I've set out to work out the following gamma-type integral where the exponential function has a complex factor multiplying the argument:
$$
\int _{0}^{\infty} x^{z-1}e^{-\alpha x} \, dx, \quad \alpha \in \mathbb{C}.
$$
This particular integral has its value reported in many places to be
$$
\frac{\Gamma(z)}{\alpha ^{z}}
$$
and what I've been attempting to do is simply reproduce this result. However, I'm running into a bit of a problem trying to work out the path integral once the substitution $u = \alpha x$ is made, since
$$
\int _{0}^{\infty}x^{z-1}e^{-\alpha x} \, dx = \frac{1}{\alpha ^z} \lim _{t \rightarrow \infty} \int _{0}^{\alpha t} u ^{z-1}e^{-u} \, du.
$$
This last integral is, as far as I understand, a path integral along a line that crosses the origin in the complex plane, and it shouldn't be simply $\Gamma (z)$, right? I mean, I may be getting this wrong, but the definition of $\Gamma (z)$ involves an integration over the real line, and not some slanted line in the complex plane.
I attempted to separate the integral in two by taking first the integration along the real line and then another one along the imaginary line, but it didn't get me anywhere. Am I missing something very obvious here? I initially thought this would be a rather straightforward result. Any and all help is appreciated! If you see any mistakes in my calculations please feel free to correct them!
Best Answer
By the Cauchy integral theorem, an integral of $f(u) = u^{z-1} e^{-u}$ on a path from $0$, on to $\alpha t$, on an arc to $|\alpha|t$ on the real line, and finally back to $0$, is zero, because the integrand is holomorphic over the whole region. Therefore $$ \int_{0}^{\alpha t} f(u) \, du = - \int_{\textrm{arc}} f(u) \, du - \int_{|\alpha| t}^0 f(u) \, du $$ The next step is to argue using the estimation lemma that in the limit $t \to \infty$, the integral along the arc vanishes -- however, this may not be true for all values of $\alpha$. After that, the result follows immediately.
To apply the estimation lemma, note that for any point $u = R e^{i \theta}$, where $R = |\alpha| t$, $$ \left| u^{z-1} e^{-u} \right| = R^{z-1} e^{-R\cos(\theta)}, $$ and then show that $$ \lim_{R \to \infty} \max_{u \textrm{ on arc}} \left| f(u) \right| = 0. $$