Computation of integral

Question

Consider the following integral: $$ I = \int_{0}^{\infty}{\int_{0}^{\infty}{x e^{-\left(2-\frac{1}{b^2}\right) x^2} e^{-\frac{1}{2}(y + a x)^2}} dy dx}$$

I was able to integrate over $y$ to get:

$$ I = \sqrt{\frac{\pi}{2}} \int_{0}^{\infty}{x e^{-\left(2-\frac{1}{b^2}\right) x^2} (1-erf(\frac{ax}{\sqrt{2}})) dx}$$

However, I find it difficult to integrate over $x$. Is it possible to compute the above integral?

Answer 1

Assuming that $b > 1/\sqrt {2}$ and using Maple I am obtaining

In order to derive the result by hand using pen and paper please write

$$I = \sqrt{\frac{\pi}{2}}( \int_{0}^{\infty}{x e^{-\left(2-\frac{1}{b^2}\right) x^2}dx- \int_{0}^{\infty}xe^{-\left(2-\frac{1}{b^2}\right) x^2} erf(\frac{ax}{\sqrt{2}}) dx})$$

The first integral is computed directly. The second integral is computed by parts with

$$u=erf(\frac{ax}{\sqrt{2}})$$

$$dv=xe^{-\left(2-\frac{1}{b^2}\right) x^2}dx$$

Answer 2

In retrospect, this takes a lot longer than integrating in $y$ first. But anyway, by completing the square of the quadratic form in the exponential one can solve it by integrating in $x$ first. We write \begin{equation} I = \int_{0}^{\infty}\int_{0}^{\infty}xe^{-\left[ (2 + a^{2}/2 - 1/b^{2})x^{2} + ayx + y^{2}/2\right]}\,dx\,dy. \end{equation} Define $\gamma^{2} := 2 + \frac{a^{2}}{2}- \frac{1}{b^{2}}$, which we assume to be strictly positive. Then we write \begin{equation} \left(2 + \frac{a^{2}}{2} - \frac{1}{b^{2}}\right)x^{2} + ayx + \frac{y^{2}}{2} = \left(\gamma x + \frac{ay}{2\gamma}\right)^{2} + \left(\frac{1}{2} - \frac{a^{2}}{4\gamma^{2}}\right)y^{2}. \end{equation}

Now use the substitution $u = \gamma x + \frac{ay}{2\gamma}$ so that $x = \frac{u}{\gamma} - \frac{ay}{2\gamma^{2}}$. We have \begin{align*} I & = \frac{1}{\gamma} \int_{y=0}^{\infty}\int_{u=\frac{ay}{2\gamma}}^{\infty}\left(\frac{u}{\gamma} - \frac{ay}{2\gamma^{2}}\right)e^{-u^{2}}e^{-\left(\frac{1}{2} - \frac{a^{2}}{4\gamma^{2}}\right)y^{2}}\,du\,dy\\ & = \frac{1}{\gamma^{2}}\int_{y=0}^{\infty}e^{\frac{-a^{2}y^{2}}{4\gamma^{2}}}e^{\frac{a^{2}y^{2}}{4\gamma^{2}}}e^{-\frac{y^{2}}{2}}\,dy - \frac{a}{2 \gamma^{3}} \int_{y=0}^{\infty}\int_{u=\frac{ay}{2\gamma}}^{\infty}ye^{-u^{2}}e^{-\left(\frac{1}{2} - \frac{a^{2}}{4\gamma^{2}}\right)y^{2}}\,du\,dy\\ & = \frac{1}{\gamma^{2}}\sqrt{\frac{\pi}{2}} - \frac{a}{2 \gamma^{3}} \int_{u=0}^{\infty}\int_{y=0}^{\frac{2\gamma u}{a}}ye^{-u^{2}}e^{-\left(\frac{1}{2} - \frac{a^{2}}{4\gamma^{2}}\right)y^{2}}\,dy\,du \end{align*} Define $\beta^{2} := \frac{1}{2} - \frac{a^{2}}{4\gamma^{2}} = \frac{2 - \frac{1}{b^{2}}}{4 + a^{2} - \frac{2}{b^{2}}}$, which is positive if $\gamma^{2}$ is. Also note the identities: \begin{align*} \gamma^{2}\beta^{2} & = 1 - \frac{1}{2b^{2}}\\ a^{2} + 4\gamma^{2}\beta^{2} & = 2\gamma^{2}. \end{align*}

Make the substitution $w = \beta y$ to get \begin{align*} I &= \frac{1}{\gamma^{2}}\sqrt{\frac{\pi}{2}} - \frac{a}{2 \gamma^{3}\beta^{2}} \int_{u=0}^{\infty}\int_{w=0}^{\frac{2\gamma \beta u}{a}}we^{-u^{2}}e^{-w^{2}}\,dw\,du\\ & = \frac{1}{\gamma^{2}}\sqrt{\frac{\pi}{2}} - \frac{a}{4\gamma^{3}\beta^{2}}\int_{u=0}^{\infty}\int_{s=0}^{\frac{4\gamma^{2}\beta^{2}u^{2}}{a^{2}}}e^{-s}\,ds\,du, \quad (s = w^{2})\\ & = \frac{1}{\gamma^{2}}\sqrt{\frac{\pi}{2}} - \frac{a}{4\gamma^{3}\beta^{2}}\int_{u=0}^{\infty}e^{-u^{2}}\left(1 - e^{-\frac{4\gamma^{2}\beta^{2}u^{2}}{a^{2}}}\right)\,du\\ & = \frac{1}{\gamma^{2}}\sqrt{\frac{\pi}{2}} - \frac{a}{4\gamma^{3}\beta^{2}}\frac{\sqrt{\pi}}{2} + \frac{a}{4\gamma^{3}\beta^{2}}\int_{u=0}^{\infty}e^{-\left(1 + \frac{4\gamma^{2}\beta^{2}}{a^{2}}\right)u^{2}}\,du\\ & = \frac{1}{\gamma^{2}}\sqrt{\frac{\pi}{2}} - \frac{a}{4\gamma^{3}\beta^{2}}\frac{\sqrt{\pi}}{2} + \frac{a}{4\gamma^{3}\beta^{2}}\sqrt{\frac{\pi}{1 + \frac{4\gamma^{2}\beta^{2}}{a^{2}}}} \end{align*}

To further simplify, we proceed: \begin{align*} I & = \frac{1}{\gamma^{2}}\sqrt{\frac{\pi}{2}} - \frac{a}{4\gamma^{3}\beta^{2}}\frac{\sqrt{\pi}}{2} + \frac{a}{4\gamma^{3}\beta^{2}}\sqrt{\frac{\pi}{1 + \frac{4\gamma^{2}\beta^{2}}{a^{2}}}}\\ & = \frac{\sqrt{\pi}}{\gamma}\left(\frac{1}{\sqrt{2}\gamma} - \frac{a}{8\gamma^{2}\beta^{2}} + \frac{a^{2}}{4\gamma^{2}\beta^{2}}\sqrt{\frac{1}{2\gamma^{2}}} \right)\\ & = \frac{\sqrt{\pi}}{\gamma}\left(\frac{1}{\sqrt{2}\gamma}\left[1 + \frac{a^{2}}{4\gamma^{2}\beta^{2}}\right] - \frac{a}{8\gamma^{2}\beta^{2}}\right)\\ & = \frac{\sqrt{\pi}}{\gamma}\left(\frac{1}{\sqrt{2}\gamma}\left[\frac{4\gamma^{2}\beta^{2} + a^{2}}{4\gamma^{2}\beta^{2}}\right] - \frac{a}{8\gamma^{2}\beta^{2}}\right)\\ & = \frac{\sqrt{\pi}}{\gamma}\left(\frac{1}{\sqrt{2}\gamma}\left[\frac{2\gamma^{2}}{4\left(1 - \frac{1}{2b^{2}}\right)}\right] - \frac{a}{8\left(1 - \frac{1}{2b^{2}}\right)}\right)\\ & = \frac{\sqrt{\pi}}{\gamma}\left(\frac{2\sqrt{2}\gamma - a}{8\left(1 - \frac{1}{2b^{2}}\right)}\right)\\ \end{align*}

Note that $\sqrt{2}b\gamma = \sqrt{4b^{2} + a^{2}b^{2} - 2}$. So \begin{align*} I & = \frac{\sqrt{\pi}\sqrt{2}b}{\sqrt{2}b\gamma}\left(\frac{2\sqrt{2}b\gamma - ab}{8b\left(\frac{2b^{2} - 1}{2b^{2}}\right)}\right)\\ & = \frac{\sqrt{\pi}\sqrt{2}}{\sqrt{4b^{2} + a^{2}b^{2} - 2}}\left(\frac{2\sqrt{4b^{2} + a^{2}b^{2} - 2} - ab}{8\left(\frac{2b^{2} - 1}{2b^{2}}\right)}\right)\\ & = \frac{\sqrt{\pi}\sqrt{2}b^{2}}{\sqrt{4b^{2} + a^{2}b^{2} - 2}}\left(\frac{2\sqrt{4b^{2} + a^{2}b^{2} - 2} - ab}{4\left(2b^{2} - 1\right)}\right)\\ \end{align*}

I can't seem to locate where the extra $2$ is coming from, which is the only difference between the final equation below and what was obtained in the other contributor's solution. Obviously, this is the longer way to do this problem, and it's more prone to mistakes. So I think I'll leave it as is.