1
$\begingroup$

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 !

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.