This post is kind of lengthy. I'm sorry
Consider the integral$$\int\limits_0^\infty\frac {\cos 2mx\, dx}{\left\{1+x^2/a^2\right\}\left\{1+x^2/(a+1)^2\right\}\left\{1+x^2/(a+2)^2\right\}\ldots}$$Where $m$ and $a$ are positive. It can be easily proved that$$\begin{align*} & \left\{1-\left(\frac ta\right)^2\right\}\left\{1-\left(\frac t{a+1}\right)^2\right\}\left\{1-\left(\frac t{a+2}\right)^2\right\}\ldots\left\{1-\left(\frac t{a+n-1}\right)^2\right\}\\\\ & =\frac {\Gamma(a+n-t)\Gamma(a+n+t)\Gamma^2(a)}{\Gamma(a-t)\Gamma(a+t)\Gamma^2(a+n)}\end{align*}$$Where $n$ is any positive integer. Hence, by splitting$$\frac 1{\left\{1+x^2/a^2\right\}\left\{1+x^2/(a+1)^2\right\}\ldots\left\{1+x^2/(a+n-1)^2\right\}}$$into partial fractions, we see that it is equal to$$\begin{align*} & \frac {2\Gamma(2a)\Gamma^2(a+n)}{\Gamma^2(a)\Gamma(n)\Gamma(2a+n)}\left\{\frac a{a^2+x^2}-\frac {2a}{1!}\frac {n-1}{n+2a}\frac {a+1}{(a+1)^2+x^2}+\\+\frac {2a(2a+1)}{2!}\frac {(n-1)(n-2)}{(n+2a)(n+2a+1)}\frac {a+2}{(a+2)^2+x^2}-\ldots\right\}\end{align*}$$ Multiplying both sides by $\cos 2mx$ and integrating from $0$ to $\infty$ with respect to $x$, we have$$\begin{align*} & \int\limits_0^\infty\frac {\cos 2mx\, dx}{\left\{1+x^2/a^2\right\}\left\{1+x^2/(a+1)^2\right\}\ldots\left\{1+x^2/(a+n-1)^2\right\}}\\\\ & =\frac {\pi\Gamma(2a)\Gamma^2(a+n)}{\Gamma^2(a)\Gamma(n)\Gamma(2a+n)}\left\{\color{blue}{e^{-2am}}-\frac {2a}{1!}\frac {n-1}{n+2a}\color{blue}{e^{-2(a+1)m}}+\ldots\right\}\end{align*}$$
There are a couple of questions I have in mind that I am not sure how the author managed to obtain.
Questions:
- How do you prove the second identity?$$\begin{align*} & \left\{1-\left(\frac ta\right)^2\right\}\left\{1-\left(\frac t{a+1}\right)^2\right\}\left\{1-\left(1-\frac t{(a+2)}\right)^2\right\}\ldots\left\{1-\left(\frac t{a+n-1}\right)^2\right\}\\\\ & =\frac {\Gamma(a+n-t)\Gamma(a+n+t)\Gamma^2(a)}{\Gamma(a-t)\Gamma(a+t)\Gamma^2(a+n)}\end{align*}$$
- How do you split the infinite fractions into partial fractions?
- How does $e^{-2am}$ show up in the last identity? (Colored in blue)
