2}}$

- 5 years ago

$Problem\quad 37$

Find $\displaystyle \int _{ 0 }^{ 1 }{ ln(x)ln(1-x)dx }$

- 5 years ago

$\int_{0}^{1} \ln(x) \ln(1-x)$ $\sum_{n=1}^{\infty} -1\frac{1}{n} \int_{0}^{1} x^n \ln(x) = \sum_{n=1}^{\infty} \frac{1}{n(n+1)^2} = \sum_{n=1}^{\infty} \frac{1}{n(n+1)} - \frac{1}{(n+1)^2} = 2 - \zeta(2)$

- 5 years ago

I think I have found a elementary way,

$x = sin^2t$

- 4 years, 11 months ago

OK, for the sake of having fun I have two easy problems but you may only answer one of them. Feel free. Of course the first one who answers correctly one of these problems (or both of them) has a right to post the next problem.

PROBLEM 36A :

If $a$ is an even positive integer and $b$ is an arbitrarily constant, then show that

$\int_{-1}^1\frac{x^a}{1+e^{bx}}\,dx=\frac{1}{a+1}$

PROBLEM 36B :

For $n>1$, prove that $\int_{0}^{\Large\frac{\pi}{2}}\frac{\tan x\sec x}{(\tan x+\sec x)^n}\,dx=\frac{1}{n^2-1}$

Good luck!! $\ddot\smile$

- 5 years ago

Solution of problem 36A :

$\displaystyle I=\int _{ -1 }^{ 1 }{ \frac { { x }^{ a } }{ 1+{ e }^{ bx } } dx }$

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

We get $\displaystyle I = \int _{ -1 }^{ 1 }{ \frac { { x }^{ a } }{ 1+{ e }^{ -bx } } dx }=\int _{ -1 }^{ 1 }{ \frac { { x }^{ a }{ e }^{ bx } }{ { e }^{ bx }+1 } dx }$

Adding these forms we get :

$\displaystyle I=\frac { 1 }{ 2 } \int _{ -1 }^{ 1 }{ { x }^{ a }dx }$

Which on evaluating gives us :

$I=\frac { 1 }{ a+1 }$

Solution to problem 36B

$\displaystyle I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sec(x)tan(x)dx }{ { (sec(x)+tan(x)) }^{ n } } }$

Put $sec(x)+tan(x)=t , dt=sec(x)(sec(x)+tan(x))dx , tan(x)=\frac{{t}^{2}-1}{2t}$

$\displaystyle I=\frac { 1 }{ 2 } \int _{ 1 }^{ \infty }{ { t }^{ 1-n }-{ t }^{ -(1+n) }dn }$

Which on evaluating gives us :

$I=\frac { 1 }{ 2 } (\frac { -1 }{ 1-n } +\frac { -1 }{ n+1 } )$

$I=\large \frac{1}{{n}^{2}-1}$

- 5 years ago

PROBLEM 35

$\displaystyle \int_0^{\infty } \frac{\arctan \left(\frac{x^2}{x^2+1}\right)}{x^4+1} \, dx$

- 5 years ago

Cleo said this to me, "While asleep, I had an unusual experience. There was a red screen formed by flowing blood, as it were. I was observing it. Suddenly a hand began to write on the screen. I became all attention. That hand wrote a number of elliptic integrals. They stuck to my mind. As soon as I woke up, I committed them to writing."

She also said that the integral evaluates to

$\color{#D61F06}{\frac{\pi}{4 \sqrt{2}}\left[\ln\sqrt{5} +\arctan2+\arctan\sqrt{2 \left(1+\sqrt{2}\right)}-\arctan\sqrt{2 \left(7+5 \sqrt{2}\right)}-\operatorname{arctanh}\left(\frac{2}{7}\sqrt{1+5 \sqrt{2}}\right)\right]}$

In my opinion, I think this integral is way too hard for kids. Can you elaborate your method on how you evaluate this integral? Preferably with a high school method (this word is really ridiculous).

- 5 years ago

Let me sketch a possible solution. I can give more details if necessary, but I think this should suffice. By substituting $x = 1/y$, we reduce it to a sum of a trivial integral and

$\displaystyle \int_0^\infty \frac{x^2 \arctan({1+x^2})}{1+x^4} = \Im \int_0^\infty \frac{x^2 \log({1+i(1+x^2)})}{1+x^4}$

Let

$\displaystyle J(a) := \Im \int_0^\infty \frac{\log({1+a(1+x^2)})}{1+x^4}.$

We can calculate $J'(a)$ by partial fractions and using the well-known formula

$\displaystyle \int_0^\infty \frac{x^{a-1}}{1+x^b} = (\pi/b) \csc(\pi a /b).$

The only complicated thing is integrating back to find $J(i)$. But no complex analysis is needed for the solution. All integrals involved can be expressed in terms of polylogs and polygamma's (but they can also be done just by integrating by parts; your answer shows that the result is elementary). Note that we could also write, for a suitable definition of $\log$,

$\displaystyle \log(1+i(1+x^2))= \log(1+i) + \log \left(1 + \frac{i}{i+1} x^2 \right) = \log(1+i) + \log \left(1 + \frac{i+1}{2} x^2 \right),$

and then introduce a parameter in the latter term.

Cleo may create the next problem if she wants.

- 5 years ago

Please don't get me wrong. I really didn't mean to cause any offence. I was just stating my opinion. I hope you don't take offence at what I said in my comment. $\ddot\smile$

Are you sure about letting me to post the next problem? If so, may I ask you (Anna is off already) to not answer about 5-6 next problems in this contest? I ask this because I only see in the past 10-15 problems this contest was dominated by you & Anastasiya. I think the other should take part in this contest too. Well, you may not agree to my suggestion. $\ddot\smile$

- 5 years ago

Don't worry, I assure you no offence was taken. I suppose I will refrain from answering the next problems unless they are not answered by anyone else. I think it is fair that Cleo can design the next problem, because she found the solution.

- 5 years ago

Problem 35 is really tedious and cumbersome by using Feynman's method. Even if I use a residue method, the answer doesn't immediately yield Mr. Tunk-Fey's answer. Anyway, he is indeed a smart guy but he is not Cleo.

- 5 years ago

PROBLEM 34 :

Show that $\large\int_{-\infty}^{\infty}\frac{\cos \left(s \arctan \left(ax\right)\right)}{(1+x^2)\left(1+a^2x^2\right)^{s/2}}\,dx=\frac{\pi}{(1+a)^s}$ where $a,s \in \mathbb{R}^{+}$.

- 5 years ago

Here is a method for problem 34 using non-contour methods. I feel like Metheusalah compared to all of you brilliant youngsters :):). I realize this site is for you brilliant young mathematicians. I post this in the event anyone finds it useful.

Abel's Theorem:

It is based on if $F(1+\alpha)$ can be written as a series of powers involving $e^{-a}$ in the form:

$P_{0}+P_{1}e^{-\alpha}+P_{2}e^{-2\alpha}+\cdot\cdot\cdot$

Then, by letting $\alpha=iax$

One has $P_{0}+P_{1}\cos(ax)+P_{2}\cos(2ax)+\cdot\cdot\cdot =1/2[F(1+iax)-F(1-iax)]$

$1/2\int_{0}^{\infty}\frac{F(1+iax)-F(1-iax)}{x^{2}+1}dx=\int_{0}^{\infty}\left(\frac{P_{0}}{x^{2}+1}+\frac{P_{1}\cos(ax)}{x^{2}+1}+\frac{P_{2}\cos(2ax)}{x^{2}+1}+\cdot\cdot\cdot \right)dx$

Notice the famous integral $\int_{0}^{\infty}\frac{\cos(ax)}{x^{2}+1}dx=\frac{\pi}{2}e^{-a}$

$=\frac{\pi}{2}[P_{0}+P_{1}e^{-a}+P_{2}e^{-2a}+\cdot\cdot\cdot ]$

$=\frac{\pi}{2}F(1+a)$

Now, let $F(z)=\frac{1}{z^{s}}$

Then, $F(1+iax)-F(1-iax)=\frac{2\cos\left(s\cdot \tan^{-1}(ax)\right)}{(x^{2}+1)^{s/2}}$

Thus:

$\int_{0}^{\infty}\frac{\cos\left(s\cdot \tan^{-1}(ax)\right)}{(x^{2}+1)^{s/2}}=\frac{\pi}{(1+a)^{s}}$

- 4 years, 11 months ago

SOLUTION 34

This integral is beautiful! Observe that

$\displaystyle \cos( s \arctan y) = \Re \exp(i s \arctan y) = \Re \left[ \left( \frac{1 + i y}{\sqrt{1+y^2}} \right)^s \right].$

Also observe that

$\displaystyle 1 + y^2 = (1 + i y)(1 - i y).$

Using these observations and putting $y = a x$, the integral reduces to

$\displaystyle I = \Re \int_{-\infty}^{\infty} \frac{(1-i a x)^{-s}}{1+x^2}.$

Now, using the fact that the integrand is an analytic function of $a$, we know that the complex conjugate of the integral is obtained by replacing $i$ by $-i$. But this is the same integral as we would obtain if we substituted $x = -y$ in the original integral. Therefore the integral is real, so we can drop the $\Re$.

It is trivial to evaluate it using residues and a semicircular contour in the upper half plane. There's a pole at $i$, which immediately gives the answer. For this contest's sake, I will give an alternative derivation. I will regularize the integral by introducing a sinusoidal convergence factor and I will also take the principal value for convenience.

$\displaystyle I = \lim \limits_{\epsilon \rightarrow 0} PV \int_{-\infty}^{\infty} \frac{(1-i a x)^{-s} e^{i \epsilon x} }{1+x^2}.$ Now let us formally expand the binomial in the integrand in an infinite series using the binomial theorem. It is clear that we cannot strictly interchange summation and integration because the integrals do not converge. Also the series expansion is only strictly valid for sufficiently small $|a x|$. However, the principal values of the integrals will turn out to exist, and it won't bother us that the radius of convergence is limited. We just want to find the coefficient of $a^k$ in the expansion around $a = 0$, which turns out to be

$\displaystyle I = \binom {-s} {k} (-i)^k \lim \limits_{\epsilon \rightarrow 0} PV \int_{-\infty}^{\infty} \frac{x^k e^{i \epsilon x} }{1+x^2} = \binom {-s} {k} (-i)^k \lim \limits_{\epsilon \rightarrow 0} \pi e^{-\epsilon} i^k= \pi \binom {-s} {k}.$ Now recognize that this is exactly the coefficient of $a^k$ in the series of $\pi/(1+a)^s$. By the uniqueness of power series, the integral equals $\pi/(1+a)^s$.

Here we have made use of the following fact: $\displaystyle PV \int_{-\infty}^{\infty} \frac{x^k e^{i \epsilon x} }{1+x^2} = \left(\frac 1 i \frac{d}{d\epsilon} \right)^k PV \int_{-\infty}^{\infty} \frac{e^{i \epsilon x} }{1+x^2} = \left(\frac 1 i \frac{d}{d\epsilon} \right)^k \pi e^{-\epsilon} = \pi e^{-\epsilon} i^k.$ I realize that this proof is not super-rigorous, but it is unnatural to do this without complex analysis. If you have a more rigorous proof, I would like to see it, Anastasiya. Oh and I left out the $dx$ everywhere because it does not improve readability, in my opinion.

- 5 years ago

I think this is too complicated. I haven't checked it yet. Here is my solution of this problem. I've just seen your problem, but I think I can't answer it now. I'm sick. I suddenly passed out at school today. Now I know why I always feel dizzy recently. The doctor told me that I should rest for a few days.

@Shivang Jindal You're correct. I reactivate that problem on M.S.E. I also answered that problem 18 days ago & I deleted it temporary for this contest's sake. I have undeleted it. You may have a look again there. Don't forget to upvote it, @Ruben too. LOL

I must also study for my college admission test after I get well, so maybe I won't take part again in this contest. You can continue this contest without me. Make sure you all obey the rules. Okay, bye guys. Cya... 👋(>‿◠)

- 5 years ago

Ah, your solution is more elegant, I agree. I hope you get well soon!

- 5 years ago

Where's your first problem @Anastasiya Romanova

- 5 years ago