3
$\begingroup$

I would like to find a "closed form" for the integral

$$I(\beta) = \int_0^\infty \frac{x^\alpha(1-x)^\beta}{(x-c)^\gamma(x-b)^\gamma(x-\bar{b})^\gamma}\mathrm{d}x$$ where $\gamma=3/2$, $\alpha,\beta>0$ , $\{b,\bar{b}\}$ is a pair of complex conjugates, and $c$ is real.

Defining $a\equiv -\alpha -\beta +7/2$ and supposing $a>0$, we can write $I$ as $[1]$ $$ I(\beta) = (-1)^\beta B(a,\alpha+1) R\left(a;-\beta,\frac32,\frac32,\frac32;-1,-c,-b,-\bar{b}\right)$$

where $B$ is the Beta function, and $R$ is a generalized Carlson elliptic function. My question is, can this expression be simplified? Is there a way to write this R function in a "more closed form"? For instance, in terms of hypergeometric functions?

In particular, I am interested in taking the derivative with respect to $\beta$ and then take the limit $\beta \rightarrow 0$ (the original integral has a log term). As noted here,

$$ \int_0^{\infty}\!\!\!\!\frac{x^{\alpha-1}\ln^n(cx+d)}{(ax+b)^\sigma(cx+d)^\rho} \! \mathrm dx=(-1)^n\left(\!\frac{d}{c}\!\right)^\alpha\!\! b^{-\sigma}\!\frac{\partial^n}{\partial \rho^n}\!\!\left[d^{-\rho}B(\alpha,\sigma+\rho-\alpha) \ _2F_1\!\!\left(\!\sigma,\alpha;\sigma+\rho;1\!-\!\frac{ad}{bc}\!\right)\!\right] $$ so maybe there is a similar formula for $I(\beta)$ in terms of the Beta function and some hypergeometric function.

$[1]$: By using equation $(\text{T}.2)$ in page $224$.

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.