# Integration involving complex singular function

I am stuck with the following integral that came up during my research and I am not sure how to correctly evaluate this expression.

$\int_{-\infty}^{\infty} dk\left[\Im[ \frac{1}{(\omega+i\eta+k)(\omega+i\eta+k+q)}] \right]$,

wher $\eta$ is an infinitesimal parameter. The integration is over the real numbers and $\Im$ is the imaginary part. This is an integral that comes from the convolution of two Greens functions, just to give the physical background.

From numerics and physical considerations I am expecting something proportional to $-\delta (q)$, but I can't for the life of me figure out how to correctly calculate the $q=0$ case. For the $q\neq 0$-case the residue theorem seems to yield the correct result (zero), but for $q=0$ it does not seem to work.

Generally, one has $\lim_{\eta \rightarrow 0} \frac{1}{\omega+i\eta} = \mathcal{P}(\frac{1}{\omega}) - i\pi \delta(\omega)$, but I am not quite sure how to do this correctly in the quadratic case at hand.

I am pretty sure I am not treating the limits correctly or ignoring the fact that these expression are only valid as distributions, but I am at a loss.

Has anybody dealt with something similar? I would really appreciate any help.

Using partial fraction decomposition $$\frac{1}{(\omega+i\eta+k)(\omega+i\eta+k+q)}=\frac 1q\left(\frac{1}{k+\omega+i \eta }-\frac{1}{k+\omega+q+i \eta }\right)$$ Multiplying by the conjugates and keeping the imaginary parts, we then end with $$I=\int \left[\Im[ \frac{1}{(\omega+i\eta+k)(\omega+i\eta+k+q)}] \right]\,dk$$ $$I=\int \frac{\eta }{q \left(\eta ^2+(k+q+\omega )^2\right)}\,dk-\int \frac{\eta }{q \left(\eta ^2+(k+\omega )^2\right)}\,dk$$ that is to say $$I=\frac 1q\left(\tan ^{-1}\left(\frac{k+q+\omega }{\eta }\right)-\tan ^{-1}\left(\frac{k+\omega }{\eta }\right) \right)$$ Combining the two terms, we the have $$I=\frac{1}{q}\tan ^{-1}\left(\frac{\eta \, q}{\eta ^2+(k+\omega ) (k+q+\omega )}\right)$$ from which I suppose that you can explore different situations.
• Thank you for your help. But you included a $k$ in your solution, which can't be there anymore after integration over $k$. Commented Dec 2, 2015 at 16:23
• @PythonSparse. How can $k$ disappear ? Commented Dec 2, 2015 at 18:45
• You integrate $k$ from $-\infty$ to $\infty$, so your integral comes out to $\pi/q$. That is what I meant with integrating over the real numbers. I guess I should have added the limits explicitly. Commented Dec 2, 2015 at 20:06
• Nonetheless, your solution works nicely in the $q\neq 0$ case. Unfortunately, for $q=0$, the partial fraction decomposition is not valid and thus one cannot use it. Commented Dec 2, 2015 at 20:29