Question

In the integral

$\mathbf{I}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}{{\left(x^2+y^2\right)}^{\mathbf{2}}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{xy}}\mathbf{dxdy}$.

Make the change of variables

$$\begin{gathered} \mathbf{u}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}} \\ \mathbf{v}\mathbf{=}\mathbf{2}\mathbf{xy} \end{gathered}$$

And evaluate I. Hint: Use (4.8) and the accompanying discussion.

Short answer

The evaluation of the given integral upon changing the variables is as follows,

$\mathrm{I}={\int }_0^{\infty }{\int }_0^{\infty }\frac{{\mathrm{x}}^2+{\mathrm{y}}^2}{{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2}{\mathrm{e}}^{-2\mathrm{xy}}\mathrm{dxdy}$

$=\frac{\mathrm{\tau \tau }}{4}$ .

Step-by-step solution

Given the condition

Theintegral given is as follows,

$\mathrm{I}={\int }_0^{\infty }{\int }_0^{\infty }\frac{{\mathrm{x}}^2+{\mathrm{y}}^2}{{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2}{\mathrm{e}}^{-2\mathrm{xy}}\mathrm{dxdy}$

The change of variables are given to be the following,

$$\begin{gathered} \mathrm{u}={\mathrm{x}}^2-{\mathrm{y}}^2 \\ \mathrm{v}=2\mathrm{xy} \end{gathered}$$

Concept of the Jacobian

The Jacobian of x,ywith respect to s,tis

$$\begin{gathered} \mathbf{J}\mathbf{=}\mathbf{J}\mathbf{\left(}\frac{\mathbf{x}\mathbf{,}\mathbf{y}}{\mathbf{s}\mathbf{,}\mathbf{t}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{\partial }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}}{\mathbf{\partial }\mathbf{(}\mathbf{s}\mathbf{,}\mathbf{t}\mathbf{)}} \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{\left|}\begin{array}{cc}\frac{\mathbf{\partial }\mathbf{x}}{\mathbf{\partial }\mathbf{s}}& \frac{\mathbf{\partial }\mathbf{x}}{\mathbf{\partial }\mathbf{t}}\\ \frac{\mathbf{\partial }\mathbf{y}}{\mathbf{\partial }\mathbf{s}}& \frac{\mathbf{\partial }\mathbf{y}}{\mathbf{\partial }\mathbf{t}}\end{array}\mathbf{\right|} \end{gathered}$$

Determination of the Integral

The substitution of variables in the given integral is as follows,

$\mathrm{u}={\mathrm{x}}^2-{\mathrm{y}}^2,\mathrm{v}=2\mathrm{xy}$

$$\begin{gathered} \frac{\partial \left(\mathrm{u},\mathrm{v}\right)}{\partial \left(\mathrm{x},\mathrm{v}\right)}=\left|\begin{array}{cc}2\mathrm{x}& -2\mathrm{y}\\ 2\mathrm{y}& 2\mathrm{y}\end{array}\right| \\ =4{\mathrm{x}}^2=4{\mathrm{y}}^2 \\ =4\left({\mathrm{x}}^2+{\mathrm{y}}^2\right) \end{gathered}$$

$\begin{array}{l}\mathrm{dxdy}=\mathrm{J}\mathrm{dudv}\\ \mathrm{dxdy}=\frac{1}{4\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}\mathrm{dudv}\\ \left({\mathrm{x}}^2+{\mathrm{y}}^2\right)\mathrm{dudv}=\frac{1}{4}\mathrm{dudv}\end{array}$

$$\begin{gathered} \mathrm{I}={\int }_{\mathrm{v}=0}^{\infty }{\int }_{\mathrm{u}=0}^{\infty }\frac{{\mathrm{e}}^{-\mathrm{v}}}{4\left(1+{\mathrm{u}}^2\right)}\mathrm{dudv} \\ =\frac{1}{4}{\int }_{\mathrm{v}=0}^{\infty }{\mathrm{e}}^{-\mathrm{v}}\mathrm{dv}{\int }_{\mathrm{u}=0}^{\infty }\frac{1}{1+{\mathrm{u}}^2}\mathrm{du} \\ =\frac{1}{4}{\left({\mathrm{e}}^{-\mathrm{v}}\right)}^{\infty }{\left({\tan }^{-1}\mathrm{u}\right)}_{-\infty }^{\infty } \\ =\frac{1}{4}\left(\frac{\mathrm{\tau \tau }}{2}+\frac{\mathrm{\tau \tau }}{2}\right) \\ =\frac{\mathrm{\tau \tau }}{4} \end{gathered}$$

The integral

$$\begin{gathered} \mathrm{I}={\int }_0^{\infty }{\int }_0^{\infty }\frac{{\mathrm{x}}^2+{\mathrm{y}}^2}{{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2}{\mathrm{e}}^{-2\mathrm{xy}}\mathrm{dxdy} \\ =\frac{\mathrm{\tau \tau }}{4} \end{gathered}$$