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August 27th, 2012, 06:24 AM   #1
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Integral Computation.

Compute the integral $\displaystyle \int_{0}^{1}\frac{1}{(x-2)\sqrt[5]{x^2(1-x)^3}}\;dx$.

Last edited by skipjack; February 18th, 2018 at 02:08 AM.
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August 27th, 2012, 09:17 AM   #2
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Re: Integral Computation.

Read the result:
integral.JPG

Last edited by skipjack; February 18th, 2018 at 02:08 AM.
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August 27th, 2012, 09:19 AM   #3
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Re: Integral Computation.

I know the answer how can we compute it? I know two ways.

Last edited by skipjack; February 18th, 2018 at 02:09 AM.
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August 27th, 2012, 10:33 AM   #4
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Re: Integral Computation.

Hey Z: Fun integral. Challenging.

When I see a (1-x) written to some power, I usually think Beta is in there somewhere.

Write as $\displaystyle \frac{-1}{2}\int_{0}^{1}\frac{1}{(1-\frac{x}{2})\sqrt[5]{x^{2}(1-x)^{3}}}dx$ by dividing by 2.

Now, write the 1-x/2 part as a geometric series:

$\displaystyle \frac{-1}{2}\int_{0}^{1}\frac{1}{\sqrt[5]{x^{2}(1-x)^{3}}}\cdot \sum_{k=0}^{\infty}(x/2)^{k}$

$\displaystyle =\frac{-1}{2}\sum_{k=0}^{\infty}\frac{1}{2^{k}}\int_{0}^{1 }x^{k-2/5}(1-x)^{-3/5}dx$

Now, use Beta:

$\displaystyle \frac{-1}{2}\sum_{k=0}^{\infty}\frac{1}{2^{k}}B(k+3/5, 2/5)$

Now, write Beta in terms of Gamma:

$\displaystyle \frac{-1}{2}\sum_{k=0}^{\infty}\frac{\Gamma(k+3/5)\Gamma(2/5)}{2^{k}\Gamma(k+1)}$

Write $\displaystyle \Gamma(k+3/5)$ in its integral form, and let $\displaystyle \Gamma(k+1)=k!$:

$\displaystyle =-\frac{\Gamma(2/5)}{2}\int_{0}^{\infty}x^{-2/5}e^{-x}\sum_{k=0}^{\infty}\frac{x^{k}}{k!2^{k}}$

The sum on the right is the series for $\displaystyle e^{x/2}$. So, we have:

$\displaystyle \frac{-\Gamma(2/5)}{2}\int_{0}^{\infty}x^{-2/5}e^{-x}e^{x/2}dx$

$\displaystyle \frac{-\Gamma(2/5)}{2}\int_{0}^{\infty}x^{-2/5}e^{-x/2}dx$

$\displaystyle =\frac{-\Gamma(2/5)\Gamma(3/5)2^{3/5}}{2}$

Note that $\displaystyle \Gamma(p)\Gamma(1-p)=\pi \csc(\pi p)$:

$\displaystyle =-2^{-2/5}\pi \csc(2\pi /5)=\boxed{\frac{-2\pi 2^{1/10}}{\sqrt{5+\sqrt{5}}}\approx -2.503407...}$

Which can be re-written in various forms.

Last edited by skipjack; February 18th, 2018 at 02:17 AM.
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August 27th, 2012, 12:25 PM   #5
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Re: Integral Computation.

Yes, nice!

Last edited by skipjack; February 18th, 2018 at 02:20 AM.
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August 27th, 2012, 01:54 PM   #6
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Re: Integral Computation.

Brief explanation of how it can be done with complex analysis.

Using the figure below,

gal.png

for $\displaystyle \displaystyle{f(z)=\frac{1}{(x-2)\sqrt[5]{z^2(1-z)^{3}}}}$, with $\displaystyle \displaystyle{-\pi\leq \arg\left(\log(z)\right)<\pi}$ and $\displaystyle \displaystyle{0\leq \arg\left(\log(1-z)\right)<2\pi}$

$\displaystyle \displaystyle\oint_{C}f(z)\;\textrm{d}z=2\pi i \textrm{Res}(f(z);z=2,\infty)$

letting the radii of the two small circles go to zero

$\displaystyle \displaystyle\oint_{C}f(z)\;\textrm{d}z= \int_{\textrm{L}_1}f(x)\;\textrm{d}x+\int_{\textrm {L}_2}f(x)\;\textrm{d}x+\int_{C_L}f(z)\; \textrm{d}z+\int_{C_R}f(z)\;\textrm{d}z= \left(1+\frac{1}{e^{\frac{\pi i }{5}}}\right)\underbrace{\int_{0}^{1}\frac{1}{(x-2)\sqrt[5]{x^2(1-x)^3}}\;\textrm{d}x}_{I}$

and so
$\displaystyle \displaystyle\left(1+\frac{1}{e^{\frac{\pi i }{5}}}\right)I=-2\pi i \lim_{z\to 2}\frac{1}{z^{\frac{2}{5}}(1-z)^{\frac{3}{5}}}\Rightarrow \left(1+e^{\frac{\pi i }{5}}\right)I=i\left[\pi 2^{\frac{3}{5}}e^{\frac{3\pi i }{5}}\right]$

and finally by equating the imaginary parts of both sides of the last equation we get

$\displaystyle \displaystyle \sin\left(\frac{\pi}{5}\right)I=\pi 2^{\frac{3}{5}}\cos\left(\frac{3}{5}\pi\right)\Rig htarrow I=2^{\frac{3}{5}}\pi\frac{\cos\left(\frac{3\pi}{5} \right)}{\sin\left(\frac{\pi}{5}\right)}$.

.

Last edited by skipjack; February 18th, 2018 at 02:34 AM.
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August 27th, 2012, 10:53 PM   #7
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Re: Integral Computation.

Quote:
Originally Posted by ZardoZ
[color=#000000]I know the answer how can we compute it? I know two ways. [/color]
Then, why asking for help ?
Anyways, it is a Gauss integral. Knowing the basic properties of the Gauss hypergeometric functions, the result is very easy to obtain directly.
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August 27th, 2012, 10:56 PM   #8
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Re: Integral Computation.

[color=#0000FF]ZardoZ[/color] posts questions like this (that he feels is interesting or instructive to work out) as a challenge to others.
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August 28th, 2012, 04:30 AM   #9
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Re: Integral Computation.

Nice Z man . I like your CA method.
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August 28th, 2012, 04:33 AM   #10
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Re: Integral Computation.

Quote:
Originally Posted by JJacquelin
Quote:
Originally Posted by ZardoZ
[color=#000000]I know the answer how can we compute it? I know two ways. [/color]
Then, why asking for help ?
Anyways, it is a Gauss integral. Knowing the bassic properties of the Gauss hypergeometric functions, the result is very easy to obtain directly.
Because it's fun for us nerds who enjoy challenging problems. Zardoz certainly needs no instruction. He posts these for others that may enjoy a challenging integral, series, etc. Also, to see various methods others may come up with.
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