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Let $f(x)=1/(1+x^{2})^{2}$, then we have

$$\int_{-\infty}^{\infty}\frac{1}{(1+x^{2})^{2}}\,dx=\int_{-\infty}^{\infty}\frac{1}{|f(x)|^{2}}\,dx$$

then by Plancherel's formula, this should yield

$$\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1}{|\hat{f}(\xi)|^{2}}\,d\xi$$

which means I must evaluate $\hat{f}(\xi)$; i.e.

$$\hat{f}(\xi)=\int_{-\infty}^{\infty}(1+x^{2})e^{ix\cdot\xi}\,dx=\int_{-\infty}^{\infty}e^{ix\cdot\xi}\,dx+\int_{-\infty}^{\infty}x^{2}e^{ix\cdot\xi}\,dx$$

Then the first addendum is equal to $\frac{1}{2\pi}\delta(\xi)$ (although this doesn't really make sense and anyway, this wasn't really my question).

My question is that the following exercise in the book I am going over, namely Strichartz' book, asks us to evaluate the Fourier transform of $1/(1+x^{2})^{2}$ on $\mathbb{R}$, which suggests to me that I should be able to evaluate the original integral using Plancherel's formula without computing the Fourier transform.

Did I perhaps cheat when I applied the Plancherel formula, which actually states that $\int_{-\infty}^{\infty}|f(x)|^{2}\,dx=\frac{1}{2\pi}\int_{-\infty}^{\infty}|\hat{f}(\xi)|^{2}\,d\xi$?

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    Well, its not even a bad integral IMO. You can handle it with a quick trig substitution, but that's not the point of the exercise, is it? – 2017-01-02
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    the result should be $$\frac{\pi}{2}$$ – 2017-01-02
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    for what stands IMO? international mathematical olympiad? – 2017-01-02
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    In My Opinion, @Dr.SonnhardGraubner. – 2017-01-02
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    If $f(x)=\frac{1}{(1+x^2)}$, $$\int_{-\infty}^{+\infty}\frac{dx}{(1+x^2)^2} = \int_{-\infty}^{+\infty}f(x)^2\,dx = 2\pi \int_{-\infty}^{+\infty}\hat{f}(\xi)^2\,d\xi, $$ so you have to compute the Fourier transform of $\frac{1}{1+x^2}$, not the Fourier transform of $(1+x^2)$. – 2017-01-02
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    ok thank you for the lesson – 2017-01-02

1 Answers 1

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Evaluation of this integral using Plancherel's formula goes something like this:

\begin{align} \int_{-\infty}^\infty\frac{dx}{\left(1+x^2\right)^2}&=\frac{1}{2\pi}\int_{-\infty}^\infty d\xi\left(\int_{-\infty}^\infty dx\frac{e^{ix\xi}}{1+x^2}\right)^2\tag1\\ &=\frac{\pi}{2}\int_{-\infty}^\infty d\xi e^{-2|\xi|}\tag2\\ &=\pi\int_0^\infty d\xi e^{-2\xi}\tag3\\ &=\frac{\pi}{2} \end{align}

where going from $(1)$ to $(2)$ is a simple application of the Residue Theorem, and $(3)$ follows from the symmetry of the integrand in $(2)$.

So yes, to answer your question, you "cheated" (I wouldn't call it that, it's simply not a justified step) when pulling the Fourier Transform $\hat{f}(\xi)$ into the denominator directly.