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Integration Contest - Season 5

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Post Mon Oct 26, 2015 11:03 am
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Very nice MCNE. Clever on that last one I posted.

This is one that looks worse than it really is.

Make a sub, then doing each integral. I think my method is pretty much what you done, MCNE.

$$\int_{0}^{1}\frac{(-\ln(x))^{m-1}}{1-x}dx-\int_{0}^{1}\frac{x^{n-1}(-\ln(x))^{m-1}}{1-x^{n}}dx$$

Now, the left one:

Make the sub $t=-\ln(x), \;\ dx=-e^{-t}dt$

$$\int_{0}^{\infty}\frac{t^{m-1}e^{-t}}{1-e^{-t}}$$

Use geometric series:

$$=\sum_{k=0}^{\infty}\int_{0}^{\infty}t^{m-1}e^{-(k+1)t}dt$$

$$=\Gamma(m)\sum_{k=0}^{\infty}\frac{1}{(k+1)^{m}}=\Gamma(m)\zeta(m)$$

Now do the right integral. Make the sub $t=x^{n}$ and we end up with the same integral as step 1.

$$\frac{1}{n^{m}}\int_{0}^{1}\frac{(-ln(t))^{m-1}}{1-t}dt$$

$$=\frac{1}{n^{m}}\Gamma(m)\zeta(m)$$

Factor out the $\Gamma(m)\zeta(m)$

$$\left(1-\frac{1}{n^{m}}\right)\Gamma(m)\zeta(m)$$

I am going to post another to keep it going. Let's not stop now. I have some goodies to post.

I do not post anymore than I do because the site has been loading real slow lately on my computer. I do not know what is causing it.

EDIT: It appears to be Firefox. I am now using IE and it is working much better. That sure is a change. :roll:

Post Mon Oct 26, 2015 11:07 am
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Evaluate

$$\int_{0}^{1}\ln(x^{2}+x+1)\ln(x^{2}-x+1)dx$$

or

$$\int_{0}^{1}\ln^{2}(x^{4}+x^{2}+1)dx$$

Post Tue Nov 03, 2015 11:06 am
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Well, since no one has bitten on the last one, maybe this one is more to your liking?.

This is an integral from that German site, but with 'ohne beweis' or 'no solution' given.

A contour should knock it out.

$$\int_{0}^{\infty}\frac{x\sin(ax)}{\cosh(\frac{\pi x}{4})(x^{2}+1)}dx$$

$$=\frac{\pi}{\sqrt{2}}e^{-a}+\frac{1}{\sqrt{2}}\sinh(a)\log\left(\frac{2\cosh(a)+\sqrt{2}}{2\cosh(a)-\sqrt{2}}\right)-\sqrt{2}\cosh(a)\tan^{-1}\left(\frac{1}{\sqrt{2}\sinh(a)}\right)$$

Post Sat Nov 07, 2015 1:19 am

Posts: 50
I have a very rough idea of how one might approach the previous 2 integrals (somewhat similar to what I tried here http://math.stackexchange.com/questions ... 67#1433467), but sadly I lack the will or time to trudge through the details. Perhaps I will attempt it again when I am less busy, or someone else might want to have a shot at it.


The function $f(z)=\dfrac{ze^{iaz}}{\cosh\left(\frac{\pi z}{4}\right)(1+z^2)}$ has simple poles at $z=i$ and $z=(4n+2)i$ in the upper half plane. Noting Jordan's lemma and summing the residues,
\begin{align}
\frac{1}{2}\int^\infty_{-\infty}\frac{x\sin(ax)}{\cosh\left(\frac{\pi x}{4}\right)(1+x^2)}\ {\rm d}x
&=\Re\pi\left(\frac{ie^{ia(i)}}{\cosh\left(\frac{\pi i}{4}\right)(2i)}+\sum^\infty_{n=0}\frac{(4n+2)ie^{ia(4n+2)i}}{\frac{\pi}{4}\sinh\left(\frac{\pi(4n+2)i}{4}\right)(1+((4n+2)i)^2)}\right)\\
&=\frac{\pi}{\sqrt{2}}e^{-a}-2e^{-2a}\sum^\infty_{n=0}(-e^{-4a})^n\left(\frac{1}{4n+1}+\frac{1}{4n+3}\right)\\
&=\frac{\pi}{\sqrt{2}}e^{-a}-2e^{-2a}\sum^\infty_{n=0}(-e^{-4a})^n\int^1_0x^{4n}(1+x^2)\ {\rm d}x\\
&=\frac{\pi}{\sqrt{2}}e^{-a}-2e^{-2a}\int^1_0\frac{1+x^2}{1+e^{-4a}x^4}\ {\rm d}x\\
\end{align}
Partial fraction decomposition (or WolframAlpha) yields
\begin{align}
\int^1_0\frac{1+x^2}{1+e^{-4a}x^4}\ {\rm d}x
&=\frac{1}{4\sqrt{2}}e^a\left[(e^{2a}-1)\ln\left(\frac{1-\sqrt{2}e^{-a}x+e^{-2a}x^2}{1+\sqrt{2}e^{-a}x+e^{-2a}x^2}\right)+2(e^{2a}+1)\arctan\left(\frac{2\sqrt{2}e^{-a}x}{1+1-2e^{-2a}x^2}\right)\right]^1_0\\
&=\frac{1}{4\sqrt{2}}e^a\left[(e^{2a}-1)\ln\left(\frac{e^a-\sqrt{2}+e^{-a}}{e^a+\sqrt{2}+e^{-a}}\right)+2(e^{2a}+1)\arctan\left(\frac{\sqrt{2}}{e^a-e^{-a}}\right)\right]\\
&=\frac{1}{4\sqrt{2}}e^a\left[(e^{2a}-1)\ln\left(\frac{2\cosh{a}-\sqrt{2}}{2\cosh{a}+\sqrt{2}}\right)+2(e^{2a}+1)\arctan\left(\frac{1}{\sqrt{2}\sinh{a}}\right)\right]\\
\end{align}
Therefore, we indeed have
\begin{align}
\int^\infty_{0}\frac{x\sin(ax)}{\cosh\left(\frac{\pi x}{4}\right)(1+x^2)}\ {\rm d}x
&=\frac{\pi}{\sqrt{2}}e^{-a}-\frac{1}{2\sqrt{2}}e^{-a}\left[(e^{2a}-1)\ln\left(\frac{2\cosh{a}-\sqrt{2}}{2\cosh{a}+\sqrt{2}}\right)+2(e^{2a}+1)\arctan\left(\frac{1}{\sqrt{2}\sinh{a}}\right)\right]\\
&=\frac{\pi}{\sqrt{2}}e^{-a}-\frac{\sinh{a}}{\sqrt{2}}\ln\left(\frac{2\cosh{a}-\sqrt{2}}{2\cosh{a}+\sqrt{2}}\right)-\sqrt{2}\cosh{a}\arctan\left(\frac{1}{\sqrt{2}\sinh{a}}\right)
\end{align}

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