Rewriting kernel using Fourier transform

Question

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.

Answer 1

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?