closed form for $\int_{0}^{\infty}\frac{ \beta(a+ix,a-ix)}{\beta(b+ix,b-ix)}\frac{dx}{(b^2+x^2)}$

Question

closed form for : $$\int_{0}^{\infty}\frac{ \beta(a+ix,a-ix)}{\beta(b+ix,b-ix)}\frac{\mathrm{dx}}{(b^2+x^2)}$$

where $\beta$ is beta function

I tried with the definition of beta and i got

$$I=\frac{\Gamma(2b)}{\Gamma(2a)}\int_{0}^{\infty}\frac{\Gamma(a-ix)\Gamma(a+ix)}{\Gamma(b-ix)\Gamma(b+ix)}\frac{\mathrm{dx}}{(b^2+x^2)}$$

I also used the product form but i failed

I think it's something related to complex analysis so if there is any help .

Thanks.

Answer 1

The integral is evaluated to be \begin{align} \int_{0}^{\infty} \, \left| \frac{\Gamma(a + i x) }{ \Gamma(b + 1 + ix)} \right|^{2} \, dx = \frac{4^{b-a}}{2 \, b} \, B\left( b - a + \frac{1}{2} , \frac{1}{2} \right) \end{align}