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$$\int_{\mathbb{R}^2} \frac{1}{\left(1+\|x\|_2^2\right)^2}; \qquad \int_{\mathbb{R}^3} \frac{1}{\left(1+\|x\|_2^2\right)^{2}}; \qquad \int_{\mathbb{R}^n}\frac{1}{\left(1+\|x\|_2^2\right)^2} $$

I am not sure what the $\mathbb{R}$s subscripted to the integral mean. Do they mean that the function is integrated over the whole of $\mathbb{R}^2, \mathbb{R}^3, \mathbb{R}^n$, as in

$$\begin{align*} \int_{\mathbb{R}^2}\frac{1}{\left(1+\|x\|_2^2\right)^2} & = \int_{\mathbb{R}\times\mathbb{R}}\frac{1}{\left(1+ x_1^2+x_2^2\right)^2}dx\\ &= \int \left(\int \frac{1}{\left(1+x_1^2+x_2^2\right)^2}dx _{1}\right)dx _{2} \end{align*}$$

and

$$\begin{align*} \int_{\mathbb{R}^3} \frac{1}{\left(1+\|x\|_2^2\right)^2} & = \int_{\mathbb{R}\times\mathbb{R}\times \mathbb{R}}\frac{1}{\left(1+x_1^2+x_2^2+x_3^2\right)^2}dx\\ &= \int \left(\int \left(\int \frac{1}{\left(1+x_1^2+x_2^2\right)^2}dx_1\right)dx_2\right)dx_3\;? \end{align*}$$

What about the case $\mathbb{R}^n$? Does anybody see how to go at it? Please, do tell.

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Yes, in each case you’re to integrate the function over the whole space $\mathbb{R}^n$. –  Brian M. Scott Dec 5 '11 at 21:54
    
For the computations, you should use the polar coordinates trick, i.e. for a radial function $f(x)=\phi(\lVert x\rVert): $$\int_{\mathbb{R}^N} f(x)\ \text{d} x =N\ \omega_N\ \int_0^\infty \phi (r)\ r^{N-1}\ \text{d} r\; ,$$ where: $$\omega_N:=\text{volume of the unit ball in } \mathbb{R}^N =\frac{\pi^{N/2}}{\Gamma (1+N/2)}$$ ($\Gamma (\cdot)$ is Euler gamma function). where: ωN:=volume of the unit ball in RN=πN/2Γ(1+N/2). –  Pacciu Dec 5 '11 at 22:24
    
For the computations, you should use the polar coordinates trick, i.e. for radial function $f(x)=\phi (\lVert x\rVert)$: $$\int_{\mathbb{R}^N} f(x)\ \text{d} x = N\ \omega_N\ \int_0^\infty \phi (r)\ r^{N-1}\ \text{d} r \; ,$$ where: $$\omega_N:=\text{volume of the unit ball in } \mathbb{R}^N = \frac{\pi^{N/2}}{\Gamma (1+N/2)}$$ (and $\Gamma (\cdot)$ is Euler gamma function). P.S.: Can someone delete my previous comment, 'cause I'm not able to do it myself... –  Pacciu Dec 5 '11 at 22:31
    
Here's a crude brute-force approach to the $\mathbb{R}^2$ case. I'd be surprised if there's not something far more elegant. And I'm suspicious since I didn't expect this to diverge to $\infty$. $\displaystyle\int_{-\infty}^\infty \int_{-\infty}^\infty \frac{dx\;dy}{1+x^2+y^2} = \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{dx\;dy}{(1+y^2)(1+\frac{x^2}{1+y^2})} = \int_{-\infty}^\infty \left( \frac{1}{1+y^2} \int_{-\infty}^\infty \frac{dx}{1+\frac{x^2}{1+y^2}}\right)\;dy$. So let....... –  Michael Hardy Dec 5 '11 at 22:40
    
....... So let $\tan\theta=x/\sqrt{1+y^2}$, so that $dx=\sqrt{1+y^2}\;\sec^2\theta\;d\theta$. Integrate to get $\sqrt{1+y^2}\cdot \pi$. Then you have $\displaystyle\pi\int_{-\infty}^\infty \frac{dy}{\sqrt{1+y^2}}$. –  Michael Hardy Dec 5 '11 at 22:40
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up vote 4 down vote accepted

Ok, I try to expand my comment into a full answer.

For the computations, you should use the polar coordinates trick, i.e. for radial function $f(x)=\phi (\lVert x\rVert)$: $$\int_{\mathbb{R}^N} f(x)\ \text{d} x = N\ \omega_N\ \int_0^\infty \phi (r)\ r^{N-1}\ \text{d} r \; ,$$ where: $$\omega_N:=\text{volume of the unit ball in } \mathbb{R}^N = \frac{\pi^{N/2}}{\Gamma (1+N/2)}$$ (and $\Gamma (\cdot)$ is Euler gamma function).

Using the polar coordinates trick mentioned above, you find: $$\tag{0} \int_{\mathbb{R}^N} \frac{1}{(1+\lVert x\rVert^2)^2}\ \text{d} x =N\ \omega_N\ \int_0^\infty \frac{r^{N-1}}{(1+r^2)^2}\ \text{d} r\; ;$$ hence you see that $(1+\lVert x\rVert^2)^{-2}$ is not integrable over $\mathbb{R}^N$ when $N\geq 4$ (for in this case the integrand in the RHside either approaches to $0$ too slowly or it doesn't approach $0$ at all when $r\to \infty$).

Thus you have to analyse only the cases $N=1,2,3$ and elementary tricks (e.g., integration by parts) do really work in these cases.


Besides, I want to try a special function approach to the answer.

The substitution $u=r^2$ in the RHside of (0) yields: $$\tag{1} \begin{split} \int_{\mathbb{R}^N} \frac{1}{(1+\lVert x\rVert^2)^2}\ \text{d} x &=N\ \omega_N\ \int_0^\infty \frac{u^{(N-1)/2}}{(1+u)^2}\ \frac{1}{2\sqrt{u}}\ \text{d} u\\ &= \frac{N}{2}\ \omega_N\ \int_0^\infty \frac{u^{N/2 -1}}{(1+u)^2}\; .\end{split}$$ Now, it is known that the integral defining the beta function $B(t,s)$ can be written as: $$\tag{2} B(t,s):=\int_0^\infty \frac{u^{t-1}}{(1+u)^{t+s}}\ \text{d} u$$ when $t,s>0$; as you can see, the rightmost side of (1) is in the form (2) with: $$t=N/2 \qquad \text{and}\qquad s=2-t=2-N/2\; ,$$ and both such $t$ and $s$ are greater than zero for $N=1,2,3$; hence: $$\tag{3} \begin{split} \int_{\mathbb{R}^N} \frac{1}{(1+\lVert x\rVert^2)^2}\ \text{d} x &= \frac{N}{2}\ \omega_N\ B\left( \frac{N}{2}, 2-\frac{N}{2}\right)\\ &= \frac{N}{2}\ \frac{\pi^{N/2}}{\Gamma (1+N/2)}\ B\left( \frac{N}{2}, 2-\frac{N}{2}\right)\; .\end{split}$$ Obviously the rightmost side of (3) gives a (correct) answer to your question.

Neverthless, you could ask if the result can be furtherly simplified in some ways. In order to answer this, you have to recall some basic identities about beta and gamma functions, e.g.: $$B(t,s)=\frac{\Gamma (t)\ \Gamma (s)}{\Gamma (t+s)}\quad \text{and} \quad \Gamma (1+t)=t\ \Gamma (t)\quad \text{(hence } \Gamma (1+n)=n! \text{ for positive integers)}\; .$$ Actually, with these tools at hand, you write: $$\begin{split} \int_{\mathbb{R}^N} \frac{1}{(1+\lVert x\rVert^2)^2}\ \text{d} x &= \frac{N}{2}\ \frac{\pi^{N/2}}{\Gamma (1+N/2)}\ B\left( \frac{N}{2}, 2-\frac{N}{2}\right)\\ &= \frac{\pi^{N/2}}{\Gamma (N/2)}\ \frac{\Gamma (N/2)\ \Gamma (2-N/2)}{\Gamma (2)}\\ &= \pi^{N/2}\ \Gamma (2-N/2)\end{split}$$ for $N=1,2,3$.

Finally, a little more computational effort (you can use tables, e.g. this one) yields: $$\int_{\mathbb{R}^N} \frac{1}{(1+\lVert x\rVert^2)^2}\ \text{d} x =\begin{cases} \pi/2 &\text{, if } N=1 \\ \pi &\text{, if } N=2 \\ \pi^2 &\text{, if } N=3\; .\end{cases}$$

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Thanks!……………………. –  VVV Dec 6 '11 at 1:58
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