August 27th, 2012, 07:24 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 $\displaystyle \int_{0}^{1}\frac{1}{(x2)\sqrt[5]{x^2(1x)^3}}\;dx$. Last edited by skipjack; February 18th, 2018 at 03:08 AM. 
August 27th, 2012, 10:17 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 03:08 AM. 
August 27th, 2012, 10:19 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 03:09 AM. 
August 27th, 2012, 11:33 AM  #4 
Senior Member Joined: May 2011 Posts: 501 Thanks: 6  Re: Integral Computation.
Hey Z: Fun integral. Challenging. When I see a (1x) written to some power, I usually think Beta is in there somewhere. Write as $\displaystyle \frac{1}{2}\int_{0}^{1}\frac{1}{(1\frac{x}{2})\sqrt[5]{x^{2}(1x)^{3}}}dx$ by dividing by 2. Now, write the 1x/2 part as a geometric series: $\displaystyle \frac{1}{2}\int_{0}^{1}\frac{1}{\sqrt[5]{x^{2}(1x)^{3}}}\cdot \sum_{k=0}^{\infty}(x/2)^{k}$ $\displaystyle =\frac{1}{2}\sum_{k=0}^{\infty}\frac{1}{2^{k}}\int_{0}^{1 }x^{k2/5}(1x)^{3/5}dx$ 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 $\displaystyle \Gamma(p)\Gamma(1p)=\pi \csc(\pi p)$: $\displaystyle =2^{2/5}\pi \csc(2\pi /5)=\boxed{\frac{2\pi 2^{1/10}}{\sqrt{5+\sqrt{5}}}\approx 2.503407...}$ Which can be rewritten in various forms. Last edited by skipjack; February 18th, 2018 at 03:17 AM. 
August 27th, 2012, 01:25 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 03:20 AM. 
August 27th, 2012, 02:54 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 $\displaystyle \displaystyle{f(z)=\frac{1}{(x2)\sqrt[5]{z^2(1z)^{3}}}}$, with $\displaystyle \displaystyle{\pi\leq \arg\left(\log(z)\right)<\pi}$ and $\displaystyle \displaystyle{0\leq \arg\left(\log(1z)\right)<2\pi}$ $\displaystyle \displaystyle\oint_{C}f(z)\;\textrm{d}z=2\pi i \textrm{Res}(f(z);z=2,\infty)$ letting the radii of the two small circles go to zero $\displaystyle \displaystyle\oint_{C}f(z)\;\textrm{d}z= \int_{\textrm{L}_1}f(x)\;\textrm{d}x+\int_{\textrm {L}_2}f(x)\;\textrm{d}x+\int_{C_L}f(z)\; \textrm{d}z+\int_{C_R}f(z)\;\textrm{d}z= \left(1+\frac{1}{e^{\frac{\pi i }{5}}}\right)\underbrace{\int_{0}^{1}\frac{1}{(x2)\sqrt[5]{x^2(1x)^3}}\;\textrm{d}x}_{I}$ and so $\displaystyle \displaystyle\left(1+\frac{1}{e^{\frac{\pi i }{5}}}\right)I=2\pi i \lim_{z\to 2}\frac{1}{z^{\frac{2}{5}}(1z)^{\frac{3}{5}}}\Rightarrow \left(1+e^{\frac{\pi i }{5}}\right)I=i\left[\pi 2^{\frac{3}{5}}e^{\frac{3\pi i }{5}}\right]$ 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 03:34 AM. 
August 27th, 2012, 11:53 PM  #7  
Senior Member Joined: Aug 2011 Posts: 334 Thanks: 8  Re: Integral Computation. Quote:
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, 11:56 PM  #8 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 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, 05:30 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, 05:33 AM  #10  
Senior Member Joined: May 2011 Posts: 501 Thanks: 6  Re: Integral Computation. Quote:
 

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