Is there a closed form for $\int_0^\infty \frac{1-\cos(tx)}{e^t-1}dt$? When $|x|<1$
$$f(x)=\int_0^\infty \frac{1-\cos(tx)}{e^t-1}dt = \sum_{n=1}^\infty \zeta(2n+1)x^{2n}$$
This is also the imaginary part of the analytic continuation of the harmonic series.
Numerically the following seem to be true:
$$\int_0^{\pi/x}\frac{1-\cos(tx)}{e^t-1}dt\to \zeta(2)$$
$$\int_{\pi/x}^{3\pi/x}\frac{1-\cos(tx)}{e^t-1}dt\to \zeta(3)$$
Does the pattern continue? And is there a closed form for $f$ ?
 A: For $x>0$,
$$ \begin{align}\int_{0}^{\infty} \frac{1- \cos(tx)}{e^{t}-1} \, dt &= \int_{0}^{\infty} \left(1- \cos (tx) \right)\frac{e^{-t}}{1-e^{-t}} \, dt \\ &= \int_{0}^{\infty} \left(1- \cos (tx) \right) \sum_{n=1}^{\infty} e^{-tn} \, dt \\ &= \sum_{n=1}^{\infty} \int_{0}^{\infty} \left(1- \cos (tx) \right) e^{-tn} \, dt \\ &= \sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{n}{n^{2}+x^{2}} \right) \\ &= \sum_{n=1}^{\infty} \frac{x^{2}}{n(n^{2}+x^{2})}. \end{align}$$
(Switching the order of summation and integration is justified by Tonelli's theorem.)
Using the series representation of the digamma function, we see that
$$\begin{align} \psi(1+ix) + \psi(1-ix) &= - 2 \gamma + \sum_{n=1}^{\infty} \left(\frac{ix}{n(n+ix)} + \frac{-ix}{n(n-ix)}\right) \\ &= -2 \gamma +2 \sum_{n=1}^{\infty} \frac{x^{2}}{n(n^{2}+x^{2})}. \end{align}$$
Therefore, $$\begin{align}\int_{0}^{\infty} \frac{1-\cos(tx)}{e^{t}-1} \, dt &= \frac{1}{2} \left(\psi(1+ix)+\psi(1-ix) \right)+ \gamma \\ &= \Re \left(\psi(1+ix) \right) + \gamma \tag{1} \\ &= \Re \left(\psi(ix) + \frac{1}{ix}\right) + \gamma \tag{2} \\ &=  \Re \left(\psi(ix)\right) + \gamma.\end{align}$$

$(1)$ https://en.wikipedia.org/wiki/Schwarz_reflection_principle
$(2)$ https://en.wikipedia.org/wiki/Digamma_function#Reflection_formula
