I'm having trouble resolving an integral which at face value seems rather simple. Consider the function:
$$ F(x, y) = \frac{1}{(i x -\lambda_l)(i(x-y)-\lambda_m)(ix + \lambda_j^\ast)(i(x-y)+\lambda_k^\ast)} $$ where $\lambda\in\mathbf{C} $. I want to find the definite integral: $$ I = \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} F(x,y) ~dx~dy $$ If I just use Mathematica's symbolic integration, it spits out $I=0$. However, for $j=l$ and $k=m$ this is not true and one finds: $$ I(j=l, k=m)=-\frac{2\pi^2}{\mathrm{Re}[\lambda_l\lambda_m])} $$ I have two questions: 1) is this integral only nonzero for $j=l$ and $k=m$? 2) and if so, is this a representation of the Kronecker delta?
EDIT:
As it turns out, by identifying that this function is an integral over a convolution, one may find:
$$ I = \left(\int\limits_{-\infty}^\infty \frac{1}{(i x-\lambda_l)(ix + \lambda_j^\ast)}dx\right)\left(\int\limits_{-\infty}^\infty \frac{1}{(i y-\lambda_m)(iy + \lambda_j^\ast)}dy\right) $$ reducing the problem to one dimension.
From Maple I get this... $$ \int_{-\infty}^\infty \frac{dx}{(ix-a)(ix+b)} = \begin{cases} \frac{-2\pi}{a+b},\qquad &\text{Re }a>0,\text{Re }b > 0 \\ 0\qquad &\text{Re }a>0,\text{Re }b < 0 \\ 0\qquad &\text{Re }a<0,\text{Re }b > 0 \\ \frac{2\pi}{a+b},\qquad &\text{Re }a<0,\text{Re }b < 0 \end{cases} $$ Of course, if one or both of $a,b$ are pure imaginary then it diverges. When $a\ne-b$ a principal value may exist; perhaps in that case the double integral can converge anyway? I did not investigate this. You would need to specify a definition for conditional convergence of an improper double integral.