Integral of the day XXVIII

Integral involving the Gaussian fucntion ${}_{2}F_{1}$

Today we show the proof of this integral posted by @integralsbot \[ \int_{-\infty}^{\infty} \frac{dx}{(1+2a+\cosh x)^s } = \frac{2^{s}\Gamma^2(s)}{\Gamma(2s)}{}_{2}F_{1}\left[{s,s \atop s+\frac{1}{2}}; -a\right] \quad |a|\lt 1 \] For the proof we will use the generalized binomial theorem, some properties of the Pochhammer symbol (risng facorial) and the duplication formula for the Gamma function

Proof:

Suppose $|a|\lt 1 $ \begin{align*} I= \int_{-\infty}^{\infty} \frac{dx}{(1+2a+\cosh x)^s } =& 2\int_{0}^{\infty} \frac{dx}{(1+2a+\cosh x)^s } \quad \textrm{(The integrand is even)}\\ =& 2\int_{0}^{\infty} \frac{dx}{\left(2a+2\cosh^2 \left(\frac{x}{2}\right)\right)^s } \quad \left( 1+\cosh(x) = 2\cosh^2\left(\frac{x}{2}\right)\right)\\ =& \frac{1}{2^{s-1}}\int_{0}^{\infty} \frac{dx}{\left(a+\cosh^2 \left(\frac{x}{2}\right)\right)^s } \\ =& \frac{1}{2^{s-1}}\int_{0}^{\infty} \sum_{j=0}^{\infty} \binom{-s}{j} \cosh^{2(-s-j)}\left(\frac{x}{2}\right)a^j dx \quad \textrm{(Generalized bionomial theorem)} \\ =& \frac{1}{2^{s-2}}\sum_{j=0}^{\infty} \binom{-s}{j}a^j \int_{0}^{\infty} \cosh^{2(-s-j)}\left(w\right)dw \\ =& \frac{1}{2^{s-2}}\sum_{j=0}^{\infty} \binom{-s}{j}a^j \int_{0}^{\infty} \operatorname{sech}^{2(s+j)}\left(w\right)dw \\ =& \frac{1}{2^{s-2}}\sum_{j=0}^{\infty} \binom{-s}{j}a^j \int_{0}^{1} t^{2(s+j)-1}(1-t^2)^{-\frac{1}{2}} dt \quad (t \mapsto \operatorname{sech}(x) )\\ =& \frac{1}{2^{s-1}}\sum_{j=0}^{\infty} \binom{-s}{j}a^j \int_{0}^{1} s^{(s+j)-1}(1-s)^{-\frac{1}{2}} ds \quad (s \mapsto t^2 )\\ =& \frac{1}{2^{s-1}}\sum_{j=0}^{\infty} \binom{-s}{j}a^j B\left(s+j,\frac{1}{2}\right)\\ \end{align*} Using the fact that \[ \binom{v}{m} = \frac{(v-m+1)_{m}}{m!} \] and the recursion formula \[(x-n+1)_{n} = (-1)^n(-x)_{n} \] where \[ (x)_{n} = x(x+1)(x+2)\cdots (x+n-1) \] we have \begin{align*} I =& \frac{1}{2^{s-1}}\sum_{j=0}^{\infty} \binom{-s}{j}a^j B\left(s+j,\frac{1}{2}\right)\\ =& \frac{1}{2^{s-1}}\sum_{j=0}^{\infty} \frac{(-s-j+1)_{j}}{j!}a^j \frac{\Gamma(s+j)\sqrt{\pi}}{\Gamma\left(s+j+\frac{1}{2}\right)}\\ =& \frac{1}{2^{s-1}}\sum_{j=0}^{\infty} \frac{(s)_{j}}{j!}(-a)^j \frac{\Gamma(s+j)\sqrt{\pi}}{\Gamma\left(s+j+\frac{1}{2}\right)}\\ \end{align*} From the definition of the Pochhammer symbol: \[ (x)_{n} = \frac{\Gamma(n+x)}{\Gamma(x)} \] we have \[I =\frac{1}{2^{s-1}}\sum_{j=0}^{\infty} \frac{(s)_{j}}{j!}(-a)^j \frac{\Gamma(s+j)\sqrt{\pi}}{\Gamma\left(s+j+\frac{1}{2}\right)} = \frac{\Gamma(s)}{2^{s-1}\sqrt{\pi}\Gamma\left(s+\frac{1}{2}\right)}\sum_{j=0}^{\infty} \frac{(s)_{j}(s)_{j}}{\left(s+\frac{1}{2}\right)_{j}}\frac{(-a)^j}{j!} \] From the duplication formula \[ \Gamma(2x) = \frac{4^x}{2\sqrt{\pi}} \Gamma(x)\Gamma\left(\frac{1}{2}+x\right)\] we have \begin{align*} I = \frac{\Gamma(s)}{2^{s-1}\sqrt{\pi}\Gamma\left(s+\frac{1}{2}\right)}\sum_{j=0}^{\infty} \frac{(s)_{j}(s)_{j}}{\left(s+\frac{1}{2}\right)_{j}}\frac{(-a)^j}{j!} & = \frac{2^{s}\Gamma^2(s)}{\Gamma(2s)}\sum_{j=0}^{\infty} \frac{(s)_{j}(s)_{j}}{\left(s+\frac{1}{2}\right)_{j}}\frac{(-a)^j}{j!}\\ =& \frac{2^{s}\Gamma^2(s)}{\Gamma(2s)}{}_{2}F_{1}\left[{s,s \atop s+\frac{1}{2}}; -a\right] \end{align*} Finally, we can conclude \[\boxed{ \int_{-\infty}^{\infty} \frac{dx}{(1+2a+\cosh x)^s } = \frac{2^{s}\Gamma^2(s)}{\Gamma(2s)}{}_{2}F_{1}\left[{s,s \atop s+\frac{1}{2}}; -a\right] \quad |a|\lt 1 }\]