I have to evaluate this integral, using properties of the Fourier Transform. $$ \int_{-\infty}^{+\infty} w^2 |\hat{g}(w)|^2 dw $$ where $ g(t) = e^{-|t|} , t\in R $

As far as I understood I cannot use Plancherel because $ g(t) \notin S(R) $ (please, correct me if I'm wrong) but the prof. told me I should not determinate the Fourier Transform and compute the integral, because I can easily evaluate it using FT properties.

Is there somebody who can help me please?


Your function $g \in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$ fails to be differentiable at $0$, but that's OK, since $$ g^\prime(t) = \begin{cases} -e^{-\lvert t \rvert} &\text{if $t > 0$,}\\ e^{-\lvert t \rvert} &\text{if $t < 0$,} \end{cases} $$ still defines a perfectly good element of $L^1(\mathbb{R}) \cap L^2(\mathbb{R})$, such that $$ g(t) = \int_{-\infty}^t g^\prime(s)\,ds $$ for almost every $t$, and this is certainly enough (see, e.g., Property 2.1 in these notes) to conclude that $$ \widehat{g^\prime}(w) = 2\pi i w \widehat{g}(w) $$ for almost every $w$. In particular, since $g^\prime$ is square integrable, so too is $\widehat{g^\prime}$, so you're perfectly free to apply Plancherel.

So, you now have everything you need to reduce the problem to elementary calculus. However, to be just a little bit more explicit:

  1. By the above discussion, your integral is precisely $\tfrac{1}{4\pi^2}\|\widehat{g^\prime}\|_2^2$.
  2. By Plancherel's theorem, you know that $\|\widehat{g^\prime}\|_2^2 = \|g^\prime\|_2^2$.
  3. By the explicit formula for $g^\prime$ above, you know that $\|g^\prime\|_2^2 = \int_{-\infty}^\infty e^{-2\lvert t \rvert}\,dt$, which you can simplify even further by splitting up this one improper integral into two (i.e., $\int_{-\infty}^0 + \int_0^{\infty}$) and observing that the function $t \mapsto e^{-2\lvert t \rvert}$ is even.

You now have everything you need to reduce your original integral to an extremely simple improper (Riemann!) integral on $(0,\infty)$.

  • $\begingroup$ Can you please also evaluate the integral? thank you $\endgroup$ – PeppeDAlterio May 26 '17 at 7:41
  • 1
    $\begingroup$ I've added a few more hints, but it really does reduce to elementary calculus. $\endgroup$ – Branimir Ćaćić May 26 '17 at 12:07

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