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Inspired in the shape of useful integrals to compute $\pi$ (see *), I've consider for each integer $k\geq 1$ $$\int_0^{\frac{1}{2}}\frac{x^{k-1}}{1-x^{2^k}}dx=\int_0^{\frac{1}{2}}x^{k-1}\sum_{n=0}^\infty \left( x^{2^k} \right)^n dx=\frac{1}{2^k}\sum_{n=0}^\infty \frac{1}{2^{n2^k}(n2^k+k)},$$ and as you see in this series doesn't appear any Pochhmammer symbol. Let the change of variable $u=x^{2^k}$ then $\log x=2^{-k}\log u$ and $\frac{dx}{x}=2^{-k}\frac{du}{u}$, thus we can write $$\frac{x^{k-1}}{1-x^{2^k}}dx=\frac{x^{k-1}}{1-u}2^{-k}\frac{du}{u}.$$ Define $\text{factor}:=x^k$ then $\log(\text{factor})=k2^{-k}\log u$ (by substitution of $\log x=2^{-k}\log u$ ) thus by exponentiation $\text{factor}:=u^{\frac{k}{2^k}}$ and we can write our integral as $$\int_0^{\frac{1}{2}}\frac{x^{k-1}}{1-x^{2^k}}dx=\frac{1}{2^k}\int_0^{\frac{1}{2^{2^k}}}\frac{u^{-1+\frac{k}{2^k}}}{1-u}du$$ for $k\geq 1$, since if $x=\frac{1}{2}$ the corresponding upper limit in previous integral is computed as was written.

I would like to know and learn more computations involving special functions and numbers.

Question. It's possible to prove that previous integral, which can be written in terms of particular values of the incomplete Beta function or the hypergeometric function, is equals to previous computed series when you simplify the corresponding Pochhmmmer symbols that appear in the expansion series of $B(z;a,b)$ or $_2F_1(a,b;c,z)$? Thanks in advance.

I say only an exercise: show your simplifications with the Beta or the hypergeometric, one of them. I will accept the best answer.

*The reference is a divulgative paper in spanish, the referenced equation is the first integral after (29) in page 171, I don't know who proved it, neither if my integral was known in the literature; Guillera, Historia de las fórmulas y algoritmos para $\pi$, La Gaceta de la RSME Vol. 10 No. 1 (2007).

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  • $\begingroup$ My code for the Wolfram Alpha online calculator is integrate (x^(k-1))/(1-x^(2^k)) dx $\endgroup$ – user243301 Jun 1 '16 at 11:04
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    $\begingroup$ In the second integral you forgot the exponents $n$, it should be $$\int_0^{\frac{1}{2}}x^{k-1}\sum_{n=0}^\infty \left( x^{2^k} \right)^{n} dx$$ $\endgroup$ – gammatester Jun 1 '16 at 11:54
  • $\begingroup$ Was fixed previous typo. Very thanks much @gammatester , and all users. $\endgroup$ – user243301 Jun 1 '16 at 12:15

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