Generalized hypergeometric functions VI

Singular integral involving the Gauss hypergeometric function

Today we show the proof of this singular integral posted by @integralsbot \[\int_{a}^{\infty} (x-a)^m\left(x-\sqrt{x^2-a^2}\right)^n dx = \frac{a^{m+n+1}}{2^{m}}\frac{n(n-m-2)!(2m+1)!}{(n+m+1)!} \] The proof relies on some properties of the Gauss hypergeometric function $F(a,b;c;x)$

Proof

\begin{align*} I = \int_{a}^{\infty} (x-a)^m\left(x-\sqrt{x^2-a^2}\right)^n dx = &\int_{a}^{\infty} a^{m+n}\left(\frac{x}{a}-1\right)^m\left(\frac{x}{a}-\sqrt{\frac{x^2 }{a^2}-1}\right)^n dx\\ =& \int_{1}^{\infty} a^{m+n+1}\left(w-1\right)^m\left(w-\sqrt{w^2-1}\right)^n dw \quad \left(w \mapsto \frac{x}{a} \right)\\ =& \int_{1}^{0} a^{m+n+1}\left(\frac{t^2-2t+1}{2t}\right)^mt^n \frac{(t^2-1)}{2t^2} dt \quad (t \mapsto w-\sqrt{w^2-1})\\ =& \frac{a^{m+n+1}}{2^{m+1}}\int_{0}^{1} (t-1)^{2m} t^{n-m-2}(1-t^2) dt\\ =& \frac{a^{m+n+1}}{2^{m+1}}\int_{0}^{1} t^{n-m-2}(1-t^2)\sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j} t^{j} dt\\ =& \frac{a^{m+n+1}}{2^{m+1}}\sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j}\int_{0}^{1} t^{j+n-m-2}(1-t^2) dt\\ =& \frac{a^{m+n+1}}{2^{m+2}}\sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j}\int_{0}^{1} s^{\frac{j+n-m-3}{2}}(1-s) ds \quad (s \mapsto t^2)\\ =& \frac{a^{m+n+1}}{2^{m+2}}\sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j}B\left(\frac{j+n-m-1}{2},2\right)\\ =& \frac{a^{m+n+1}}{2^{m}}\sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j}\frac{1}{(j-m+n-1)(j-m+n+1)}\\ =& \frac{a^{m+n+1}}{2^{m+1}}\sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j}\frac{1}{(j-m+n-1)}-\frac{a^{m+n+1}}{2^{m+1}}\sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j}\frac{1}{(j-m+n+1)}\\ \end{align*} Now using the recursion formula for Pochhammer symbols (rising factorial): \[ (n+x)(x)_{n} = x(x+1)_{n}\] Then: \[ \frac{1}{j-m+n-1} = \frac{(n-m-1)_{j}}{(n-m-1)(n-m)_{j}}\] \[ \frac{1}{j-m+n+1} = \frac{(n-m+1)_{j}}{(n-m+1)(n-m+2)_{j}}\] Hence \[ I =\frac{a^{m+n+1}}{2^{m+1}(n-m-1)}\sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j}\frac{(n-m-1)_{j}}{(n-m)_{j}}-\frac{a^{m+n+1}}{2^{m+1}(n-m+1)}\sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j} \frac{(n-m+1)_{j}}{(n-m+2)_{j}}\] Now, recall that if either of the parameters in the Gauss hypergeometric function is a nonpositive integer, then the function reduces to a polynomial function of degree n: \[ F(a,-n;c;x) = \sum_{j=0}^{n}\binom{n}{j} \frac{(a)_{j}}{(c)_{j}}(-x)^j\] Then \[ I =\frac{a^{m+n+1}}{2^{m+1}(n-m-1)}F(n-m-1,-2m;n-m;1)-\frac{a^{m+n+1}}{2^{m+1}(n-m+1)}F(n-m+1,-2m;n-m+2;1) \] Using the formula: \[ F(a,b;c;1) = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}\quad c>a+b\] \begin{align*} I =&\frac{a^{m+n+1}}{2^{m+1}(n-m-1)}\frac{\Gamma(n-m)\Gamma(2m+1)}{\Gamma(n+m)}-\frac{a^{m+n+1}}{2^{m+1}(n-m+1)}\frac{\Gamma(n-m+2)\Gamma(2m+1)}{\Gamma(n+m+2)}\\ =& \frac{a^{m+n+1}}{2^{m+1}}\left[\frac{\Gamma(n-m-1)\Gamma(2m+1)}{\Gamma(n+m)}-\frac{\Gamma(n-m+1)\Gamma(2m+1)}{\Gamma(n+m+2)}\right]\\ =& \frac{a^{m+n+1}}{2^{m+1}}\left[\frac{\Gamma(n-m-1)\Gamma(2m+1)\Gamma(n+m+2)-\Gamma(n-m+1)\Gamma(2m+1)\Gamma(n+m)}{\Gamma(n+m)\Gamma(n+m+2}\right]\\ =& \frac{a^{m+n+1}}{2^{m+1}}\Gamma(2m+1)\left[\frac{\Gamma(n-m-1)\Gamma(n+m)(n+m+1)(n+m)-\Gamma(n-m+1)\Gamma(n+m)}{\Gamma(n+m)\Gamma(n+m+2)}\right]\\ =& \frac{a^{m+n+1}}{2^{m+1}}\Gamma(2m+1)\left[\frac{\Gamma(n-m-1)(n+m+1)(n+m)-\Gamma(n-m-1)(n-m)(n-m-1)}{\Gamma(n+m+2)}\right]\\ =& \frac{a^{m+n+1}}{2^{m+1}}\Gamma(2m+1)\Gamma(n-m-1)\left[\frac{2(2m+1)n}{\Gamma(n+m+2)}\right]\\ =& \frac{a^{m+n+1}}{2^{m}}\frac{n\Gamma(2m+2)\Gamma(n-m-1)}{\Gamma(n+m+2)}\\ =& \frac{a^{m+n+1}}{2^{m}}\frac{n(n-m-2)!(2m+1)!}{(n+m+1)!}\\ \end{align*} Hence, we can conclude \[\boxed{\int_{a}^{\infty} (x-a)^m\left(x-\sqrt{x^2-a^2}\right)^n dx = \frac{a^{m+n+1}}{2^{m}}\frac{n(n-m-2)!(2m+1)!}{(n+m+1)!} }\]