I need Help to evaluate : $$I=\int^1_0\frac{1}{a^2+\ln^2(x)}\cdot\frac{dx}{1+x^2}$$ Let:$x=e^{-t}$ $$I=\int^{\infty}_0\frac{e^{-t}}{(a^2+t^2)(1+e^{-2t})}dt=\int^{\infty}_0\sum_{n=0}^{\infty}(-1)^n\frac{e^{-t(2n+1)}}{a^2+t^2}dt$$
let:$x=t(2n+1)$ $$I=\int^{\infty}_0\sum_{n=0}^{\infty}(-1)^n\frac{e^{-x}(2n+1)}{x^2+(2na+a)^2}dx$$
How can I evaluate this integration?
$$I=\int^1_0\frac{1}{a^2+\ln^2(x)}\cdot\frac{dx}{1+x^2}\overset{x=e^{-t}}{=}\frac{1}{2}\int^{\infty}_0\frac{dt}{(a^2+t^2)\cosh(t)}\overset{t=\pi x}{=}\frac{1}{2\pi}\int^{\infty}_0\frac{dx}{(\left({\frac{a}{\pi}}\right)^2+x^2)\cosh(\pi x)}$$
we have : $$\int^{\infty}_0\frac{1}{(a^2+x^2)\cosh(\pi x)}dx=\frac{1}{2a}\left({\psi\left({\frac{3}{4}+\frac{a}{2}}\right)-\psi\left({\frac{1}{4}+\frac{a}{2}}\right)}\right)$$
Therfore: $$I=\frac{1}{4a\pi}\left({\psi\left({\frac{3}{4}+\frac{a}{2\pi}}\right)-\psi\left({\frac{1}{4}+\frac{a}{2\pi}}\right)}\right)$$