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Integration Contest - Season 4

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Post Mon Feb 17, 2014 7:36 pm
galactus User avatar
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Random Variable wrote:
Going back to problem 27,

\(\displaystyle \int_{0}^{1} \frac{\log(1-x)}{\sqrt{1-x^{2}}} \ dx = \int_{0}^{\pi /2} \log(1- \sin u ) \ d u\)


In general,

\(\displaystyle \int_{0}^{\theta} \log(1- \sin x) \ dx = \int_{0}^{\theta} \Big( 2 \ \text{Re} \log(1+ie^{ix}) - \log 2 \Big) \ d x\)

\(\displaystyle = 2 \ \text{Re} \int_{0}^{\theta} \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(ie^{ix})^{n}}{n} \ dx - \theta \log 2\)

\(\displaystyle = 2 \ \text{Re} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}i^{n}}{n} \int_{0}^{\theta} e^{inx} \ dx - \theta \log 2\)

\(\displaystyle = 2 \ \text{Re} \ i \sum_{n=1}^{\infty} \frac{(-ie^{i \theta})^{n}}{n^{2}} - 2 \ \text{Re} \sum_{n=1}^{\infty} \frac{(i)^{n-1}}{n^{2}} - \theta \log 2\)

\(\displaystyle = 2 \ \text{Re} \ i \ \text{Li}_{2} (-ie^{i \theta}) - 2 G - \theta \log 2\)

\(\displaystyle = 2 \sum_{n=0}^{\infty} \frac{(-1)^{n} \cos(2n+1) \theta}{(2n+1)^{2}} + \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \sin 2 n \theta}{n^{2}} - 2 G - \theta \log 2\)


So

\(\displaystyle \int_{0}^{1} \frac{\log(1-x)}{\sqrt{1-x^{2}}} \ dx = - 2 G - \frac{\pi \log 2}{2}\)


that's a nice method, RV.

Post Tue Feb 18, 2014 7:09 am
Shobhit Site Admin
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The Problem 31 is a special case of the formula:
$$\operatorname{Li}_s(z) = \tfrac{1}{2}z + z \int_0^\infty \frac{\sin[s \tan^{-1} t \,- \,t \log(-z)]} {(1+t^2)^{s/2} \,\sinh(\pi t)} \,\mathrm{d}t$$ Surprisingly this holds for all complex \(s\).

Post Wed Feb 19, 2014 11:43 am
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Hey S man. Sorry for not being around. Busy through the week, especially on Tuesday and Thursday.

But, it would appear your problem 31 is related to \(\displaystyle Li_{2}(-\phi)=\frac{-\pi^{2}}{10}-log^{2}(\phi)\)?.

I will look at it more this afternoon.

Post Tue Mar 04, 2014 10:06 am
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Problem 32 (Restored)

\(\displaystyle \int_{0}^{\infty}\frac{(1-x^{2})\tan^{-1}(x^{2})}{x^{4}+4x^{2}+1}dx=\frac{-\pi^{2}}{12\sqrt{2}}\)

Problem 33 (New)

Here's another integral because the contest has been kind of stagnant lately.

\(\displaystyle \int_{0}^{1}\frac{\log(1-x)}{\sqrt{x(1-x^{2})}}dx=\frac{\Gamma^{2}(1/4)(\log(2)-\pi)}{4\sqrt{2\pi}}\)

Post Sat Mar 08, 2014 10:58 am
Shobhit Site Admin
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galactus wrote:
Problem 32

Here's another integral because the contest has been kind of stagnant lately.

\(\displaystyle \int_{0}^{\infty}\frac{(1-x^{2})\tan^{-1}(x^{2})}{x^{4}+4x^{2}+1}dx=\frac{-\pi^{2}}{12\sqrt{2}}\)


For \(\displaystyle \alpha>0\) let

\(\displaystyle \; f(z)=\frac{\log(1+i z^2)}{z^2+\alpha^2}\)

\(\displaystyle I(\alpha) =\int_{0}^{\infty}\frac{\tan^{-1}(x^{2})}{x^2+\alpha^2}dx\)

Then using a sand dollar with with branch point at $e^{i\pi/4}$ we get

\(\displaystyle \begin{align*}
I(\alpha) &=\frac{1}{2}\Im \int_{-\infty}^{\infty}f(x) dx \\
&=\Im \left\{ \pi i \Big( \text{Res}_{z=i\alpha}f(z)\Big)+\pi i \int_{e^{i\pi/4}}^{e^{i\pi/4} \infty }\frac{1}{x^2+\alpha^2}dx \right\} \\
&= -\frac{\pi}{2\alpha} \tan^{-1}\alpha^2 +\Im \left\{ \pi i \int_{e^{i\pi/4}}^{e^{i\pi/4} \infty }\frac{1}{x^2+\alpha^2}dx \right\}\\
&= \frac{\pi^2}{2\alpha}-\frac{\pi}{2\alpha} \tan^{-1}\alpha^2-\Im \left\{ \frac{\pi i}{\alpha}\tan^{-1}\left(\frac{e^{i\pi/4}}{\alpha} \right) \right\} \\
&= \frac{\pi^2}{2\alpha}-\frac{\pi}{2\alpha} \tan^{-1}\alpha^2+\frac{\pi}{2\alpha} \arg \left( \frac{\alpha-i e^{i\pi/4}}{\alpha+i e^{i\pi/4}}\right) \\
&= \frac{\pi}{2\alpha}\left\{\pi - \tan^{-1}(\alpha^2)-\tan^{-1}\left(\frac{\sqrt{2}\alpha}{\alpha^2-1} \right) \right\}
\end{align*}\)

Now use the partial fraction decomposition

\(\displaystyle \frac{1-x^2}{x^4+4x^2+1}=\frac{\sqrt{3}-1}{2}\frac{1}{x^2+2-\sqrt{3}}-\frac{\sqrt{3}+1}{2}\frac{1}{x^2+2+\sqrt{3}}\)

to get

\(\displaystyle \begin{align*}
\int_0^\infty \frac{(1-x^2)\tan^{-1}(x^2)}{x^4+4x^2+1}dx &= \frac{\sqrt{3}-1}{2}I\left(\frac{\sqrt{3}-1}{\sqrt{2}} \right)-\frac{\sqrt{3}+1}{2}I\left(\frac{\sqrt{3}+1}{\sqrt{2}} \right) \\
&= \frac{\pi}{2\sqrt{2}}\left\{ \tan^{-1}\left( 2+\sqrt{3}\right)-\tan^{-1}\left( 2-\sqrt{3}\right)-\frac{\pi}{2} \right\} \\
&= \frac{\pi}{2\sqrt{2}} \left\{ \frac{\pi}{3}-\frac{\pi}{2} \right\} \\
&= -\frac{\pi^2}{12\sqrt{2}}
\end{align*}\)

Post Sun Mar 09, 2014 2:42 am
Shobhit Site Admin
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And for problem 33,

\(\displaystyle I(a)=\int_0^1 \frac{(1-x)^a}{\sqrt{x(1+x)}}dx=\sqrt{\pi}\Gamma(1+a) {\;}_2F_1 \left(\frac{1}{2},\frac{1}{2};\frac{3}{2}+a;-1 \right)\)

Then the required integral is $I'(-1)$ but I have no idea how to calculate it.

Post Sun Mar 09, 2014 8:28 am
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Yes, I noticed the same thing S. If it can only be expressed in terms of hypergeometric, then I am not too interested in it. Perhaps that hypergeo has an equivalent nicer closed form.

I managed to find something that is numerically close that is easier to solve.

\(\displaystyle 1/4\int_{0}^{1}\frac{log(x)}{x^{3/4}(1-x)^{1/2}}dx-1/4\int_{0}^{1}\frac{log(x)}{x^{1/2}(1-x)^{1/2}}dx=\frac{-\pi^{5/2}\sqrt{2}}{4\Gamma^{2}(3/4)}+\frac{\pi}{2}log(2)\approx -3.20992533...\)

compare to the actual numerical for the integral is -3.20998247....

Just a thing. I ran it through Maple and all I got was MeijerG crap.

If you have something better for problem 33...go ahead and post it instead. I just posted it because I didn't have anything else.

How about \(\displaystyle \int_{0}^{\infty}\frac{x^{2}\tan^{-1}(x)}{e^{2\pi x}-1}dx=\frac{-11}{36}+1/4log(2\pi)+\zeta'(-1)+\frac{\zeta(3)}{8\pi^{2}}\) ?.

assuming we have not done it before at some point.

Post Mon Mar 10, 2014 8:16 am
Shobhit Site Admin
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How did you guess that answer? Inverse symbolic calculator?

Anyway, I successfully solved problem 33 :)

http://math.stackexchange.com/questions/702681/integral-int-01-frac-log1-x-sqrtx-x3dx/706447#706447

Post Mon Mar 10, 2014 8:46 am
Shobhit Site Admin
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I will copy my solution here:

Rewrite the integral as

$$
\begin{align*}
I &= \int_0^1 \frac{\log(1-x)}{\sqrt{x}\sqrt{1-x^2}}dx \\
&= \int_0^1 \frac{\log(1-x^2)-\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx \\
&= \int_0^1 \frac{\log(1-x^2)}{\sqrt{x}\sqrt{1-x^2}}dx-\int_0^1 \frac{\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx
\end{align*}
$$
The first integral can be computed by calculating a derivative of the beta function.
$$
\begin{align*}
&\;\int_0^1 \frac{\log(1-x^2)}{\sqrt{x}\sqrt{1-x^2}}dx \\
&\stackrel{y=x^2}{=}\frac{1}{2}\int_0^1 \frac{\log(1-y)}{y^{\frac{3}{4}}\sqrt{1-y}}dy \\
&= \frac{1}{2}\frac{\partial}{\partial \alpha}\left\{B\left(\frac{1}{4},\alpha \right)\right\}_{\alpha=\frac{1}{2}}\\
&= \frac{\sqrt{\pi}\Gamma\left(\frac{1}{4}\right)}{2\Gamma\left(\frac{3}{4} \right)}\left\{ \psi_0\left(\frac{1}{2} \right)-\psi_0\left(\frac{3}{4} \right)\right\} \\
&= \frac{\Gamma\left(\frac{1}{4} \right)^2}{4\sqrt{2\pi}}\left(2\log 2-\pi \right)
\end{align*}
$$

The other integral can be evaluated by using equation $(3.22)$ ofthis paper.

$$
\begin{align*}
&\;\int_0^1 \frac{\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx \\
&\stackrel{x=\sin^2 t}{=}2\int_0^{\frac{\pi}{2}}\frac{\log(1+\sin^2 t)}{\sqrt{1+\sin^2 t}}dt\\
&= \log(2) K(\sqrt{-1}) \\
&= \log(2)\frac{\Gamma\left(\frac{1}{4} \right)^2}{4\sqrt{2\pi}}
\end{align*}
$$
where $K(k)$ is the complete elliptic integral of the first kind. After combining everything, one gets

$$I=\frac{\Gamma\left(\frac{1}{4} \right)^2}{4\sqrt{2\pi}}\left(\log 2-\pi \right)\approx -3.20998$$

Post Mon Mar 10, 2014 4:55 pm
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Shobhit wrote:
How did you guess that answer? Inverse symbolic calculator?

Anyway, I successfully solved problem 33 :)

http://math.stackexchange.com/questions/702681/integral-int-01-frac-log1-x-sqrtx-x3dx/706447#706447


No, no symbolic thingie. I just kind of noticed it by playing around.


anyway, wonderful finale to that problem. I got hung up on the elliptic part.

here is a couple fun ones to possibly do with contours(how I done the second one). But, whatever method you wish.

\(\displaystyle \int_{0}^{1}\frac{1}{\sqrt{1-\sqrt[n]{x}}}dx=\frac{4^{n}}{\binom{2n}{n}}\)

or

Problem 34 (Declared)

\(\displaystyle \int_{0}^{1}\frac{\sqrt[3]{x^{2}(1-x)}}{(1+x)^{3}}dx=\frac{\pi\sqrt[3]{2}}{18\sqrt{3}}\)

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