Well, this is a physics class problem, and I did not learn anything about the contour integral. But I wish to show that:
Use the contour integration to show that the transformation of $$f(x)={{1}\over{x^2+a^2}}$$ is $$f(k)={\pi\over a}e^{-|k|a}.$$
Could anyone give me any hint about this?
I can only get: $f(k)=\int_{-\infty}^{+\infty}e^{-ikx}{{1}\over{x^2+a^2}}dx$,
but I have no idea how to move on to integrate this. Any help, please.
This is done using residue theorem and Jordan's lemma. For example, if $k<0$, you can close the contour of integration by a semicircle of infinite radius in the upper half plane, and then the integral becomes $$ \int_{-\infty}^{\infty}\frac{e^{-ikx}dx}{x^2+a^2}=2\pi i \sum_k\mathrm{res}_{z=z_k}\,\frac{e^{-ikz}}{z^2+a^2},$$ where the sum is taken over the singularities of $f(z)=\frac{e^{-ikz}}{z^2+a^2}$ in the upper half plane. In your case there is only one such point - a simple pole $z=ia$ (we assume $a>0$), and the residue is simply $$\mathrm{res}_{z=ia}f(z)=\lim_{z\rightarrow ia}(z-ia)f(z)=\frac{e^{ka}}{2ia},$$ so that $$ \int_{-\infty}^{\infty}\frac{e^{-ikx}dx}{x^2+a^2}=2\pi i \frac{e^{ka}}{2ia}=\frac{\pi e^{ka}}{a}.$$ When $k>0$, Jordan lemma allows you to similarly close the contour in the lower half plane and the rest of the calculation is analogous (except that now you will have minus sum of the residues due to the change of contour orientation).
