# Hypergeometric Function Proof

Supposing $p+2 < q$ and $\mu > 0$, I am trying to show with formal integration that $$\int^{\infty}_{0}\ _pF_q(a_1\dots a_p,b_1\dots b_q;\beta x)e^{-\alpha x}x^{\mu-1}dx$$ $$=\frac{\Gamma(\mu)}{\alpha^\mu}\ _{p+1}F_q \Bigr(a_1\dots a_p \mu,b_1\dots b_q;\frac{\beta}{\alpha}\Bigr).$$

This does seem quite difficult to me but I am only beginning these topics so I would like to learn really. My first idea is to replace the geometric function in the first equation using the relation $$\ _pF_q(a_1\dots a_p,b_1\dots b_q;\beta x) = \sum^{\infty}_{k=0}\frac{(a_1)_k\dots(a_p)_k}{(b_1)_k\dots(b_q)_k} \frac{(\beta x)^k}{k!}.$$

This then gives us $$\int^{\infty}_{0}\sum^{\infty}_{k=0}\frac{(a_1)_k\dots(a_p)_k}{(b_1)_k\dots(b_q)_k} \frac{(\beta x)^k}{k!}e^{-\alpha x}x^{\mu-1}dx.$$ My next guess would've been to take the summation out of the integral formally and then perhaps try to simplify, however I cannot remove the $(\beta x)^k$ term as this depends on $x$ but ofcourse it also depends on $k$ so I guess this means I cannot take the summation outside. In which case I am a little stumped on what the next step should be...

Any help is greatly appreciated !

In keeping with what is commonly used in the literature on the hypergeometric function, we will use the following notation $(a)_n$ to denote the Pochhammer symbol for the rising factorial.
Following your idea of replacing the hypergeometric function $_p F_q (a_1,\ldots,a_p;b_1,\ldots,b_q; \beta x)$ by its Maclaurin series representation of $$_p F_q(a_1,\dots, a_p;b_1,\dots, b_q;\beta x) = \sum^{\infty}_{k=0}\frac{(a_1)_k\cdots(a_p)_k}{(b_1)_k \cdots (b_q)_k} \frac{(\beta x)^k}{k!},$$ substituting this form into the integral, after interchanging the summation with the integral sign, one has $$I = \sum_{k = 0}^\infty \frac{(a_1)_k \cdots (a_p)_k}{(b_1)_k \cdots (b_q)_k} \frac{\beta^k}{k!} \int_0^\infty e^{-\alpha x} x^{k + \mu - 1} \, dx.$$ Enforcing a substitution of $x \mapsto x/\alpha$ in the integral yields \begin{align*} I &= \sum_{k = 0}^\infty \frac{(a_1)_k \cdots (a_p)_k}{(b_1)_k \cdots (b_q)_k} \frac{\beta^k}{k!} \frac{1}{\alpha^{k + \mu}} \int_0^\infty e^{-x} x^{k + \mu - 1} \, dx\\ &= \frac{1}{\alpha^\mu} \sum_{k = 0}^\infty \frac{(a_1)_k \cdots (a_p)_k}{(b_1)_k \cdots (b_q)_k} \left (\frac{\beta}{\alpha} \right )^k \frac{1}{k!} \Gamma (k + \mu). \tag1 \end{align*}
Now from properties for the rising factorial it is known that in terms of Gamma functions it can be expressed as $$(x)_n = \frac{\Gamma (x + n)}{\Gamma (x)}.$$ Thus we can write $$(\mu)_k = \frac{\Gamma (k + \mu)}{\Gamma (\mu)} \quad \text{or} \quad \Gamma (k + \mu) = \Gamma (\mu) (\mu)_k.$$
So we may rewrite (1) as $$I = \frac{\Gamma (\mu)}{\alpha^\mu} \sum_{k = 0}^\infty \frac{(a_1)_k \cdots (a_p)_k (\mu)_k}{(b_1) \cdots (b_q)_k} \left (\frac{\beta}{\alpha} \right )^k \frac{1}{k!},$$ or $$\int^{\infty}_{0}\ _pF_q(a_1,\dots,a_p;b_1,\dots,b_q;\beta x)e^{-\alpha x}x^{\mu-1}dx = \frac{\Gamma (\mu)}{\alpha^\mu} \ _{p + 1}F_q \left (a_1, \ldots, a_p, \mu; b_1, \ldots ,b_q; \frac{\beta}{\alpha} \right ),$$ as required to show.