My Math Forum Challenging integral marathon!

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 October 6th, 2018, 06:18 PM #11 Global Moderator   Joined: Dec 2006 Posts: 21,119 Thanks: 2331 You slipped up when doing the initial substitution. Also, you forgot to include a constant of integration.
 October 6th, 2018, 08:31 PM #12 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,700 Thanks: 2682 Math Focus: Mainly analysis and algebra Apart from skipjack's comments, it looks good. But if you are going to use a substitution $x=\sec u$, why not do it on the integral $\displaystyle \int \sqrt{x^2-1} \,\mathrm dx$? Admittedly it leads to the mildly challenging $$\int \sec^3 u \, \mathrm du$$ I approached that one via \begin{align}\int \sec{(u)} \, \mathrm du &= \int \cos^2{(u)}\sec^3{(u)}\, \mathrm du \\ &= \int (1-\sin^2{(u)})\sec^3{(u)}\, \mathrm du \\ &= \int \sec^3{(u)}\,\mathrm du - \int \underbrace{\sin{(u)}}_{u} \underbrace{\sin{(u)} \cos^{-3}{(u)}}_{\mathrm dv}\, \mathrm du \end{align}
 October 6th, 2018, 09:05 PM #13 Global Moderator   Joined: Dec 2006 Posts: 21,119 Thanks: 2331 Substituting $x = \cosh u$ is simpler for that (or for integrating $x^2 /\sqrt{x^2 - 1}$) if you know about the hyperbolic functions.
October 21st, 2019, 03:27 PM   #14
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 Originally Posted by SDK Anyways, I remembered the following integral which kept me busy for a few hours (or more) in grad school. Enjoy. Compute $\int_0^{2\pi} \log \Gamma (\frac{x}{2 \pi}) \exp(\cos x) \sin(x + \sin x) \ dx$ where $\Gamma$ is the usual gamma function.
Anyone end up solving this?

October 24th, 2019, 09:19 PM   #15
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 Originally Posted by Joppy Anyone end up solving this?
I guess not. I might post the solution if I get some time. In any case I think a year is probably plenty long and we can safely skip it. You should feel free to continue the marathon by posting another integral.

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