The following integral appears in Gradshteyn & Ryzhik (8th ed.) 3.462.1

\begin{equation} \int_{0}^{\infty}x^{\nu-1}\, e^{-\beta x^{2}-\gamma x} \ \mathrm{d}x=\frac{\Gamma(\nu)}{(2\beta)^{\nu/2}}\, \exp\left\{\frac{\gamma^{2}}{8\beta}\right\}\, D_{-\nu}\left(\frac{\gamma}{\sqrt{2\beta}}\right) \quad\quad [\mathrm{Re} \ \beta>0, \ \ \mathrm{Re} \ \nu>0] \end{equation}

where, $D_{\nu}(z)$ is the parabolic cylinder function.

Does anyone know where a proof of this solution exists? I am interested in how this integral can be solved.

Additional Information:

The integral

\begin{equation} \int e^{-a^{2}x^{2}+bx} \ \mathrm{d}x=\frac{\sqrt{\pi}}{2a}\,\exp\left(\frac{b^{2}}{4a^{2}}\right)\, \mathrm{erf}\left(ax-\frac{b}{2a}\right) \end{equation}

Source: http://nvlpubs.nist.gov/nistpubs/jres/73B/jresv73Bn1p1_A1b.pdf

This could perhaps lead to an integration by parts solution.

  • $\begingroup$ Would it suffice to differentiate the right hand side (and obtain the integrand)? $\endgroup$ – hardmath Aug 12 '16 at 15:41
  • 1
    $\begingroup$ @hardnath It's a definite integral. $\endgroup$ – Robert Israel Aug 12 '16 at 15:48
  • $\begingroup$ Maybe you could add the definition of parabolic cylinder functions here, since they are not universally known $\endgroup$ – Yuriy S Aug 27 '16 at 21:58

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