# Integral $\int_0^1 \zeta^H(2,x+1)sin(2 \pi x)$

Good day everybody.

I would like to compute

$\int_0^1 \zeta^H(2,x+1)\sin(2\pi x)$,

where $\zeta^H$ is Hurwitz zeta function.

Here are my hints, which make me believe this integral may exist in terms of special functions:

One can write:

$\zeta^H(z,x+1)=\zeta^H(z,x)- x^{-z}$

For z=2 a converging integral is split into two diverging (bad news), but hopefully the divergences may cancel in the final summation:

$\int_0^1 \zeta^H(z,x+1)\sin(2\pi x) = \int_0^1 \zeta^H(z,x)\sin(2\pi x) - \int_0^1 x^{-z}\sin(2\pi x)$

General expression for the first one exists (http://129.81.170.14/~vhm/papers_html/hurwitz1.pdf):

$\int_0^1 \zeta^H(z,x)\sin(2\pi x) = \frac{\Gamma(1-z)}{(2\pi)^{1-z}}\cos(\frac{\pi z}{2})$

The second one I try with "wolfram mathematica online integator". I am not successful to find the integral with general a $z$, however I am able to find a primitive function and its limit (value) for $x \to 1$:

http://www.wolframalpha.com/input/?i=integrate+x%5E(-z)sin(2+pi+x)dx

http://www.wolframalpha.com/input/?i=limit+-2%5E(-2+%2B+z)+%CF%80%5E(-1+%2B+z)+x%5E(-z)+((-i+x)%5Ez+%CE%93(1+-+z,+-2+i+%CF%80+x)+%2B+(i+x)%5Ez+%CE%93(1+-+z,+2+i+%CF%80+x))+as+x-%3E1

But I am unable to evaluate the primitive function at zero, i.e. to find the limit of the primitive function $x \to 0$ with general $z$. Maybe a full version of wolfram mathematica could help (I do not have it).

Then, maybe, two integrals expressed with a general $z$ may allow for a finite limit when $z \to 2$.

Thank you.

PS: Let me recall that $\zeta^H(2,x)$ is identical to the trigamma function (maybe it helps).

• $\sin(2\pi x)$ is $1$ periodic so when everything converges $\int_0^1 \zeta^H(z,x)\sin(2\pi x) = \int_0^\infty x^{-z} \sin(2\pi x)dx$ which reduces to $\frac{\Gamma(1-z)}{(2\pi)^{1-z}}\cos(\frac{\pi z}{2})$ after a change of contour, see the Cauchy integral theorem
• and $\int_0^1 \zeta^H(z,x+1)\sin(2\pi x) = \int_1^\infty x^{-z} \sin(2\pi x)dx$ which reduces with the same method to something like $\displaystyle\frac{\Gamma(1-z,i/2\pi)e^{i\pi z/2}+\Gamma(1-z,-i/2\pi)e^{-i\pi z/2}}{2(2\pi)^{1-z}}$ where $\Gamma(z,a) = \int_a^\infty x^{z-1}e^{-x}dx$ is the incomplete gamma function
• finally $$\int_0^1 \zeta^H(2,x+1)\sin(2\pi x)dx =\int_1^\infty \frac{\sin(2\pi x)}{x^2}dx= \int_1^\infty \frac{2\pi \cos(2\pi x)}{x}dx$$ a cosine integral which doesn't seem to have a closed-form