I came across the following "proof" of the Eulerian integrals
$$ \int_0^{\infty} t^{s-1} \cos(bt) e^{-at} \, dt = \frac{\Gamma(s) \cos \left(s\arctan \left(\frac{b}{a} \right) \right)}{(a^2+b^2)^{s/2}}, \int_0^{\infty} t^{s-1} \sin(bt) e^{-at} \, dt = \frac{\Gamma(s) \sin \left(s\arctan \left(\frac{b}{a} \right) \right)}{(a^2+b^2)^{s/2}} $$
where $s>0, b \in \mathbb{R}, a>0 $ and I'm wondering if it is rigorous and if not, what details need to be filled in to make it rigorous.
Begin with the gamma integral
$$ \Gamma(s) = \int_0^{\infty} x^{s-1} e^{-x} \, dx $$
and make the substitution $x=zt$ to get
$$ \frac{\Gamma(s)}{z^s} = \int_0^{\infty} t^{s-1} e^{-zt} \, dt. $$
This is fine as long as we assume $z$ is real and positive. But now the "proof" asserts that the formula also holds for complex $z$ with positive real part* (this is the step I'm unsure about) and substitutes $z=a-ib$ for $a>0$ and $b \in \mathbb{R}$ and simplifies the result with the aid of polar coordinates
$$ \int_0^{\infty} t^{s-1} e^{ibt} e^{-at} \, dt = \frac{\Gamma(s)}{(a-ib)^s} = \frac{\Gamma(s)}{\left(\sqrt{a^2+b^2} e^{i\arctan \left(-\frac{b}{a} \right)} \right)^s} = \frac{\Gamma(s)}{(a^2+b^2)^{s/2}} e^{is \arctan \left(\frac{b}{a} \right)} $$
and the integrals above follow by equating real and imaginary parts. Can someone please explain why the complex number substitution is valid here, and how to justify it? In general when is it ok to do this in an integral? Thanks
Best Answer
Svyatoslav’s comment is good, but I would like to point out another way to see that the identity holds. Consider the function $F: z \longmapsto z^s\int_0^{\infty}{t^{s-1}e^{-zt}\,dt}$, where $s$ is a fixed number with positive real part, and $z$ has positive real part. It’s easy to see that $F$ is holomorphic, and you just stated that $F(z)=\Gamma(s)$ as soon as $z$ is a positive real number.
Thus the holomorphic function $F-\Gamma(s)$ has a nonisolated zero so must vanish everywhere. QED.