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Integration Contest - Season 4

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Post Tue Mar 11, 2014 9:03 pm
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Well, RV, I decided to try this one using this method. I will probably screw it up, but see if it's OK.

\(\displaystyle \int_{0}^{1}\frac{\sqrt{x(x-1)}}{(x^{2}+1)^{2}}dx\)

The poles are at \(\displaystyle \pm i\)

order 2, so diffing is in order.

\(\displaystyle 2\pi i Res(i)=\frac{-1}{8\sqrt{i-1}}\)

\(\displaystyle 2\pi i Res(-i)=\frac{-1}{8\sqrt{-1-i}}\)

add them gives \(\displaystyle \frac{-\pi}{4}\sqrt{\sqrt{2}-1}i\)

the residue at infinity is 0.

I tried the same args.

below the cut: \(\displaystyle arg(z)=2\pi, \;\ arg(z-1)=\pi\)

above the cut: \(\displaystyle arg(z)=0, \;\ arg(z-1)=\pi\)

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

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

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

\(\displaystyle \int_{0}^{1}\frac{\sqrt{x(x-1)}}{(x^{2}+1)^{2}}dx=\frac{\pi}{8}\sqrt{\sqrt{2}-1}\)

Post Thu Mar 13, 2014 1:37 pm
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Here's another integral in case anyone is interested. Have we done this one before?. It sure seems familiar. I saw it on AoPS.

This one sure looks familiar, but I can find no reference to it. Strange.

\(\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}(ax)}{\sin^{2}(x)}dx=\frac{\pi a}{2}+\frac{\sin(\pi a)}{2}+\frac{a\cdot \sin(\pi a)}{2}\left(\psi(a/2)-\psi\left(\frac{a+1}{2}\right)\right)\)

when 'a' in an integer, then this has a general form of just \(\displaystyle \frac{\pi a}{2}\). It is the non-integer values of 'a' that require more work and the formula involving digamma.

EDIT:
I made progress on this one by using the rather obscure identity:

\(\displaystyle \int_{0}^{\pi}\sin^{p}(x)e^{iqx}dx=\frac{\pi}{2^{p}}\left[\frac{\Gamma(1+x)e^{\pi i q/2}}{\Gamma(1+\frac{p+q}{2})\Gamma(1+\frac{p-q}{2})}\right]...[1]\)

I have to assemble the pieces though, but it's there. One may write:

\(\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}(ax)}{\sin^{2}(x)}dx=1/2\int_{0}^{\pi}\frac{1-\cos(ax)}{1-\cos(x)}dx\)

But first, we can evaluate \(\displaystyle \int_{0}^{\pi}\frac{\sin(ax)}{\tan(x/2)}dx=\int_{0}^{\pi}\frac{\sin(ax)(1+\cos(x))}{\sin(x)}dx=\int_{0}^{\pi}\frac{\sin(ax)\sin(x)}{1-\cos(x)}dx=\int_{0}^{\pi}\frac{2\sin(ax)+\sin(a+1)x+\sin(a-1)x}{2\sin(x)}dx\)




to be continued...............I have got the solution assembled. Please ask if you would like to see it. I had trouble with a nasty/tedious limit, but it appears to work out rather nicely. The integral identity in [1] can be shown by using contours and considering:

\(\displaystyle f(x)=\frac{1}{(u+ix)^{a}(v-ix)^{b}}\) and make the sub \(\displaystyle x=cot(t)\). This is one way to go about it.

EDIT:

I was playing around with this when u=v=1.

\(\displaystyle \int_{-\infty}^{\infty}\frac{1}{(1+ix)^{a}(1-ix)^{b}}dx\)

Factor out i:

\(\displaystyle \frac{1}{i^{a}}\int_{-\infty}^{\infty}\frac{1}{(x-i)^{a}(1-ix)^{b}}dx\)

There are poles at \(\displaystyle x=\pm i\). Only \(\displaystyle i\) is in the contour.

But, it has order 'a', so diff a-1 times to find the residue.

\(\displaystyle =\frac{2\pi i}{i^{a}}\lim_{z\to i}\frac{1}{(a-1)!}\frac{d^{a-1}}{dz^{a-1}}\frac{1}{(1-iz)^{b}}\)

do this a-1 times and note that \(\displaystyle (a-1)!=\Gamma(a)\)

\(\displaystyle =2\pi\frac{b(b+1)(b+2)\cdot\cdot\cdot (b+a-2)}{\Gamma(a)}\lim_{z\to i}(1-iz)^{1-a-b}\)

\(\displaystyle =\pi 2^{2-a-b}\frac{b(b+1)\cdot\cdot\cdot (b+a-2)}{\Gamma(a)}\)

\(\displaystyle =2^{2-a-b}\pi \frac{\Gamma(a+b-1)}{\Gamma(a)\Gamma(b)}\)

Post Sat Mar 22, 2014 10:15 am
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Evaluate


\(\displaystyle \int_{0}^{\infty}\frac{\cos(ax)}{b^{2n}+x^{2n}}dx\)

Post Sat Mar 22, 2014 3:51 pm
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First we need to find the partial fraction decomposition of \(\displaystyle \frac{1}{x^{2n}+b^{2n}}\).


\(\displaystyle \frac{1}{x^{2n}+b^{2n}} = \sum_{k=0}^{2n-1} \frac{A_{k}}{x-be^{i \pi (2k+1)/(2n)}}\)

where \(\displaystyle A_{k} = \text{Res} \left[ \frac{1}{z^{2n}+b^{2n}}, be^{i \pi (2k+1) /(2n)} \right] = - \frac{1}{2nb^{2n-1}} e^{i \pi (2k+1)/(2n)}\)


I'm going to denote \(\displaystyle \frac{2k+1}{2n} \pi\) by \(\displaystyle \lambda_{k}\)


Then

\(\displaystyle \frac{1}{x^{2n}+b^{2n}} = - \frac{1}{2nb^{2n-1}} \sum_{k=0}^{2n-1} \frac{e^{i \lambda_{k}}}{x-be^{i \lambda_{k}}}= - \frac{1}{2nb^{2n-1}} \sum_{k=0}^{n-1} \Big( \frac{e^{i \lambda_{k}}}{x-be^{i \lambda_{k}}} + \frac{e^{-i \lambda_{k}}}{x-be^{-i \lambda_{k}}} \Big)\)

\(\displaystyle = \frac{1}{nb^{2n-1}} \sum_{k=0}^{n-1} \frac{b- x \cos (\lambda_{k})}{x^{2}-2bx \cos( \lambda_{k})+b^{2}}\)

\(\displaystyle = \frac{1}{nb^{2n-1}} \sum_{k=0}^{n-1} \frac{b - x \cos (\lambda_{k})}{(x-b \cos \lambda_{k})^{2} + b^{2} \sin^{2} (\lambda_{k})}\)




We also need the following two results which are fairly easy to derive using other known results.

\(\displaystyle \int_{-\infty}^{\infty} \frac{\cos rx}{(x-b)^{2}+a^{2}} \ dx = \frac{\pi}{a} e^{-ar} \cos(br)\)

\(\displaystyle \int_{-\infty}^{\infty} \frac{x \cos (rx)}{(x-b)+a^{2}} \ dx = \frac{\pi}{a} e^{-ar} \left( b \cos br - a \sin br \right)\)


Then

\(\displaystyle \int_{-\infty}^{\infty} \frac{\cos (ax)}{x^{2n}+b^{2n}} \ dx = \frac{1}{nb^{2n-1}} \sum_{k=0}^{n-1} \left[ b \int_{-\infty}^{\infty} \frac{\cos (ax)}{(x-b \cos \lambda_{k})^{2}+ b^{2} \sin^{2} (\lambda_{k})} - \cos \lambda_{k} \int_{-\infty}^{\infty} \frac{ x \cos(ax)}{(x-b \cos \lambda_{k})^{2} + b^{2} \sin^{2} \lambda_{k}} \ dx \right]\)

\(\displaystyle = \frac{1}{nb^{2n-1}} \sum_{k=0}^{n-1} \Big[ \frac{\pi}{\sin \lambda_{k}} e^{-ab \sin \lambda_{k}} \cos(ab \cos \lambda_{k}) - \frac{\pi \cos \lambda_{k} }{b \sin \lambda_{k}} e^{-ab \sin \lambda_{k}} \Big( b \cos(\lambda_{k}) \cos( ab \cos \lambda_{k}) - b \sin (\lambda_{k}) \sin (ab \cos \lambda_{k} \Big) \Big]\)

\(\displaystyle = \frac{\pi}{nb^{2n-1}} \sum_{k=0}^{n-1} \frac{1}{\sin \lambda_{k}} e^{-ab \sin \lambda_{k}} \Big( \cos(ab \cos \lambda_{k}) - \cos(\lambda_{k}) \cos(ab \cos \lambda_{k}+\lambda_{k}) \Big)\)

\(\displaystyle = \frac{\pi}{nb^{2n-1}} \sum_{k=0}^{n-1} \frac{1}{\sin \lambda_{k}} e^{-ab \sin \lambda_{k}} \Big( \cos( ab \cos \lambda_{k}) - \cos^{2}(\lambda_{k}) \cos(ab \cos \lambda_{k}) + \cos (\lambda_{k}) \sin(\lambda_{k}) \sin(ab \cos \lambda_{k}) \Big)\)

\(\displaystyle = \frac{\pi}{nb^{2n-1}} \sum_{k=0}^{n-1} e^{-ab \sin \lambda_{k}} \Big( \sin(\lambda_{k}) \cos(ab \cos \lambda_{k}) + \cos(\lambda_{k}) \sin(ab \cos \lambda_{k}) \Big)\)

\(\displaystyle = \frac{\pi}{nb^{2n-1}} \sum_{k=0}^{n-1} e^{-ab \sin (\lambda_{k})} \sin \left(ab \cos \lambda_{k} + \lambda_{k} \right)\)

Post Sat Mar 22, 2014 4:42 pm
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Very nice RV meister :)

I tried that one pretty much the same way. But man it was tedious :!:

I will post my workings. It's quite a lot of typing though. Looks like you done a lot of typing too. Nice job.

Can I ask you something?. Have you ever done this one using contours?.

\(\displaystyle \int_{0}^{\infty}\frac{\cos(x^{2})-\cos(x)}{x}dx\)

Post Sat Mar 22, 2014 9:00 pm
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Cody

I've never even considered trying to evaluate that integral using contour integration.

Can you get Maple or Mathematica to evaluate \(\displaystyle \int_{0}^{\infty} \frac{\cos x}{1+x^{2n}}\) for a value of $n$ greater than or equal to $4$ and return something that doesn't look totally ridiculous?

Post Sun Mar 23, 2014 11:18 am
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Hey RV:

I done as you asked and ran that through Maple and Wolfram (I do not have regular Mathematica on my computer).

Wolfram returned nothing. (But, then again, it does not have the power as that of the regular computer-installed Mathematica).

Maple returned, as you said, gobbledy-gook in terms of the sum of the roots of some 2n degree polynomial and the Si and Ci functions. In other words, a mess. It may as well return nothing as far as a closed form goes. It returns numerical answers, but that's not what we want.

i.e. for n = 4, Maple gave \(\displaystyle -\sum_{n=Root \;\ of \;\ z^{8}+1}\left[\frac{\pi \sin(n)}{16n^{7}}+\frac{Si(n)\sin(n)+Ci(-n)\cos(n)}{n^{7}}\right]\).

The larger n becomes, the ickier it gets.

This is what you worked out with a = 1 and b = 1 in the problem you just posted the solution to. See, it's cool when this happens. You solve a problem the fancy math engines can not. :)

I will try to take the time to post my workings for that one. It is lengthy. I tried contours as well.

Post Sun Mar 23, 2014 12:37 pm
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Well, RV, I will go ahead and post my workings to that one. All of the little intricacies were cumbersome. For me anyway. But, I got it assembled.

What I have here is for a = 1.

Write \(\displaystyle \int_{C}\frac{e^{iz}}{b^{2n}+z^{2n}}dz\), where C is a typical semicircle in the UHP.

The poles are at \(\displaystyle z=be^{\frac{(2k+1)\pi i}{2n}}\) and are equally spaced at \(\displaystyle \frac{\pi}{n}\) intervals.

The poles that contribute are in the UHP. So, I am not going to bother writing out the others because they don't contribute anything anyway.

\(\displaystyle \sum_{k=0}^{n-1}\frac{1}{2n\left(be^{\frac{(2k+1)}{2n}\pi i}\right)^{2n-1}\left(z-be^{\frac{(2k+1)\pi i}{2n}}\right)}\)

The poles in the contour then contribute, due to the sum of the residues, in total (I hope this is correct. It appears to be. One itty-bitty negative or whatever and it's blown out of the water :)):

\(\displaystyle -\sum_{k=0}^{n-1}\frac{\pi i}{nb^{2n-1}}e^{-b\sin(\frac{2k+1}{2n}\pi}\left[\cos\left(\frac{2k+1}{2n}\pi +b\cos(\frac{2k+1}{2n}\pi)\right)+i\cdot \sin\left(\frac{2k+1}{2n}\pi +b\cos(\frac{2k+1}{2n}\pi)\right)\right]\).

The outside arc goes to 0 as R goes to infinity because of the \(\displaystyle e^{-R\sin(t)}\) thing. So, I will not write it up.

Assembling:

\(\displaystyle \int_{-\infty}^{0}\frac{e^{ix}}{b^{2n}+x^{2n}}dx+\int_{0}^{\infty}\frac{e^{ix}}{b^{2n}+x^{2n}}dx+\int_{0}^{\pi}\frac{e^{iRe^{it}}iRe^{it}}{b^{2n}+R^{2n}e^{2nit}}dt=\)
\(\displaystyle \sum_{k=0}^{n-1}\frac{\pi i}{nb^{2n-1}}e^{-b\sin(\frac{2k+1}{2n}\pi}\left[\cos\left(\frac{2k+1}{2n}\pi +b\cos(\frac{2k+1}{2n}\pi)\right)+i\cdot \sin\left(\frac{2k+1}{2n}\pi +b\cos(\frac{2k+1}{2n}\pi)\right)\right]\)

Now, on the left, let \(\displaystyle x\to -x\) in the first integral and add to the second one. Thus, giving:

\(\displaystyle \int_{0}^{\infty}\frac{\cos(x)}{b^{2n}+x^{2n}}dx=\sum_{k=0}^{n-1}\frac{\pi}{2nb^{2n-1}}e^{-b\sin(\frac{2k+1}{2n}\pi)}\sin\left(\frac{2k+1}{2n}\pi +b\cos(\frac{2k+1}{2n}\pi)\right)\)

There certainly could be a typo in all of that anyway.

For example, if n = 4, then the result is:

\(\displaystyle \frac{\pi}{4}e^{-\sqrt{\sqrt{2}+2}\sin(\frac{\sqrt{2-\sqrt{2}}}{2}+\frac{3\pi}{8})/2}+\frac{\pi}{4}e^{-\sqrt{2-\sqrt{2}}\sin(\frac{\sqrt{2+\sqrt{2}}}{2}+\frac{\pi}{8})/2}\)

which, I am sure, can be written in various forms, and appears to check numerically.


BTW, here is another general one to try with or without contours. Whatever you wish.

\(\displaystyle \int_{0}^{\infty}\frac{\cos(x)}{(b^{2}+x^{2})^{n+1}}dx\)

Maybe a little tougher?. Anyway, the next time an integral like this pops up on SE or wherever, one can just show the general case and maybe show off a little bit :):):).

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