Residue theorem for $ I=\int_{-\infty}^{+\infty}\frac{e^{\mathrm{i}\,t\,z}}{(z-z_1)(z-z_2)} \, \mathrm{d}z$

Question

If I use the residue theorem to evaluate the integral

$$ I(t)=\int_{-\infty}^{+\infty}\frac{e^{\mathrm{i}\,t\,z}}{(z-z_1)(z-z_2)} \, \mathrm{d}z$$

with $t>0$, $\mathrm{Im}(z_1)>0$ and $\mathrm{Im}(z_2)<0$, I would have thought to get

$$ I(t)=2\,\pi\,\mathrm{i}\,\frac{e^{\,\mathrm{i}\,t\,z_1}}{z_1-z_2}$$

since only the pol in the upper half plane contributes to the integral. If I solve the integral with Mathematica 12.0 it evaluates to

$$ I(t)=2\,\pi\,\mathrm{i}\,\frac{e^{\,\mathrm{i}\,t\,z_1}-e^{\,\mathrm{i}\,t\,z_2}}{z_1-z_2}$$

even though I set the correct assumptions on $z_1$ and $z_2$ and allowed for the calculatoin of the Cauchy principal value.

Now I am wondering if I misunderstood the residue theorem or Mathematica evaluates the integral incorrectly.

Answer 1

You are right. Note that if you compute$$\int_{-\infty}^\infty\frac{e^{itz}}{z^2+1}\,\mathrm dz,\tag1$$with $t>0$, you will get $\pi e^{-t}$. But $(1)$ is equal to$$\int_{-\infty}^\infty\frac{e^{itz}}{(z-z_1)(z-z_2)}\,\mathrm dz,$$with $z_1=i$ and $z_2=-i$. Your answer will then be$$2\pi i\frac{e^{-t}}{2i},$$which is correct. But that answer provided by Mathematica 12.0 will then be$$2\pi i\frac{e^{-t}-e^t}{2i},$$which is wrong.