1
$\begingroup$

Question: How do you split$$\dfrac {1}{\left\{1+\dfrac {x^2}{a^2}\right\}\left\{1+\dfrac {x^2}{(a+1)^2}\right\}\left\{1+\dfrac {x^2}{(a+1)^2}\right\}\cdots\left\{1+\dfrac {x^2}{(a+n-1)^2}\right\}}\tag{1}$$ Into a sum of partial fractions?

I was never formally taught how to split a fraction into different components. Apparently, $(1)$ can be expressed as$$\dfrac {2\Gamma(2a)\{\Gamma(a+n)\}^2}{\{\Gamma(a)\}^2\Gamma(n)\Gamma(2a+n)}\left\{\dfrac a{a^2+x^2}-\dfrac {2a}{1!}\dfrac {n-1}{n+2a}\dfrac {a+1}{(a+1)^2+x^2}+\dfrac {2a(2a+1)}{2!}\dfrac {(n-1)(n-2)}{(n+2a)(n+2a+1)}\dfrac {a+2}{(a+2)^2+x^2}\right\}$$ Which I do not understand how the Gamma function made its way into the expansion.

  • 0
    Hint: They multiplied through such that there were no fractions in the denominators. After you get through that hint, Noticed that what we multiplied through was a [multifactorial](https://en.wikipedia.org/wiki/Factorial#Multifactorials) squared. – 2017-02-04

0 Answers 0