# Thread: Fourier transform of a hyperbolic secant

1. ## Fourier transform of a hyperbolic secant

Hello.

I'm trying to manually calculate the spectrum of a hyperbolic secant function. It's well known that
$\int_{-\infty }^{\infty } \text{sech} (t \alpha) \exp (- i \omega t)d t = \frac{\pi}{\alpha} \text{sech} \left(\frac{\pi^2}{\alpha} \omega\right).$
How exactly is this expression obtained? All of my atempts to integrate this equation were unsuccessful. What measures should be taken in this particular case?

2. I've never done that, but may result if you integrate by parts two times putting $du=e^{-iwt}$ and then use hyperbolic identities.

3. Originally Posted by Nouvelle Vague
Hello.

I'm trying to manually calculate the spectrum of a hyperbolic secant function. It's well known that
$\int_{-\infty }^{\infty } \text{sech} (t \alpha) \exp (- i \omega t)d t = \frac{\pi}{\alpha} \text{sech} \left(\frac{\pi^2}{\alpha} \omega\right).$
How exactly is this expression obtained? All of my atempts to integrate this equation were unsuccessful. What measures should be taken in this particular case?
Hi. I think you mean:

$\int_{-\infty}^{\infty} \text{sech}(\alpha x) e^{-2\pi i x \omega} dx=\frac{\pi}{\alpha}\text{sech}\left(\frac{\pi^2 \omega}{\alpha}\right)$

which we can evaluate via the Residue Theorem. Consider the contour integral:

$\mathop\oint\limits_{D_u} \text{sech}(\alpha z)e^{-2\pi i z\omega}dz$

where $D_u$ is the path along the real-axis, then circling back in a large upper half-circle. As the radius of this contour approaches infinity, it encloses all the poles of the integrand in the upper half-plane at $\frac{(2n+1)\pi i}{2\alpha}$.

Now, the integral tends to zero along the circular half-circle as the radius tends to infinity and we're left with just the line segment along the real axis or:

$\int_{-\infty}^{\infty} \text{sech}(\alpha z)e^{-2\pi i z\omega}dz=-\frac{2 \pi }{\alpha }\sum _{n=0}^{\infty } (-1)^{n+1}\text{Exp}\left[\frac{\pi ^2 \omega }{\alpha }(2n+1)\right]=\frac{\pi}{\alpha}\text{sech}\left(\frac{\pi^2 \omega}{\alpha}\right)$

4. ## last step...?

I can get to the alternating exponential series in your last post (albeit in a different way by calculating the residues by differentiating the denominator of a rational function and evaluating at the poles). However, I can't seem to follow your final step. The Taylor series expansion of a sech function, while alternating as this one does, has the Euler numbers as coefficients and I don't see how they are equivalent. Thanks in advance.

5. Nevermind, I figured it out--its just a simple infinite geometric series with the common ratio r = -exp[b*Pi/a] so that it goes to (coeff)*series = (coeff)/(1-r) = (coeff)/(1+exp[b*Pi/a]) and then rearrange to get in canonical form for cosh as sum of exponentials in the denominator. Thanks anyway.