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When I was messing around with some integrals I got the result that

$$\sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \int_0^{\frac{\pi}{2}} (1+\csc(x))^{k+\frac{1}{2}} \, \,dx = \frac{\pi^2}{6}$$

Now I would like to solve the integral, but I don't really have any idea (I wonder whether it's possible at all?). I tried to change $1+\csc(x)$ in terms of sines and cosines (using this) so that one gets

$$\int_0^{\frac{\pi}{2}} (1+\csc(x))^{k+\frac{1}{2}} \, \,dx = \int_0^{\frac{\pi}{2}} \left(\frac{\cos^2(x)}{\sin(x)(1-\sin(x))}\right)^{k+\frac{1}{2}} \, \,dx$$

...but I can't see how this would lead anywhere.

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  • $\begingroup$ Just out of curiosity : how did you get the nice $\frac {\pi^2}6$ ? $\endgroup$ – Claude Leibovici Aug 4 '20 at 7:38
  • $\begingroup$ @ClaudeLeibovici I like messing around with integrals that have a known value ("Integral milking"), and so I used this result and used the taylor series for $\cot^{-1}(x)$. I might have done some error tho... $\endgroup$ – Casimir Rönnlöf Aug 4 '20 at 8:07
  • $\begingroup$ $\large\int$ diverges for $\large k = 1,2,3,\ldots$. $\endgroup$ – Felix Marin Aug 5 '20 at 4:15
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I think that there is a problem around $x=0$ except for $k=0$. Using Taylor series and bionmial expansion $$(1+\csc(x))^{k+\frac{1}{2}}=\frac 1 {x^k}\left(\frac{1}{\sqrt{x}}+\left(k+\frac{1}{2}\right) \sqrt{x}+O\left(x^{3/2}\right)\right) $$

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Since $\sum_{k\ge0}\frac{(-1)^kz^{2k+1}}{2k+1}=\arctan z$, your series is $\int_0^{\pi/2}\arctan\sqrt{1+\csc x}dx$. This is $\infty$ as @ClaudeLeibovici noted, because for small $x$ the integrand is $\sim x^{-1/2}$. It's more interesting if we delete the $k=0$ case, giving $\int_0^{\pi/2}(\arctan\sqrt{1+\csc x}-\sqrt{1+\csc x})dx$, which WA can only numerically evaluate.

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  • $\begingroup$ Your WA link provides a numerical computation but no closed form. However, your WA link is (apparently) to a definite rather than indefinite integral. How do you know, based on the results given by your WA link, that there is no closed form answer to the corresponding indefinite integral? $\endgroup$ – user2661923 Aug 4 '20 at 8:22
  • $\begingroup$ @user2661923 Good question. $\endgroup$ – J.G. Aug 4 '20 at 9:31

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