Size of a Mellin integral

Question

I am trying to estimate the size (in $X$) of $$\int_{(1/2)} \frac{\Gamma(s)}{\Gamma(1/2-s)} X^{-2s} ds$$ where the integration is on the vertical line of real part $1/2+\varepsilon$. I would have liked to express it in terms of Bessel function, but the formulas I find only work for a denominator with larger shifts (larger than $1$).

Answer 1

This doesn't converge absolutely, since Stirling's formula shows that the absolute value of the integrand, for $s = \sigma + i\tau$, is $\asymp (1 + |\tau|)^{2\sigma - 1/2} X^{-2\sigma}$. However, if we use a piecewise contour that is to the right of the pole at $s = 0$ when $|\tau|$ is small but to the left of the line $\sigma = -1/4$ when $|\tau|$ is large, then this is integrable. By the reflection and dublication formulae for the gamma function, the integral is equal to $$\frac{1}{\sqrt{\pi}} \int \Gamma(s) \cos \frac{\pi s}{2} (2X)^{-s} \, ds,$$ which is $2\sqrt{\pi} i \cos(2X)$ by 12.43.3 of Gradshteyn-Ryzhik.