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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} $ ??

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up vote 3 down vote accepted

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

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thanks a lot for your replies :) thank you very much in fact i needed this result and the result involving the logarithmic derivative for the zeta function to use it in one idea i had anyway how can i citate a reference of this forum ?? thanks. – Jose Garcia Oct 10 '11 at 10:34
i am trying to use Riemann-Weyl explicit formula to get that the sum over ALL the imaginary part of the zeros satisfy $ \sum_{t} \frac{4}{1+4t^{2}} =2+ \gamma -log(4\pi) $ however a missing term '2' can not be obtained – Jose Garcia Oct 10 '11 at 13:51
@Jose: As a reference, you should provide the permalink to this page <>;. That URL should persist as long as is around. – robjohn Oct 10 '11 at 14:37
has a question if we replace inside the integral the factor $ x^{2}+1/4 $ by a general $ x^{2}+a^{2} $ is then the integral equal to $ \Psi (1/4+a/2) $ so in the case a=1/2 the integral reduces to the previous one could someone numerically tested for 0<a<1 ?? – Jose Garcia Oct 13 '11 at 14:25

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
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