Computing $\int_0^{\infty} \dfrac{1}{(1+x^2)^2}dx$ using Plancherel I want to compute $\int_0^{\infty} \dfrac{1}{(1+x^2)^2}dx$, using Plancherel.
So define $f(x) = \dfrac{1}{1+x^2}$ , then $f\in L_1\cap L_2$ so by Plancherel we know that $||f||_2 =||\hat{f}||_2$ where $\hat{f}(t) = \dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ixt}dx$. 
So we have that $$\int_0^{\infty} \dfrac{1}{(1+x^2)^2}dx = ||f||_2^2 = ||\hat{f}||_2^2 = \int_0^{\infty}( \dfrac{1}{\sqrt{2\pi}}\int_{0}^{\infty}f(x)e^{-ixt}dx)^2dt = $$
$$\dfrac{1}{2\pi}\int_0^\infty(\int_0^\infty\dfrac{e^{-ixt}}{1+x^{2}}dx)^2dt$$ 
This was the guidance in the exercise (using Plancherel) and it seems a bit difficult (or maybe I'm missing something). Is the inner integral easy to compute? If so, how? 
Thanks for helping!
 A: This isn't the way that you asked for the integral to be computed, but it's a way that it could be computed. 
Consider the indefinite integral 
$$I_n=\int\frac{\mathrm{d}x}{(ax^2+b)^n}$$
Integrating by parts with $\mathrm{d}v=\mathrm{d}x$ gives
$$I_n=\frac{x}{(ax^2+b)^n}+2n\int\frac{ax^2\ \mathrm{d}x}{(ax^2+b)^{n+1}}$$
$$I_n=\frac{x}{(ax^2+b)^n}+2n\int\frac{ax^2+b}{(ax^2+b)^{n+1}}\mathrm{d}x-2bn\int\frac{\mathrm{d}x}{(ax^2+b)^{n+1}}$$
$$I_n=\frac{x}{(ax^2+b)^n}+2nI_n-2bnI_{n+1}$$
$$2bnI_{n+1}=\frac{x}{(ax^2+b)^n}+(2n-1)I_n$$
$$I_{n+1}=\frac{x}{2bn(ax^2+b)^n}+\frac{2n-1}{2bn}I_n$$
$$I_n=\frac{x}{2b(n-1)(ax^2+b)^{n-1}}+\frac{2n-3}{2b(n-1)}I_{n-1}$$
Plugging in $a=1,\,b=1,\,n=2$, and evaluating form $x=0$ to $x=\infty$ gives your integral as
$$\frac12\int_0^\infty \frac{\mathrm dx}{1+x^2}=\frac\pi4$$
A: Let me define $f$ as you have defined, namely $f(x) =(1+x^2)^{-1}$. In this case Plancherel doesn't make the evaluation much easier. However, in these kind of exercises they usually assume that you know that
$$\hat f(t) = \frac{1}{\sqrt{2\pi} }\int_\mathbb{R} e^{-itx} f(x) \, dx =\sqrt{\frac\pi 2}e^{-|t|}. $$
For example, finding the Fourier transform of that particular $f$ is asked in a previous exercise or it is given as an example in your notes. Anyways you can find a proof in this link. 
Let us now use Plancherel to conclude 
\begin{align} 
\int_0^\infty \frac 1 {(1+x^2)^2}\,dx&=\frac 1 2 \int_\mathbb{R} |f(x) |^2\,dx \\
&=\frac 1 2 \int_\mathbb{R} |\hat f(x) |^2\,dx \\
&=\frac 1 2 \int_\mathbb{R} \frac\pi 2e^{-2|t|}\,dt \end{align} 
The latter integral is elementary 
$$\frac 1 2 \int_\mathbb{R} \frac\pi 2e^{-2|t|}\,dt = \int_0^\infty\frac\pi 2e^{-2t}\,dt = \frac \pi 4$$
