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I want to understand the formula 8.315.2 in Gradshteyn and Ryzhik. It reads $$ \int_{-\infty}^{\infty}{ \frac{e^{ibt}}{(a+it)^k}dt} = \frac{ 2\pi e^{-ab}b^{k-1}}{\Gamma(k)}, $$ where $a,b,k$ are real with $a,b > 0$ and $k \geq 2$ (these are the ranges I am interested in, but the formula might hold more generally). Here, $(a+it)^k = e^{\log(a+it)k}$, where the we take the standard branch of the logarithm (the one that agrees with the usual real logarithm on the real axis). If I did things correctly, the proof reduces to $$ \frac{1}{\Gamma(k)} = \frac{1}{2\pi i}{\int_{a -i\infty}^{a+ i \infty}\frac{e^z}{z^k} dz} $$ (and the RHS is indeed independent of $ a > 0$.). If $k$ is an integer, the above is easy to verify, by shifting the contour far to the left picking up the residue at zero (also, $\Gamma(k) = (k-1)!$ in this case). How can we prove the formulas for general $k$?

metamorphy
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m.s
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Essentially this is Hankel's formula for $\Gamma$. It suffices to prove it for $\Re k<1$ (after deforming the contour; the rest is done by analytic continuation). We deform the contour so that it now encircles the negative real axis (used as the branch cut of $z^{-k}$) closely. In the limit of "closely", the integral becomes $$\frac{1}{2\pi i}\int_\infty^0\frac{e^{-x}\,d(-x)}{(xe^{-\pi i})^k}+\frac{1}{2\pi i}\int_0^\infty\frac{e^{-x}\,d(-x)}{(xe^{\pi i})^k}=\frac{\sin k\pi}{\pi}\Gamma(1-k)=\frac{1}{\Gamma(k)}$$ by (the integral definition of, and) the reflection formula for $\Gamma$.

metamorphy
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