Integral of error-like function

Question

I have some trouble evaluating the following integral:

$$\int_{v=0}^{\infty}\int_{u=0}^{\infty}\frac{\sqrt{uv}}{(u+v)^2}e^{-\frac{(u+v)}{2}}dudv.$$

I know that the value of this double integral is $\dfrac{\pi}{4}$.

However, I tried to solve it by using the change of variables: $t=u+v$ and separate the variables by bringing $v$ out. Doing so yields

$$\int_{v=0}^{\infty}\sqrt{v}\int_{t=v}^{\infty}\frac{\sqrt{t-v}}{t^2}e^{-\frac{t}{2}}dtdv,$$

but it does not seem to get into the form of Gaussian's error function.

Answer 1

$\def\d{\mathrm{d}} \def\e{\mathrm{e}}$By Tonelli's theorem,\begin{align*} \int_0^{+\infty} \int_v^{+\infty} \frac{\sqrt{v(t - v)}}{t^2} \e^{-\frac{t}{2}} \,\d t \d v &= \int_0^{+\infty} \int_0^t \frac{\sqrt{v(t - v)}}{t^2} \e^{-\frac{t}{2}} \,\d v \d t\\ &= \int_0^{+\infty} \e^{-\frac{t}{2}} \d t \int_0^t \frac{\sqrt{v(t - v)}}{t^2} \,\d v. \tag{1} \end{align*} For a fixed $t > 0$, make substitution $\displaystyle v = t \sin^2 θ \ \left( 0 < θ < \frac{π}{2} \right)$ to get\begin{align*} \int_0^t \frac{\sqrt{v(t - v)}}{t^2} \,\d v &= \int_0^{\frac{π}{2}} \frac{\sqrt{t \sin^2 θ(t - t \sin^2 θ)}}{t^2} \cdot 2t \sin θ \cos θ \,\d θ\\ &= 2 \int_0^{\frac{π}{2}} \sin^2 θ \cos^2 θ \,\d θ. \end{align*} Therefore\begin{align*} (1) &= 2 \int_0^{+\infty} \e^{-\frac{t}{2}} \d t \int_0^{\frac{π}{2}} \sin^2 θ \cos^2 θ \,\d θ\\ &= 2 \left( \int_0^{+\infty} \e^{-\frac{t}{2}} \d t \right) \left( \int_0^{\frac{π}{2}} \sin^2 θ \cos^2 θ \,\d θ \right). \end{align*}

Answer 2

Using the integral representation $$\frac1{(u+v)^2}e^{-\frac{(u+v)}{2}}=\frac12\int^\infty_{1/2}t\,e^{-(u+v)t}\,dt$$ and the theorem of Fubini, we obtain \begin{align}\int_0^{\infty}\int_0^{\infty}\frac{\sqrt{uv}}{(u+v)^2}e^{-\frac{(u+v)}{2}}\,du\,dv&=\frac12\int_0^{\infty}\int_0^{\infty}\sqrt{uv}\int^\infty_{1/2}t\,e^{-(u+v)t}\,dt\,du\,dv\\&=\frac12\int^\infty_{1/2}t\int_0^{\infty}\int_0^{\infty}\sqrt{uv}\,e^{-(u+v)t}\,du\,dv\,dt \\&=\frac12\int^\infty_{1/2}t\left(\Gamma(3/2)\,t^{-3/2}\right)^2\,dt \\&=\frac{\pi}8\int^\infty_{1/2}t^{-2}\,dt=\frac{\pi}4\end{align} where we used the integral representation of the Gamma function and the well-known $\Gamma(3/2)=\dfrac{\sqrt{\pi}}2$.

Answer 3

Here is a slight variation to @Alex Francisco's solution that uses a change of variables for the double integral directly.

Let $$I = \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{\sqrt{uv}}{(u + v)^2} \exp \left (-\frac{u + v}{2} \right ) \, du dv.$$

Consider the change of variables $u(x,y) = x \cos^2 y$ and $v(x,y) = x \sin^2 y$. The Jacobian for this change of variables is $$\frac{\partial (u,v)}{\partial (x,y)} = 2x \sin y \cos y,$$ while for the semi-infinite region in the first quadrant ($x,y \geqslant 0$) under such a transformation the limits of integration become $0 \leqslant x < \infty$ and $0 \leqslant y \leqslant \pi/2$. Finally, noting that $$u + v = x \quad \text{and} \quad uv = x^2 \sin^2 y \cos^2 y,$$ since $$\int \!\!\!\int_{\mathcal{R}} f(u,v) \, d(u,v) = \int \!\!\!\int_{\mathfrak{R}} f(x,y) \left |\frac{\partial (u,v)}{\partial (x,y)} \right | \, d(x,y),$$ where $\mathcal{R}$ is the region in the $uv$-plane while $\mathfrak{R}$ is the region in the $xy$-plane, the integral becomes $$I = 2 \int_0^\infty e^{-x/2} \, dx \int_0^{\pi/2} \sin^2 y \cos^2 y \, dy,$$ as given by @Alex Francisco.