Calculating Residues

Question

I want to calculate this integral $$I:=\int dk^{0}\frac{e^{-ik^{0}(x^{0}-x'^{0})}}{\left(\left(k^{0}\right)^{2}-|\vec{k}|^{2}\right)} $$ for that I recall the Residue Theorem: $$I=2\pi i \left\{ \mathrm{Res}\left[\frac{e^{-ik^{0}(x^{0}-x'^{0})}}{\left(\left(k^{0}\right)^{2}-|\vec{k}|^{2}\right)},k^{0}=|\vec{k}|\right]+\mathrm{Res}\left[\frac{e^{-ik^{0}(x^{0}-x'^{0})}}{\left(\left(k^{0}\right)^{2}-|\vec{k}|^{2}\right)},k^{0}=-|\vec{k}|\right]\right\} $$ My question is, how do I calculate those residues?

Answer 1

Let us try first to write that integral in a little less horrible , more usual, way:

$$I:=\int\,dz\frac{e^{-iz(x_0-x'_0)}}{(z^2-t^2)}=\int f(z)\,dz$$

Now:

$$Res_{z=\pm t}=\lim_{z\to \pm t}(z\pm t)f(z)=\frac{e^{\mp it(x_0-x'_0)}}{\pm 2t}$$

so the integral equals

$$I=\frac{2\pi i}{2t}\left(e^{-it(x_0-x'_0)}-e^{it(x_0-x_0')}\right)=\frac{2\pi}{t}\sin\left(t(x_0-x_0')\right)$$

and this agrees with your result, so I think your book's wrong .

The above is, of course, assuming the integral is around a path enclosing both poles of $\,f(z)\,$ .