4
$\begingroup$

As per the title, I am looking for a closed-form expression for the integral

$$\frac{1}{B(\alpha,\beta)}\int_{0}^{1}e^{-ax(1 - bx )}x^{\alpha-1}(1-x)^{\beta - 1}dx$$

where $a,\alpha,\beta>0$ and $b \in [0,\frac{1}{2}]$. The factor $\frac{1}{B(\alpha,\beta)} x^{\alpha-1}(1-x)^{\beta - 1}$ is the beta probability density function where $B$ denotes the beta function. For $b=0$, the above is an integral representation of the confluent hypergeometric function $_{1}F_{1}(\alpha,\alpha+\beta,-a)$.

My first attempts at evaluating this by numerical integration in MATLAB indicate that it will be problematic for $\alpha,\beta < 1$, i.e. when the beta function goes to infinity at the endpoints of the interval. If there is a closed-form expression in terms of special functions, it may be easier to evaluate numerically.

$\endgroup$
1
$\begingroup$

The problematic component of the proposed integral is the term $e^{ab x^2}$. This can be eliminated by expanding this exponential into the corresponding power series for which the integral in question becomes \begin{align} I &= \sum_{n=0}^{\infty} \frac{(ab)^{n}}{n!} \, \int_{0}^{1} \, e^{-ax} \, x^{2n + \mu-1} \, (1-x)^{\nu - 1} \, dx \\ &= \sum_{n=0}^{\infty} \frac{(ab)^{n}}{n!} \, B(2n+\mu, \nu) \, {}_{1}F_{1}(2n+\mu; 2n+\mu+\nu; -a) \\ &= B(\mu, \nu) \, \sum_{n=0}^{\infty} \sum_{r=0}^{\infty} \frac{(\mu)_{2n} (2n+\mu)_{r} }{ (\mu + \nu)_{2n} (2n+\mu + \nu)_{r} } \, \frac{(ab)^{n}}{n!} \, \frac{(-a)^{r}}{r!} \\ &= B(\mu, \nu) \, \sum_{n=0}^{\infty} \sum_{r=0}^{\infty} \frac{(\mu)_{2n+r}} { (\mu + \nu)_{2n+r} } \, \frac{(ab)^{n}}{n!} \, \frac{(-a)^{r}}{r!} \\ &= B(\mu, \nu) \, F_{1:0}^{1:0} \left[ \begin{array}{cc} [(\mu):2,1]: - ; \\ [(\mu + \nu):2,1]: - ; \end{array} \hspace{3mm} ab , \, -a \right] \end{align} where the last function, $F$, is the Srivastava-Daoust function. From this it can be stated that \begin{align} \frac{1}{B(\mu, \nu)} \, \int_{0}^{1} \, e^{-ax(1-bx)} \, x^{\mu-1} \, (1-x)^{\nu - 1} \, dx = F_{1:0}^{1:0} \left[ \begin{array}{cc} [(\mu):2,1]: - ; \\ [(\mu + \nu):2,1]: - ; \end{array} \hspace{3mm} ab , \, -a \right] \end{align}


The Srivastava-Daoust Function is defined by formula (1.1) in the article by Rekha Panda

$\endgroup$
  • 2
    $\begingroup$ Nice! I've never heard of the Srivastava-Daoust function and it appears that neither did MATLAB... But since I can evaluate the beta function and $_{1}F_{1}$ (slow, but reliable), I'll hopefully be able to truncate the series in the second line to obtain a good approximation. $\endgroup$ – vgnils May 4 '15 at 19:33
  • $\begingroup$ What is Srivastava-Daoust function? $\endgroup$ – Harry Peter May 6 '15 at 13:23
  • $\begingroup$ A note has been added that defines what the Srivastava-Daoust function is and some applications. $\endgroup$ – Leucippus May 6 '15 at 17:15

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.