How to prove that $ \int_0^{\infty } \frac{\psi ^{(0)}(z+1)+\gamma }{z^{15/8}} \, dz = \frac{2 \pi}{\sqrt{2-\sqrt{2}}} \zeta(\frac{15}{8}) $?

Question

This integral is from an exercise in an old calc book by Edwards from the 20s.

The $\psi$ is the Digamma function and the $\zeta$ is the Riemann Zeta function. How would one go about trying to prove this? Does this require just basic substitution? Thank you for your time.

Mathematica shows that the values are equivalent and is about $14.611879190740$

Answer 1

Being careful with the branch of $z^{15/8}$ look at $$(e^{2i\pi 15/8}-1) \int_0^{\infty } \frac{\psi(x+1)+\gamma }{x^{15/8}}dx=\int_C \frac{\psi(z+1)+\gamma }{z^{15/8}}dz$$ where $C$ is a contour $+\infty\to -\epsilon\to +\infty$ enclosing $[0,\infty)$, then apply the residue theorem by adding an infinite circle to the contour so that it encloses the region $\Bbb{C}-[0,\infty)$.

The $\sqrt{2-\sqrt2}$ comes from $\sin(\pi 15/8)$.