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Do you have any idea how to solve this integral? $$\int\limits_0^\infty {\frac{{\cos \left( mx \right)}}{{x + {x_0}}}\left( {1 + n\left( x \right)} \right)} - \int\limits_0^\infty {\frac{{\cos \left( mx \right)}}{{x - {x_0}}}n\left( x \right)} $$ where $n$ is the Bose-Einstein distribution: $$n\left( x \right) = \frac{1}{{{e^{bx}} - 1}}$$ and $x_0>0,m>0,b>0$.

--Notes:

Note that $n(-x)=-n(x)-1$. $n(x)$ exhibits poles in the upper half plane for purely imaginary frequencies $i{x_j}$, where ${x_j} = \frac{{2\pi }}{b}j$, $j=0,1,2,...$ are the Matsubara frequencies. In these poles: $$n\left( x \right) \simeq \frac{1}{b}\frac{1}{{x - i{x_j}}}$$

--Other notes:

If instead of $\cos(x)$, the $\sin(x)$ function appears, the integral is trivial because we can write $\sin mx = ({e^{imx}} - {e^{ - imx}})/2i$, and make the substitution $x \to -x$ in the term ${e^{ - imx}}$. Using the property $n(-x)=-n(x)-1$ the integral can be extended from $- \infty$ to $\infty$: $$\frac{1}{{2i}}\int\limits_{ - \infty }^\infty {\frac{{{e^{imx}}}}{{x + {x_0}}}\left( {1 + n\left( x \right)} \right)} - \frac{1}{{2i}}\int\limits_{ - \infty }^\infty {\frac{{{e^{imx}}}}{{x - {x_0}}}n\left( x \right)} $$
We can take the residuum in the Matsubara frequencies, and $ \pm x_0$. For the cosine function this trick is not valid anymore, because we can not extend the integral to $[-\infty,\infty]$. Any ideas???

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$\int\limits_0^\infty {\frac{{\cos \left( mx \right)}}{{x + {x_0}}}\left( {1 + \frac{1}{{{e^{bx}} - 1}} } \right)} - \int\limits_0^\infty {\frac{{\cos \left( mx \right)}}{{x - {x_0}}} \frac{1}{{{e^{bx}} - 1}} } $ is not convergent :

For $x$ tending to $0$ $$ {\frac{{\cos \left( mx \right)}}{{x + {x_0}}}\left( {1 + \frac{1}{{{e^{bx}} - 1}} } \right)} - {\frac{{\cos \left( mx \right)}}{{x - {x_0}}} \frac{1}{{{e^{bx}} - 1}} }\sim \frac{2}{b x_0}\frac{1}{x} $$ The lower bound ($x=0$) of the integral generates a logarithmic term $-\frac{2}{b x_0}\ln(0)=\infty$

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  • $\begingroup$ Nice work as always. Are the two integrals not the same since we can relate $n(-x) =-(1+n(x))$? $\endgroup$
    – Chinny84
    Jun 1, 2015 at 7:04
  • $\begingroup$ The two integrals are not the same : after remplacing $1+n(x)$ by $-n(-x)$ in the first one and changing of variable $x$ to $-x$ the bounds become ($-\infty,0$) instead of ($0,\infty$). The singularity at $x=0$ should be removable if the sign of the second integral would be $+$. The sum of the two integrals would be convergent if :$\int\limits_0^\infty {\frac{{\cos \left( mx \right)}}{{x + {x_0}}}\left( {1 + \frac{1}{{{e^{bx}} - 1}} } \right)} +\int\limits_0^\infty {\frac{{\cos \left( mx \right)}}{{x - {x_0}}} \frac{1}{{{e^{bx}} - 1}} } $. Is there a typo in the Pablo's equation ? $\endgroup$
    – JJacquelin
    Jun 1, 2015 at 8:24
  • $\begingroup$ I am definitely not questioning the mathematics :)!! Furthermore to my previous comment, I have learnt quite a lot from your answers in the past..I just thought that you could replace $1+n(x)$ in his first integral and change from $x\to -x$? But to be honest I haven't tried. $\endgroup$
    – Chinny84
    Jun 1, 2015 at 9:16

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