What is the closed form of $ \int_0^1 \frac{x^{a/2} }{(x+1)(x^a+1)} dx $

Question

What is the closed form of $$ \int_0^1 \frac{x^{a/2} }{(x+1)(x^a+1)}dx $$, where $0<a<1$

I need its closed form to solve another hard problem.

Answer 1

Via $x\mapsto 1/x$ we have $$ I(a) = \int_{0}^{1}\frac{dx}{(x+1)(x^{a/2}+x^{-a/2})} = \int_{1}^{+\infty}\frac{dx}{x(x+1)(x^{a/2}+x^{-a/2})}=\int_{0}^{+\infty}\frac{e^{z} dz}{2\cosh(z)\cosh(az)}$$ hence $$ I(a)=\frac{1}{2}\int_{0}^{+\infty}\frac{dz}{\cosh(az)}+\frac{1}{2}\int_{0}^{+\infty}\frac{\tanh(z)}{\cosh(az)}\,dz =\frac{\pi}{4a}+\frac{1}{2a}\int_{0}^{+\infty}\frac{\tanh(z/a)}{\cosh(z)}\,dz.$$ The Laplace transform of $\tanh(z/a)$ is a Dirac comb and the Laplace transform of $\frac{1}{\cosh(z)}$ is $\frac{1}{2}\left[\psi\left(\frac{3+s}{4}\right)-\psi\left(\frac{1+s}{4}\right)\right]$, so $I(a)$ is non-elementary, it depends on $\log\Gamma$ and the Hurwitz $\zeta$-function.

Special values are $$I\left(\tfrac{1}{2}\right)=\frac{\pi}{2}-\sqrt{2}\log(1+\sqrt{2}),$$ $$I\left(\tfrac{2}{3}\right)=\frac{3\pi}{8}-\frac{5}{4}\log(2),$$ $$I\left(\tfrac{1}{3}\right)=\frac{3\pi}{4}-\frac{1}{2}-\frac{2}{\sqrt{3}}\log(2+\sqrt{3}).$$