# Rewriting kernel using Fourier transform

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?

EDIT:

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.

HINT:

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?

• Additional hint, $$\sin\phi=\frac{e^{i\phi}-e^{-i\phi}}{2i}$$ Sep 23, 2020 at 15:50
• I'm afraid I'm still quite confused - How does computing the inverse Fourier transform help? Sep 24, 2020 at 8:31
• @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? Sep 24, 2020 at 22:36
• @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. Sep 25, 2020 at 8:13
• 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? Sep 25, 2020 at 22:16