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While working on a problem I came across the following interesting result.

Let: $$ H_{nm}(x)=\int_0^\infty t^{x-1}e^{-t}\log t\;F(-n;x;t)F(-m;x;t)\;dt, $$ where $n,m$ are non-negative integer numbers, $x$ is positive real number and $$F(a;b;t)=\sum_{k\ge0}\frac{a^{\overline k}}{b^{\overline k}}\frac{t^k}{k!}$$ is the confluent hypergeometric function.

Since the integral is symmetric with respect to permutation of $n$ and $m$ in what follows $n\ge m$ is assumed.

By numerical evidence the integral evaluates to the following simple values: $$ H_{nm}(x)=\frac{n!\;\Gamma^2(x)}{\Gamma(x+n)}\times \begin{cases} \displaystyle\psi(x+n),& n=m; \\ \displaystyle\frac1{m-n},&n\ne m. \end{cases}\tag1 $$ where $\Gamma(x)$ and $\psi(x)$ are the gamma and digamma functions, respectively.

Is there a simple way to prove the relations $(1)$?

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The confluent hypergeometric functions are related to the generalized Laguerre polynomials: \begin{align} F(-n;x;t)&=\frac{\Gamma(n+1)\Gamma(x)}{\Gamma(x+n)}L_n^{(x-1)}(t) \end{align} so \begin{equation} H_{n,m}(x)=\frac{n!m!\Gamma^2(x)}{\Gamma(x+n)\Gamma(x+m)}\int_0^\infty t^{x-1}e^{-t}\ln t L_n^{(x-1)}(t)L_m^{(x-1)}(t)\,dt \end{equation} The orthogonality relation for the Laguerre polynomials reads \begin{equation} \int_0^\infty t^{x-1}e^{-t}L_n^{(x-1)}(t)L_m^{(x-1)}(t)\,dt=\frac{\Gamma(n+x)}{n!}\delta_{n,m} \end{equation} It can be differentiated with respect to $x$ to obtain \begin{equation} \int_0^\infty t^{x-1}e^{-t}\ln t L_n^{(x-1)}(t)L_m^{(x-1)}(t)\,dt+\int_0^\infty t^{x-1}e^{-t}\frac{d}{dx}\left[L_n^{(x-1)}(t)L_m^{(x-1)}(t)\right]\,dt=\frac{\Psi(n+x)\Gamma(n+x)}{n!}\delta_{n,m} \end{equation} From the differentiation relation \begin{equation} \frac{d}{dx}L_n^{(x-1)}(t)=\sum_{k=0}^{n-1}\frac{L_k^{(x-1)}(t)}{n-k} \end{equation} and recognizing the definition of $H_{n,m}(x)$, we have thus \begin{align} \frac{\Gamma(n+x)\Gamma(m+x)}{n!m!\Gamma^2(x)}H_{n,m}(x)&+\sum_{k=0}^{n-1}\frac{1}{n-k}\int_0^\infty t^{x-1}e^{-t}L_k^{(x-1)}(t)L_m^{(x-1)}(t)\,dt\\ &+\sum_{k=0}^{m-1}\frac{1}{m-k}\int_0^\infty t^{x-1}e^{-t}L_k^{(x-1)}(t)L_n^{(x-1)}(t)\,dt\\ &=\frac{\Psi(n+x)\Gamma(n+x)}{n!}\delta_{n,m} \end{align} using the orthogonality relation and supposing that $n> m$, only one term in the sums survives, while there is no if $n=m$: \begin{equation} \frac{\Gamma(n+x)\Gamma(m+x)}{n!m!\Gamma^2(x)}H_{n,m}+\frac{1}{n-m}\frac{\Gamma(m+x)}{m!} \left( 1-\delta_{n,m} \right)=\frac{\Psi(n+x)\Gamma(n+x)}{n!}\delta_{n,m} \end{equation} which is the proposed expression.

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  • $\begingroup$ This is a very nice proof. Thank you very much! I have corrected some small inaccuracies which had mutually cancelled in the derivation. Hopefully you don't mind. $\endgroup$ – user Nov 30 '19 at 23:24
  • $\begingroup$ You're welcome. Sorry for the mistakes and thanks for correcting them! $\endgroup$ – Paul Enta Nov 30 '19 at 23:30

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