Show (using the Fourier transform) that $$ K(u) = \frac{1}{2} \exp \left(- \frac{|u|}{\sqrt{2}} \right) \sin \left(\frac{|u|}{\sqrt{2}} + \frac{\pi}{4} \right) $$ can also be written $$ K(u) = \int_{-\infty}^\infty \frac{\cos(2\pi t u)}{1+(2\pi t)^4} \, \mathrm{d}t. $$

I have no idea how to proceed. Any thoughts or hints?


I've accepted Mark Viola's answer since it was very helpful and, I assume, even more so if you're familiar with complex integrals. I ended up using a different approach myself though:

Noting that $K$ is symmetric, the Fourier transform can be written $$ \mathcal{F}(K) = \int K(u) \cos(u \omega) \, \mathrm{d}u . $$

Using symmetry of $K(u)\cos(u\omega)$ and the product-to-sum trigonometric identity: $2\sin x \cos y = \sin( x+y )+ \sin (x-y)$, we get $$ \mathcal{F}(K) = \frac{1}{2} \int_0^\infty \exp \left(-\frac{u}{\sqrt{2}} \right) \sin \left( \left[ \frac{1}{\sqrt{2}} + \omega \right] u + \frac{\pi}{4} \right)\, \mathrm{d}u \\ + \frac{1}{2} \int_0^\infty \exp \left(-\frac{u}{\sqrt{2}} \right) \sin \left( \left[ \frac{1}{\sqrt{2}} - \omega \right] u + \frac{\pi}{4} \right)\, \mathrm{d}u . $$

Using the integral formula, derived by change-of-variables and partial integration, $$ \int_0^\infty \exp(-ax)\sin(bx+c) \, \mathrm{d}x = \frac{\cos(c)b+\sin(c)a}{a^2+b^2} $$ we get that $$ \mathcal{F}(K) = \frac{1}{1+\omega^4}. $$

Now applying the inverse Fourier transform yields the desired result.


1 Answer 1



Note from even symmetry that $K(u)$ is the Fourier Transform of $\frac1{1+(2\pi t)^4}$ with Fourier kernel $e^{i(2\pi u)t}$. That is to say that

$$\int_{-\infty}^\infty \frac{\cos(2\pi ut)}{1+(2\pi t)^4}\,dt=\int_{-\infty}^\infty \frac{e^{i2\pi ut}}{1+(2\pi t)^4}\,dt$$

Now, determine the inverse Fourier Transform of $K(u)=\frac12 e^{-|u|/\sqrt 2}\sin\left(\frac{|u|}{\sqrt 2} +\frac\pi4\right)$.

Can you proceed now?

  • $\begingroup$ Additional hint, $$\sin\phi=\frac{e^{i\phi}-e^{-i\phi}}{2i}$$ $\endgroup$
    – Andrei
    Sep 23, 2020 at 15:50
  • $\begingroup$ I'm afraid I'm still quite confused - How does computing the inverse Fourier transform help? $\endgroup$
    – Lundborg
    Sep 24, 2020 at 8:31
  • $\begingroup$ @Lundborg The Fourier Transform pair are unique. So, if you can show that the inverse transform of $K(u)$ is $\frac{1}{1+(2\pi t)^4}$, then it must be true that the Fourier Transform of $\frac1{1+(2\pi t)^4}$ is $K(u)$. Is that clearer now? $\endgroup$
    – Mark Viola
    Sep 24, 2020 at 22:36
  • $\begingroup$ @MarkViola You've been incredibly helpful but I think my problem is perhaps more fundamental, I'm not really used to computing Fourier transforms at all and get very confused by the resulting integrals over complex exponential functions. $\endgroup$
    – Lundborg
    Sep 25, 2020 at 8:13
  • $\begingroup$ Well, note that $$\begin{align} \int_{-\infty }^\infty e^{-(|u|/\sqrt 2)}e^{\pm i(|u|/\sqrt 2+\pi/4)}e^{i2\pi t u}\,du&=e^{\pm i\pi/4}\int_{-\infty }^\infty e^{-\frac{|u|}{\sqrt 2}(1\pm i)+i2\pi t u}\,du\\\\ &=e^{\pm i\pi/4}\int_{-\infty }^0 e^{\frac{|u|}{\sqrt 2}(1\pm i)+i2\pi t u}\,du+e^{\pm i\pi/4}\int_{-\infty }^\infty e^{-\frac{u}{\sqrt 2}(1\pm i)+i2\pi t u}\,du \end{align}$$Can you proceed now? $\endgroup$
    – Mark Viola
    Sep 25, 2020 at 22:16

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