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October 6th, 2018, 06:18 PM   #11
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You slipped up when doing the initial substitution.

Also, you forgot to include a constant of integration.
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October 6th, 2018, 08:31 PM   #12
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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}
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October 6th, 2018, 09:05 PM   #13
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Substituting $x = \cosh u$ is simpler for that (or for integrating $x^2 /\sqrt{x^2 - 1}$) if you know about the hyperbolic functions.
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October 21st, 2019, 03:27 PM   #14
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Quote:
Originally Posted by SDK View Post
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?
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October 24th, 2019, 09:19 PM   #15
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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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