Prove that $\int_0^1 \psi{(x) \sin(2 n \pi x)} \space\mathrm{dx}=-\frac{\pi}{2}$ Prove that
$$\int_0^1 \psi{(x) \sin(2 n \pi x)} \space\mathrm{dx}=-\frac{\pi}{2}, \space n\ge1$$
where $\psi(x)$ - digamma function
 A: Here's another approach using the integral representation for $\psi$.
We assume $n$ is an integer greater than or equal to one. 
Then 
$$\begin{eqnarray*}
\int_0^1 dx\, \sin(2n\pi x)  \psi(x)
&=& \int_0^1 dx\, \sin(2n\pi x) 
\int_0^\infty dt\, 
\left(
\frac{e^{-t}}{t} - \frac{e^{-x t}}{1-e^{-t}}
\right) \\
&=& \int_0^\infty dt\,
\left(
\frac{e^{-t}}{t} \int_0^1 dx\, \sin(2n\pi x)
- \frac{1}{1-e^{-t}} \int_0^1 dx\, \sin(2n\pi x)e^{-x t}
\right).
\end{eqnarray*}$$
But $\int_0^1 dx\, \sin(2n\pi x) = 0$ and 
$$\int_0^1 dx\, \sin(2n\pi x)e^{-x t} = \frac{2n\pi}{t^2+4n^2\pi^2}(1-e^{-t}).$$
(Details for the second integral can be given if necessary.)
Therefore 
$$\begin{eqnarray*}
\int_0^1 dx\, \sin(2n\pi x)  \psi(x)
&=& -\int_0^\infty dt\, \frac{2n\pi}{t^2+4n^2\pi^2} \\
&=& -\frac{\pi}{2}.
\end{eqnarray*}$$
A: By the log-differentiation of the Euler's reflection formula, we have
$$ \psi_0(x) - \psi_0(1-x) = -\pi \cot (\pi x). $$
Thus we have
\begin{align*}
\int_{0}^{1}\psi_0(x) \sin (2\pi n x) \, dx
&= \frac{1}{2}\int_{0}^{1}\psi_0(x) \sin (2\pi n x) \, dx - \frac{1}{2}\int_{0}^{1}\psi_0(1-x) \sin (2\pi n x) \, dx \\
&= -\frac{\pi}{2} \int_{0}^{1} \frac{\sin (2\pi n x)}{\sin (\pi x)} \, \cos (\pi x) \, dx \\
&= -\frac{1}{2} \int_{0}^{\pi} \frac{\sin (2 n \theta)}{\sin \theta} \, \cos \theta \, d\theta \\
&= - \int_{0}^{\frac{\pi}{2}} \frac{\sin (2 n \theta)}{\sin \theta} \, \cos \theta \, d\theta.
\end{align*}
Now the rest follows by my blog posting.
