My Math Forum Integral Computation!

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 September 4th, 2011, 12:55 PM #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 $\displaystyle \int_{0}^{+\infty}\frac{x^{a-1}}{x+1}\;dx$ for $\displaystyle 0<\Re \mathfrak{e}(a)<1$. Last edited by skipjack; February 18th, 2018 at 12:55 AM.
 September 8th, 2011, 04:21 AM #2 Newbie   Joined: Sep 2011 Posts: 7 Thanks: 0 Re: Integral Computation! I'm not entirely sure I'm right, but I would start with taking the numerator as u, and denominator as dv. After performing I.B.P you end up with x^(a-1)ln abs(x+1)- int(ln abs(x+1)*(a-1)x^(a-2)). Now take (a-1)x^(a-2) as dv and perform I.B.P again. You should then see that you again get the original unknown integral on RHS. By moving it over you can see that you have 0=0, so your unknown integral should equal [x^(a-1)ln abs(x+1)] from 0 to infinity. That integral diverges.
 September 8th, 2011, 04:36 AM #3 Newbie   Joined: Sep 2011 Posts: 7 Thanks: 0 Re: Integral Computation! Sorry, it converges.
 September 8th, 2011, 04:49 AM #4 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! The result you should come up with is $\displaystyle \pi \cdot\csc(\pi \cdot a )$. Learn how to use $\displaystyle \rm L^{a}T_{e}X$ in order for your posts to be more readable! Last edited by skipjack; February 18th, 2018 at 12:54 AM.

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