My Math Forum Integral Computation.

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 August 27th, 2012, 06:24 AM #1 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus 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.
 August 27th, 2012, 09:17 AM #2 Senior Member   Joined: Aug 2011 Posts: 334 Thanks: 8 Re: Integral Computation. Read the result: integral.JPG Last edited by skipjack; February 18th, 2018 at 02:08 AM.
 August 27th, 2012, 09:19 AM #3 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus 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.
 August 27th, 2012, 10:33 AM #4 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 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.
 August 27th, 2012, 12:25 PM #5 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Integral Computation. Yes, nice! Last edited by skipjack; February 18th, 2018 at 02:20 AM.
 August 27th, 2012, 01:54 PM #6 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus 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.
August 27th, 2012, 10:53 PM   #7
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Joined: Aug 2011

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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.

 August 27th, 2012, 10:56 PM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 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.
 August 28th, 2012, 04:30 AM #9 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: Integral Computation. Nice Z man . I like your CA method.
August 28th, 2012, 04:33 AM   #10
Senior Member

Joined: May 2011

Posts: 501
Thanks: 6

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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