definite integral, regularized hypergeometric function I am currently stuck at the following equation: $\frac{\sqrt{m} \sec \left( (m+1)\pi \right)}{4\Gamma(m)} \int\limits_{0}^{\infty} \sqrt{z}\: {}_1\mathcal{M}_1\left(\frac{1}{2},\frac{3}{2}-m,\frac{-mz}{4} \right) dz=1$ $\forall$ $m \in \mathbb{Z}^+$, 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.
Given that I evaluated the integral using Mathematica, currently I'm interested in how did Mathematica obtain this solution.
What I have been able to obtain after some manipulations and exploiting properties of the Gamma function is the following: $\frac{\sqrt{m} \sec \left( (m+1)\pi \right)}{4\Gamma(m)} \int\limits_{0}^{\infty} \sqrt{z}\: {}_1\mathcal{M}_1\left(\frac{1}{2},\frac{3}{2}-m,\frac{-mz}{4} \right) dz=\frac{2\Gamma(m-\frac{1}{2})}{m\pi\Gamma(m)}\int\limits_{0}^{\infty}t^{\frac{1}{2}}{}_1F_1\left(\frac{1}{2},\frac{3}{2}-m,-t \right) dt$, where we apply the transformation $\frac{mz}{4}\rightarrow t$ and ${}_1F_1(a,b,z)={}_1\mathcal{M}_1(a,b,z)\Gamma(b)$ is the standard confluent hypergeometric function.
There exists a standard integral in the book "Table of Integrals, Series, and Products" by I. S. Gradshteyn and I. M. Ryzhik, which is as follows: $\int\limits_{0}^{\infty}t^{b-1}{}_1F_1(a,c,-t)dt=\frac{\Gamma(b)\Gamma(c)\Gamma(a-b)}{\Gamma(a)\Gamma(c-b)}$ [7.612.1]. But the problem is that this result holds only when $b<a$, which is not the case in my problem. This is precisely the point where I am stuck and cannot arrive at the solution that is provided by Mathematica.
Any suggestion will be helpful.
 A: To evaluate
\begin{equation}
I=\frac{2\Gamma(m-\frac{1}{2})}{m\pi\Gamma(m)}\int\limits_{0}^{\infty}t^{\frac{1}{2}}{}_1F_1\left(\frac{1}{2},\frac{3}{2}-m,-t
\right)\,dt
\end{equation}
we replace the hypergeometric function by its representation in terms of
Laguerre polynomials (see
here)
\begin{equation}
{}_1F_1(a, a - n, z)=\frac{(-1)^nn!}{(1-a)_n}e^zL_n^{a-n-1}(-z)
\end{equation}
with $a=1/2,n=m-1,z=-t$, to express, after several simplifications,
\begin{equation}
I=\frac{2(-1)^{m-1}}{m\sqrt{\pi}}\int\limits_{0}^{\infty}t^{1/2}e^{-t}L_{m-1}^{1/2-m}(t)\,dt
\end{equation}
From the Rodrigues-type expression
\begin{equation}
  L_n^\lambda(z)=\frac{e^zz^{-\lambda}}{n!}\frac{\partial^n}{\partial z^n}\left( z^{n+\lambda}  e^{-z}\right)
 \end{equation}
with $n=m-1,\lambda=1/2-m$,
\begin{equation}
 L_{m-1}^{1/2-m}(z)=\frac{e^zz^{m-1/2}}{(m-1)!}\frac{\partial^{m-1}}{\partial z^{m-1}}\left( z^{-1/2}  e^{-z}\right)
 \end{equation}
Thus
\begin{equation}
  I=\frac{2(-1)^{m-1}}{m!\sqrt{\pi}}\int\limits_{0}^{\infty}t^{m}\frac{\partial^{m-1}}{\partial t^{m-1}}\left( t^{-1/2}  e^{-t}\right)\,dt
 \end{equation}
By performing integrations by parts $m-2$ times,
\begin{equation}
\int\limits_{0}^{\infty}t^{m}\frac{\partial^{m-1}}{\partial t^{m-1}}\left( t^{-1/2}  e^{-t}\right)\,dt=(-1)^{m-1}m!\int_0^\infty t^{1/2} e^{-t}\,dt
 \end{equation}
Finally
\begin{equation}
  I=1
 \end{equation}
