MHB Evaluating Double Integral $II_{5d}$

[karush]

$\textsf{d. Evaluate :}\\$
\begin{align*}\displaystyle
II_{5d}&=\int_{-\infty}^{+\infty}
\int_{-\infty}^{+\infty}
\frac{1}{(x^2+1)(y^2+1)}
\, dy dx
\end{align*}

 

[MarkFL]

Re: double Integral

First, I would begin by computing:

$$I_1=\int_{-\infty}^{\infty}\frac{1}{u^2+1}\,du=\lim_{t\to\infty}\left(\int_{-t}^{t}\frac{1}{u^2+1}\,du\right)=\lim_{t\to\infty}\left(\left[\arctan(t)-\arctan(-t)\right]_{-t}^{t}\right)=2\lim_{t\to\infty}\left(\arctan(t)\right)=2\cdot\frac{\pi}{2}=\pi$$

And now we can write:

$$I=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{\left(x^2+1\right)\left(y^2+1\right)}\,dy\,dx=\int_{-\infty}^{\infty}\frac{1}{x^2+1}\int_{-\infty}^{\infty}\frac{1}{y^2+1}\,dy\,dx$$

Can you now express $I$ as a function of $I_1$?

 

[karush]

Re: double Integral

so would this simply be:

$$\pi \cdot \pi = \pi^2$$

 

[MarkFL]

Re: double Integral

so would this simply be:

$$\pi \cdot \pi = \pi^2$$

Yes, that's correct. :)

 

[karush]

Re: double Integral

ahhh progress