Brilliant Integration Contest - Season 1 (Part 3)

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, $f(a) = \frac{\pi e^{-a}}{\sqrt{2}}$

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).

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:

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

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.

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.

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.

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.

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

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

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

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

From my answer on Math SE (see also other answers there), we have

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

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.

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.

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

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.

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 $\sqrt{2}\,E\left(\frac{\pi}{2},\frac{1}{\sqrt{2}}\right)$

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

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

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

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

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

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

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

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

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

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.

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 :

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

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

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