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

Question

Good day everybody.

I would like to compute

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

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:

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

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

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

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 $x \to 1$:

WA1

WA2

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).

Answer 1

• $\sin(2\pi x)$ 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