Digamma function integral Does anyone how to get a finite value to this integral ?
$ \int\nolimits_{0}^{\infty} dx \frac{ \Psi (1/4+ix/2) +\Psi (1/4-ix/2)}{x^{2}+1/4} $
i have tried residue theorem but i got nonsenses :( can anyone help 
or give some advice 
i believe that the integral should be equal to the $ \Psi (3/2) $ or similar although i can not prove it :(
anyway, is this integral equal to · $ \int\nolimits_{-\infty}^{\infty} dx \frac{ \Psi (1/4+ix/2)}{x^{2}+1/4} $ ??
 A: Use $\sum_{n=1}^\infty \left( \frac{1}{n+a-1} - \frac{1}{n}\right) = -\gamma - \Psi(a)$, where $\Psi(a)$ is di-gamma function. Using this representation for the di-gamma function and integrating term-wise:
$$ \begin{multline}
 \mathcal{I} = \int_0^\infty \frac{ \Psi(\frac{1}{4} + i \frac{x}{2}) + \Psi(\frac{1}{4} - i \frac{x}{2}) }{x^2 + 1/4} \, \mathrm{d} x = \\ 
  \int_0^\infty \frac{ -2 \gamma }{x^2 + 1/4} \, \mathrm{d} x  - 
  \sum_{n=1}^\infty \int_0^\infty \frac{1}{x^2 + 1/4} \left( \frac{8 (4 n-3)}{(3-4 n)^2+4 x^2}-\frac{2}{n} \right) \, \mathrm{d} x
 \end{multline}
$$
Thus
$$
  \mathcal{I} = -2 \pi \gamma - \sum_{n=1}^\infty \frac{2 \pi }{n (2 n-1)} = -2 \pi \left(  \gamma + 2 \log 2 \right) = 2\pi \left( \Psi\left(\frac{3}{2}\right) - 2\right)
$$
Verify numerically:
In[55]:= NIntegrate[(
  PolyGamma[1/4 + I x/2] + PolyGamma[1/4 - I x/2])/(
  x^2 + 1/4), {x, 0, \[Infinity]}, WorkingPrecision -> 50] == 
 2 Pi (PolyGamma[3/2] - 2)

Out[55]= True

A: Sasha's answer is very good for the first part.  For the second part, a simple change of variables $x\to-x$ yields
$$
\begin{align}
&\int_{0}^{\infty} \frac{ \Psi (1/4+ix/2) +\Psi (1/4-ix/2)}{x^{2}+1/4}\mathrm{d}x\\
&=\int_{0}^{\infty} \frac{ \Psi (1/4+ix/2)}{x^{2}+1/4}\mathrm{d}x+\int_{0}^{\infty} \frac{\Psi (1/4-ix/2)}{x^{2}+1/4}\mathrm{d}x\\
&=\int_{0}^{\infty} \frac{ \Psi (1/4+ix/2)}{x^{2}+1/4}\mathrm{d}x+\int_{-\infty}^{0} \frac{\Psi (1/4+ix/2)}{x^{2}+1/4}\mathrm{d}x\\
&=\int_{-\infty}^{\infty} \frac{ \Psi (1/4+ix/2)}{x^{2}+1/4}\mathrm{d}x
\end{align}
$$
