In a physics course, this integral showed up:
$\int_0^\infty dp\, p^2\, e^{ibp}\; , \quad b \in \mathbb{R}$
My prof proceeded with the seemingly insane substitution $x=-ib$ to yield $\Gamma(3)x^{-3} = 2 (-ib)^{-3}$ as if $x$ were real.
But immediately he justified it: we're actually doing a contour integral the shape of a counterclockwise quarter cake slice in the upper right quadrant: $\int_0^\infty + \int_{i \infty}^0 + \int_0^{\pi/2}d\phi$
Which, I checked, if the arc vanishes yields the same result after all. But the arc doesn't seem to vanish:
$\lim_{R \rightarrow \infty}\int_0^{\pi/2}d\phi \,i R e^{i \phi} R^2 e^{2 i \phi} e^{i b R \cos \phi} e^{-b R \sin \phi}$
because of the $\phi = 0$ part. Unless I'm missing something. Two questions:
1) what did I miss? I trust my prof but not with my life.
2) Even then, I don't understand how the crazy substitution is justified (other than proving it a posteriori). He said it's equivalent to doing $-i\Gamma$.
Thanks.
Okay I guess I did miss the obvious. For instance, I have to distinguish between $b > 0$ and $b < 0$ (with $b = 0$ it's hopeless, but fortunately I don't need it): but this just means doing the vertically flipped version of the above contour and the result is identical. Then, I noticed that the arc vanishes after all (with $-\pi/2$ for the flipped version):
$\lim_{a \rightarrow 0^+}\lim_{R \rightarrow \infty}\int_a^{\pi/2}d\phi \,i R e^{i \phi} R^2 e^{2 i \phi} e^{i b R \cos \phi} e^{-b R \sin \phi} = 0$
so unless switching the order of limits is illegal for some reason, point 1) should be settled.
On point 2), revisiting it, it seems not so simple as I made it to be:
$\int_0^\infty dp\, p^2\, e^{ibp} = \int_0^\infty dp\, p^2 e^{-xp} = x^{-3}\int_0^{x\infty}dt\, t^2 e^{-t}$
which can't immediately be a Gamma. Again, distinguishing between $x$ making the integral go to positive or negative imaginary infinity, it is just the counterpart of the above, solving for the imaginary branch of the contour. More than a naive substitution, it seems my prof was just skipping passages or providing a mnemonic. But he is right that the result is a Gamma times an imaginary term.