This problem is supposed to be from an old Acta Mathematica volume I circa 1880's, and is attributed to Bourguet.
By using a parabola with its focus on the origin as a contour, show that:
$$\int_{0}^{\infty}e^{-ax^{2}}(1+x^{2})^{z-1/2}\cos(2ax+(2z-1)\tan^{-1}(x))dx=\frac{\sin(\pi z)\Gamma(z)}{2a^{z}e^{a}}$$
I am not a total cabbage head with contour integration, yet certainly not along the lines of Ron G, RV, robjohn, et. al.
But, this thing looks nasty. Sometimes looks can be deceiving.
I have never used a parabola as a contour. The closest thing I can think of is a Hankel contour. That is not exactly a parabola, but is commonly used with the Gamma function integral. A Hankel looks more like a bobby pin. But, if anyone would care to exert their contour integration prowess, it would be nice to see this one worked out.
Best Answer
Let $\mathcal{I}$ be the nasty integral we want to calculate. For $x \in \mathbb{R}$, we have
$$e^{\pm i\tan^{-1}x} = \frac{1\pm i x}{\sqrt{1+x^2}}$$
This leads to $$\cos(2ax + (2z-1)\tan^{-1}x) = \frac12\left[ e^{2iax} \left(\frac{1 + ix}{\sqrt{1+x^2}}\right)^{2z-1} + e^{-2iax} \left(\frac{1-ix}{\sqrt{1+x^2}}\right)^{2z-1} \right] $$ As a result, we can rewrite integral $\mathcal{I}$ as
$$\begin{align} \mathcal{I} &= \frac{1}{2e^a}\int_0^\infty \left( e^{a(1+ix)^2} (1+ix)^{2z-1} + e^{a(1-ix)^2}(1-ix)^{2z-1}\right) dx\\ &= \frac{1}{2i e^a} \int_{1-i\infty}^{1+i\infty} e^{at^2} t^{2z-1}dt\\ &= \frac{1}{4i e^a a^z} \int_C e^t t^{z-1} dt \end{align}$$ where $C$ is the contour $\mathbb{R} \ni x \mapsto a(1+ix)^2 \in \mathbb{C}$. It is easy to see $C$ is a parabola which start from infinity at third quadrant. It first move towards the origin, circle around it counterclockwisely in the fourth and then first quadrant. Finally, move away to infinity in the second quadrant.
To evaluate $\mathcal{I}$, we can deform $C$ to a keyhole contour along the negative axis. The keyhole contour start at $-\infty -i\epsilon$, move towards the $-i\epsilon$, circle around the origin counterclockwisely to $i\epsilon$ and then move away to $-\infty + i\epsilon$. This give us $$\mathcal{I} = \frac{1}{4i e^a a^z}\int_C e^{t} t^{z-1} dt = \frac{ e^{i\pi z} - e^{-i\pi z} }{4i e^a a^z}\Gamma(z) = \frac{\sin(\pi z)}{2 e^a a^z}\Gamma(z)$$