I am currently struggling to find out how we arrive at the following equation: $\int\limits_{s/2}^{\infty}\frac{\alpha^{2m-1}e^{-m\alpha^2}}{\sqrt{4\alpha^2-s^2}}d\alpha= \frac{\pi \sec (m\pi)}{4}\left\lbrace \frac{2^{1-2m} \sqrt{\pi}s^{2m-1} {}_1\mathcal{M}_1\left(m,\frac{1}{2}+m,\frac{-ms^2}{4} \right) }{\Gamma(1-m)} -m^{\frac{1}{2}-m}{}_1\mathcal{M}_1\left(\frac{1}{2},\frac{3}{2}-m,\frac{-ms^2}{4} \right) \right\rbrace$, where ${}_1\mathcal{M}_1(a,b,z)=\frac{1}{\Gamma(b-a)\Gamma(a)}\int\limits_0^1e^{z\alpha}\alpha^{a-1}(1-\alpha)^{b-a-1}d\alpha$ is the regularized confluent hypergeometric function.
Actually I needed to evaluate the above integral for my research work and hence, I evaluated it using Mathematica. The result works perfectly fine for me. But now I want to know (at least the intuition behind it) how did Mathematica obtain this solution. I have been thinking over this for quite some time now, but unfortunately haven't succeeded yet.
What I have been able to obtain after some manipulations is the following: $\int\limits_{s/2}^{\infty}\frac{\alpha^{2m-1}e^{-m\alpha^2}}{\sqrt{4\alpha^2-s^2}}d\alpha=\frac{1}{4}(\frac{s}{2})^{2m-1}\int\limits_{0}^{\pi/2}(\sec \theta)^{2m}e^{-\frac{ms^2}{4}\sec^2\theta}d\theta$. Even giving this as the input, Mathematica gives the same result as above. But again, I cannot understand...how.
Can anyone please help?
