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Can someone help me evaluate $G_g(z)=\int_0^{\infty}x^{z-1}e^{igx}dx$, where $g$ is real and $z$ is complex?

By closing the contour in the upper half plane, I've managed to prove that if $0<Re(z)<1$ and $Im(z)>0$ (we can then ignore these conditions by analytic continuation) and $g>0$, then $G_g(z)=e^{i\pi z}G_{-g}(z)$. But it doesn't look as though contour integration will be enough to get the whole answer unless $G_g(z)\equiv0$, since the integrand has no poles, only a branch cut.

Can we use the gamma function somewhere? The expression $\Gamma(z)=\int_0^{\infty}t^{z-1}e^{-t}dt$ looks very similar to what we've got for $G$, and the first half of this problem was about the gamma function. Maybe we could substitute $t=-igx$ and then deform the contour from the imaginary axis to the real?

Many thanks for any help with this!

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Answering my own question here - it's much easier than I thought!

Assume $Re(z)>1$ (we can extend by analytic continuation later). Use Jordan's lemma in the upper half plane if $g>0$ and in the lower half plane if $g<0$, changing the contour of integration to the upper or lower (respectively) imaginary axis. Then the integral is a $z$-dependent constant times the gamma function. The final answer is:

$G(z)=\exp\big(sgn(g)\frac{zi\pi}{2}\big)|g|^{-z}\Gamma(z)$.

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