I try to express this integral for the squared lower incomplete gamma function in terms of the same lower incomplete gamma and other common functions $$\int x^{b-1}\gamma(a,x)^2dx\;\;\;(1)$$
When there is no square it is relatively easy to prove $$\int x^{b-1}\gamma(a,x)dx = \frac{1}{b}[x^b\gamma(a,x)-\gamma(a+b,x)]$$ by integrating by parts, starting with change of variables $$u=\gamma(a,x); dv=x^{b-1}dx$$ and then calculating $$\int udv = uv-\int vdu$$
Similar way I'm trying to apply for the first integral (1) by variables change $$u=\gamma(a,x)^2;dv=x^{b-1}dx$$Then I get $$\int x^{b-1}\gamma(a,x)^2dx = \frac{1}{b}[x^b\gamma(a,x)^2-2\int\gamma(a,x)e^{-x}x^{a+b-1}dx]\;\;\;(2)$$Next step again I use change of variables for the integral from right side $$u=\gamma(a,x); dv=e^{-x}x^{a+b-1}dx$$ and then I get$$\int x^{b-1}\gamma(a,x)^2dx = \frac{1}{b}[x^b\gamma(a,x)^2-2(\gamma(a,b)\gamma(a+b,x)-\int\gamma(a+b,x)e^{-x}x^{a-1}dx)]$$ and here I'm stuck. If I try to apply similar integration by parts to the integral from right side I'm getting back the equation (2) Is there a way to get rid of integral on right hand side?
Integral of squared lower incomplete gamma function
gamma functionintegration
Best Answer
Let’s try a series representation and see if it can be simplified. The series seems to work for a large radius of convergence when integrating by each term:
$$γ(a,z)=-z^a \sum_{n=0}^\infty \frac{(-z)^n}{(a+n)n!},\Gamma(a,z)=\int e^{-x} x^{a-1}dx$$
Therefore: $$\int γ(a+b,x)e^{-x} x^{a-1}dx=-\int e^{-x} x^{a-1}x^{a+b}\sum_{n=0}^\infty \frac{(-1)^nx^n}{(a+b+n)n!}= -\sum_{n=0}^\infty \frac{(-1)^n}{(a+b+n)n!} \int e^{-x}x^{2a+b+n-1}dx= C+\sum_{n=0}^\infty \frac{(-1)^n\Gamma(2a+b+n,x)}{(a+b+n)n!} $$
Here is a technically closed form with the Incomplete Gauss Hypergeometric function which has many papers like these about it, so it is an official special function:
$$\,_2\Gamma_1((a_1,k),a_2;b_1;z)\mathop=^\text{def} \sum_{n=0}^\infty \frac{\Gamma(a_1+n,k)(a_2)_nz^n}{\Gamma(a_1)(b_1)n!}\implies \sum_{n=0}^\infty \frac{\Gamma(a_1+n,k)\Gamma(a_2+n)z^n}{\Gamma(b_1+n)n!}= \frac{\Gamma(a_1)\Gamma(a_2)}{\Gamma(b_1)} \,_2\Gamma_1((a_1,k),a_2;b_1;z),(0)_0\to 1$$
where $(a)_n=\frac{\Gamma(a+n)}{\Gamma(a)}$ is a Pochhammer Symbol:
$$\int γ(a+b,x)e^{-x} x^{a-1}dx= C+\sum_{n=0}^\infty \frac{(-1)^n\Gamma(2a+b+n,x)}{(a+b+n)n!} = C+\sum_{n=0}^\infty \frac{(-1)^n\Gamma(2a+b+n,x) \Gamma(a+b+n)}{\Gamma(a+b+1+n)n!} =\frac{\Gamma(2a+b)}{a+b}\,_2\Gamma_1((2a+b,x),a+b;a+b+1;-1)+C$$ Please correct me and give me feedback!