Proof of 3.461.5 of Gradshteyn Book needed

Question

The 3.461.5 Equation of Gradshteyn and Ryzhik book is as given below $$\int_u^{\infty}e^{-\mu x^2}x^{-2}\, dx=\frac{1}{u}e^{-\mu u^2}-\sqrt{\mu \pi}\left[1-\Phi(\sqrt{\mu u})\right]$$ where $\Phi(x)$ is defined as $$\Phi(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt$$ I want to know how to proof this equation? Any hint or reference for this will be very helpful. I will be very thankful to you for your help. Thanks in advance.

Answer 1

Let $v = e^{-\mu x^2}$ and $dw = dx/x^2$. Then integrating by parts we have \begin{align} \int_u^\infty e^{-\mu x^2} \frac{dx}{x^2} &= \left[ e^{-\mu x^2} \left( - \frac{1}{x} \right) \right]_u^\infty - \int_u^\infty \left( - \frac{1}{x} \right) \left( - 2 \mu x e^{-\mu x^2} dx \right) \\ &= \left[ 0 - \left( -\frac{1}{u} e^{-\mu u^2} \right) \right] - 2 \mu \int_u^\infty e^{-\mu x^2} dx \\ &= \frac{1}{u} e^{-\mu u^2} - 2 \mu \int_{\sqrt{\mu} u}^\infty e^{-t^2} \left( \frac{dt}{\sqrt{\mu}} \right) & (t \equiv \sqrt{\mu} x) \\ &= \frac{1}{u} e^{-\mu u^2} - 2 \sqrt{\mu} \left[ \int_0^\infty e^{-t^2} dt - \int_0^{\sqrt{\mu} u} e^{-t^2} dt \right] \end{align} The first integral that remains is well-known to be equal to $\sqrt{\pi}/2$, and the second one is equal to $\frac{1}{2}\sqrt{\pi} \Phi(\sqrt{\mu} u)$. Plugging these in, we achieve the desired result.

(Note that the argument of $\Phi$ in the integral should be $\sqrt{\mu} u$, rather than $\sqrt{\mu u}$. I assume this was a typo you made in entering the equation—it's also one I made a few times while typing this up.)

Answer 2

\begin{align} I(\mu) &= \int\limits_{u}^{\infty} \mathrm{e}^{-\mu x^{2}} \mathrm{d}x \\ &= \frac{1}{\sqrt{\mu}} \int\limits_{u\sqrt{\mu}}^{\infty} \mathrm{e}^{-y^{2}} \mathrm{d}y \\ &= \frac{1}{2}\sqrt{\frac{\pi}{\mu}} \mathrm{erfc}(u\sqrt{\mu}) \\ &= \frac{1}{2}\sqrt{\frac{\pi}{\mu}} - \frac{1}{2}\sqrt{\frac{\pi}{\mu}}\mathrm{erf}(u\sqrt{\mu}) \end{align} We let $y^{2} = \mu x^{2}$.

Integrating with respect to $\mu$ we have \begin{equation} \int\limits_{u}^{\infty} \int \mathrm{e}^{-\mu x^{2}} \mathrm{d}\mu \mathrm{d}x = -\int\limits_{u}^{\infty} x^{-2} \mathrm{e}^{-\mu x^{2}} \mathrm{d}x \tag{1} \label{eq:1} \end{equation} and \begin{equation} \frac{\sqrt{\pi}}{2}\int\frac{1}{\sqrt{\mu}} \mathrm{d}\mu - \frac{\sqrt{\pi}}{2} \int\frac{1}{\sqrt{\mu}}\mathrm{erf}(u\sqrt{\mu}) \mathrm{d}\mu = \frac{\sqrt{\pi}}{2} I_{1} - \frac{\sqrt{\pi}}{2} I_{2} \tag{2} \label{eq:2} \end{equation}

\begin{equation} I_{1} = \int \frac{1}{\sqrt{\mu}} \mathrm{d}\mu = 2\sqrt{\mu} \end{equation}

\begin{align} I_{2} &= \int \frac{1}{\sqrt{\mu}} \mathrm{erf}(u\sqrt{\mu}) \mathrm{d}\mu \\ &= \frac{2}{u} \int \mathrm{erf}(x) \mathrm{d}x \\ &= \frac{2}{u} x\, \mathrm{erf}(x) + \frac{2}{u} \frac{1}{\sqrt{\pi}} \mathrm{e}^{-x^{2}} \\ &= 2\sqrt{\mu} \mathrm{erf}(u\sqrt{\mu}) + \frac{2}{u} \frac{1}{\sqrt{\pi}} \mathrm{e}^{-\mu u^{2}} \end{align} We let $x=u\sqrt{\mu}$.

Inserting $I_{1}$ and $I_{2}$ into equation \eqref{eq:2}, equating the result with equation \eqref{eq:1}, and simplifying, we have

\begin{align} \int\limits_{u}^{\infty} x^{-2} \mathrm{e}^{-\mu x^{2}} \mathrm{d}x &= \frac{1}{u} \mathrm{e}^{-\mu u^{2}} - \sqrt{\pi \mu} [1 - \mathrm{erf}(u\sqrt{\mu})] \\ &= \frac{1}{u} \mathrm{e}^{-\mu u^{2}} - \sqrt{\pi \mu} \mathrm{erfc}(u\sqrt{\mu}) \end{align}