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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.

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    $\begingroup$ We have $${_1 \hspace {-1 px} F_1} {\left( \frac 1 2; \frac 3 2 - m; -\frac {m z} 4 \right)} = e^{-m z/4} \hspace {1 px} {_1 \hspace {-1. px} F_1} {\left( 1 - m; \frac 3 2 - m; \frac {m z} 4 \right)},$$ which is a finite sum. Multiplying the $k$th term by $\sqrt z$ and integrating adds another $(a)_k$ factor, so gives the $k$th term in the series expansion of a ${_2 F_1}$ function. It remains to apply the identity $${_2 F_1}(a, 1 - m; c; 1) = \frac {(c - a)_{m - 1}} {(c)_{m - 1}}.$$ $\endgroup$
    – Maxim
    Sep 22, 2020 at 13:45

1 Answer 1

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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}

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  • $\begingroup$ I did not even think that Laguerre polynomials may be required for the proof. Thanks a lot! $\endgroup$ Sep 8, 2020 at 16:37
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    $\begingroup$ You're welcome. I think that other proof are possible without these polynomials. $\endgroup$
    – Paul Enta
    Sep 8, 2020 at 16:46

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