Integral with digamma

Let $\psi(x)$ denote the digamma function. Evaluate the integral:

$$\int_0^1 \psi(x) \sin 2n \pi x \, {\rm d}x \quad , \qquad n \in \mathbb{N} $$

Solution

We are using the definition of the digamma function:

$$\psi(x)= -\gamma +\sum_{m=0}^{\infty}\left ( \frac{1}{m+1}- \frac{1}{m+x} \right )$$

and the integral is transformed successively:

$$\begin{aligned}
\int_{0}^{1}\psi (x)\sin 2n \pi x\,{\rm d}x &=\int_{0}^{1}\left [ -\gamma+ \sum_{m=0}^{\infty}\left ( \frac{1}{m+1}- \frac{1}{m+x} \right ) \right ]\sin 2n \pi x\, {\rm d}x \\
 &= - \sum_{m=0}^{\infty}\int_{0}^{1}\frac{\sin 2n \pi x}{m+x}\, {\rm d}x\\
 &= - \sum_{m=0}^{\infty}\int_{m}^{m+1}\frac{\sin 2n \pi \left ( x-m \right )}{x}\, {\rm d}x\\
 &= - \sum_{m=0}^{\infty}\int_{m}^{m+1}\frac{\sin 2n \pi x}{x}\, {\rm d}x\\
 &= - \int_{0}^{\infty}\frac{\sin 2 n \pi x}{x}\, {\rm d}x \\
 &= - \frac{\pi}{2}
\end{aligned}$$

It clearly holds that $\displaystyle \int_{0}^{1}c \sin 2n \pi x \, {\rm d}x =0 $.