Improper integral; exponential divided by polynomial How can I evaluate $$\int_{-\infty}^\infty {\exp(ixk)\over -x^2+2ixa+a^2+b^2} dx,$$ where $k\in \mathbb R, a>0$? Would Fourier transforms simplify anything? I know very little about complex analysis, so I am guessing there is a rather simple way to evaluate this? Thanks.

Assume $b \neq 0$ and $k\neq 0$. 
Write
 $\dfrac{\exp(ixk)}{-x^2+2iax+a^2+b^2}= \dfrac{\exp(ixk)}{-(x-ia)^2+b^2}=\dfrac{\exp(i(x-ia)k)}{-(x-ia)^2+b^2} \exp(-ak)$ hence 
the integral becomes $I=\int_{-\infty}^\infty \dfrac{\exp(i(x-ia)k)}{-(x-ia)^2+b^2} \exp(-ak)dx=\int_{-\infty-ia}^{\infty -ia} \dfrac{\exp(izk)}{-z^2+b^2} \exp(-ak)dz$ on the contour the straight line parallel the $x$-axis and intercepting the $y$-axis (imaginary) at $-ia$.
 We need to close the contour by a great semicircle in the upper half plane if $k>0$ and in this case there are two poles at $z=b$ and $z=-b$ enclosed in the contour. 
Now we will use the residue theorem. The poles of the fraction $\dfrac{\exp(izk)}{-z^2+b^2}$ are $-b,+b$ and then the integral will be $I=(2\pi i)\exp(-ak)\lbrace Res(z=b)\exp(ibk)+Res(z=-b)\exp(-ibk)\rbrace$.
 Now observe that $Res(z=b)=1/2b$ and $Res(z=-b)=-1/2b$ hence $I=2 \pi i \exp(-ak) \frac{\sin bk}{b}$.
 If $k<0$ then we close the contour by a semicircle in the lower halfplane and in this case there are no poles enclosed and so the integral becomes zero.
Now assume $b=0$ and $K\neq 0$: in this case the residue (at $z=0$) becomes $-ik$ and the integral becomes $2\pi k$ (if $k>0$) and $0$ if $k<0$.
Finally let $k=0$: then in this case the result will be easy and I leave it for you as an exersice.
