A contour integral for Fourier transform

Question

How does one show the following, preferably with contour integral on the complex plane?

$$\frac{\Gamma(\alpha)}{2\pi}\int_{-\infty}^\infty (ik)^{-\alpha}e^{-ikx}dk = (-x)_+^{\alpha-1},$$ where $x$ is a real number, constant Re$(\alpha)>0$, the integrating contour of $k$ is on the lower half of the complex plane.

I am deforming the contours to the imaginary axis but I can not seem to get a residue which will prevent the whole thing from vanishing or not vanishing at all. I must have missed something...

Answer 1

This is one of those unusual places where it is convenient to turn the path of integration clockwise 90 degrees in the complex plane, at least for $x > 0$. Turn the contour counterclockwise 90 degrees if $x < 0$. Then, for example, $$ \begin{align} \int_{0}^{\infty}(ik)^{-\alpha}e^{-ikx}dk & = \int_{0}^{\infty}(i(-ik))^{-\alpha}e^{-i(-ik)x}dk \\ & = \int_{0}^{\infty}k^{-\alpha}e^{-kx}dk \\ & = \int_{0}^{\infty}(kx)^{-\alpha}e^{-kx}d(kx) x^{-1+\alpha} \\ & = \int_{0}^{\infty}u^{-\alpha}e^{-u}du\; x^{-1+\alpha} = \Gamma(1-\alpha)x^{-1+\alpha}. \end{align} $$ The above holds for $\Re \alpha < 1$. Once you've finished rotating and evaluating the contour integrals, then you can work on moving the contour away from the origin in order to extend the definition for $\Re\alpha > 1$. All of the functions of $\alpha$ are holomorphic and, so, you can use the identity theorem for Complex Variables to extend to all $\alpha$ for which the expressions are defined and holomorphic in $\alpha$. The factor $\Gamma(\alpha)$ in front of the integral is probably needed to deal with the jump across the branch cut of $(\cdot)^{-\alpha}$, in which case it is useful to know that $$ \Gamma(\alpha)\Gamma(1-\alpha)\sin\pi\alpha = \pi. $$ This may well come up in the process of dealing with the branch cut of $(ik)^{\alpha}$.

Answer 2

$$I(x) := \int_{-i\infty}^{i\infty} z^{-\alpha}e^{-zx}dz,$$ where the integrating contour is on the imaginary axis. We proceed just as the original question suggests, only this time we take better care of the branch cut. Consider first real $\alpha>0$. Let the branch cut of $z$ be arg$(z)=\pm\pi$.

Let $z=re^{i\theta}$, where $r\ge 0$ and $\theta\in\mathbf R$. $|e^{-zx}|=e^{-xr\cos\theta}$. To evaluate this integral, we shall choose a closed contour, depending on the signs of $x$, that keep the exponent of $e$ negative. For $x>0$, we pick a closed contour running through the imaginary axis from $-iR$ to $iR$ and from there describing a semicircle with radius $r$ clockwise back to $-iR$. For the semicircular contour

$$|I_C(x)|\le \int_C |z|^{-\alpha}|e^{-zx}|d|z|=\int_{-\frac{\pi}{2}}^\frac{\pi}{2} r^{1-\alpha}e^{-xr\cos\theta}d\theta<2r^{1-\alpha}\int_0^\frac{\pi}{2} e^{-xr(1-\frac{2}{\pi}\theta)}d\theta=\pi\frac{1-e^{-xr}}{xr^\alpha}.$$

For $x<0$, we choose a contour running through the imaginary axis from $0$ to iR and from there describing a quarter circle with radius $r$ counterclockwise to $-R$ and back along the real axis to $0$. (to be continued)