Fourier Transform of Hyperbolic Secant is itself!

$\DeclareMathOperator{\sech}{sech}$ $\DeclareMathOperator{\cot}{cot}$ $\DeclareMathOperator{\csc}{csc}$ $\DeclareMathOperator{\Re}{Re}$

 

Fourier transforms are ubiquitous in signal analysis, and thus neuroscience. Neurons after all are oscillators that sends signals whose properties are encoded in firing rate (frequency), and spike time (phase). Thus analyzing the signals of the brain with Fourier analysis is an oft used technique. While the below problem doesn't involve neuroscience, its a fun exercise that uses a few integration and Taylor series techniques that are useful in Fourier analysis.  I highly encourage you to try the example yourself before looking at the answer below.

 Can you show that the Fourier transform of  $\sech(\pi t)$ is $\sech(\pi \omega)$.

More specifically show $$\sech(\pi \omega)= \int^\infty_{-\infty} e^{-2\pi i \omega t} \sech(\pi t) dt$$

 

This is an example of finding a fixed point for the Fourier transform. As it turns out, Fourier Transforms have many different fixed points, that is, functions whose Fourier Transform is the same as itself. A canonical examples of such a fixed point is  $e^{-\pi x^2}$. There is actually a infinite family of such fixed points (see this stack exchange post for a detailed explanation). As we have already mentioned,  $\sech(\pi t)$ is also a fixed point of the Fourier transform. And using a some math it is not to hard to show. 

First, recall that $$\sech(\pi t) = \frac{1}{\cosh(\pi t)} = \frac{2}{e^{\pi t} + e^{- \pi t}}$$

Moreover, we can write this in two other forms, that will become convenient.

$$\frac{e^{\pi t}}{e^{\pi t}} \frac{2}{e^{\pi t} + e^{- \pi t}} = \frac{2e^{\pi t}}{e^{2 \pi t} +1}$$

as well as 

$$\frac{e^{-\pi t}}{e^{-\pi t}} \frac{2}{e^{\pi t} + e^{- \pi t}} = \frac{2e^{-\pi t}}{e^{-2 \pi t} +1}$$

Lets split our integral into two parts

$$ \int^\infty_{-\infty}  \frac{2e^{-2\pi i \omega t}}{e^{\pi t} + e^{- \pi t}}dt= \int^\infty_{0}  \frac{2e^{-2\pi i \omega t}}{e^{\pi t} + e^{- \pi t}}dt+\int^0_{-\infty}  \frac{2e^{-2\pi i \omega t}}{e^{\pi t} + e^{- \pi t}}dt $$

$$ \int^\infty_{-\infty}  \frac{2e^{-2\pi i \omega t}}{e^{\pi t} + e^{- \pi t}}dt=-\int^{0}_{\infty}  \frac{2e^{-2\pi i \omega t}}{e^{\pi t} + e^{- \pi t}}dt+\int^0_{-\infty}  \frac{2e^{-2\pi i \omega t}}{e^{\pi t} + e^{- \pi t}} dt$$

 

Next we can now rewrite our integral into two different forms

$$-\int^{0}_{\infty}  \frac{2 e^{-2\pi i \omega t} e^{-\pi t}}{e^{-2 \pi t} +1}dt+\int^0_{-\infty}  \frac{2 e^{-2\pi i \omega t} e^{\pi t}}{e^{2 \pi t} +1}dt $$

$$-\int^{0}_{\infty}  \frac{2 e^{-2\pi i \omega t-\pi t}}{e^{-2 \pi t} +1}dt+\int^0_{-\infty}  \frac{2 e^{-2\pi i \omega t+\pi t}}{e^{2 \pi t} +1}dt $$

$$-\int^{0}_{\infty}  \frac{2 e^{\pi (-1-2 i \omega ) t}}{e^{-2 \pi t} +1}dt+\int^0_{-\infty}  \frac{2 e^{\pi (1-2i \omega) t}}{e^{2 \pi t} +1}dt $$

While this may be strange to set it up, this is going to help when we turn this fraction into a series. Recall that 

$$\frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n $$ 

if $|x|<1$ also note that if  $ e^{2\pi t} <1$ if $t<0$ and $e^{-2\pi t} <1$ if $t>0$. Therefore we can replace the bottom fraction with this series to get

$$-\int^{0}_{\infty}  2 e^{\pi (-1-2 i \omega ) t} \sum_{n=0}^{\infty} (-1)^n {e^{-2 \pi t n}}dt+\int^0_{-\infty}  2 e^{\pi (1-2i \omega) t} \sum_{n=0}^{\infty} (-1)^n {e^{2 \pi t n}} dt$$

 We next can tidy up this equation to get  

$$-2 \sum_{n=0}^{\infty} \int^{0}_{\infty} (-1)^n {e^{\pi (-1-2i \omega-2  n)t}}dt+2 \sum_{n=0}^{\infty} \int^0_{-\infty}   (-1)^n {e^{\pi (1-2i \omega+2  n) t}} dt$$

Now we can integrate every term in the sum.  

$$-2 \sum_{n=0}^{\infty} \left. \frac{(-1)^n {e^{\pi (-1-2i \omega-2  n)t}}}{\pi (-1-2i \omega-2  n)}\right|^{0}_{\infty}+2 \sum_{n=0}^{\infty}  \left.  \frac{  (-1)^n {e^{\pi (1-2i \omega+2  n) t}}}{\pi (1-2i \omega+2  n)}\right|^{0}_{-\infty} $$

Conveniently $e^{0} = 1$ and $e^z$ where $\Re(z) = -\infty$ then $e^z = 0$

 Thus we can simplify the following series to

$$f(\omega) = -2 \sum_{n=0}^{\infty} \frac{(-1)^n}{\pi (-1-2i \omega-2  n)}  +2 \sum_{n=0}^{\infty}  \frac{  (-1)^n}{\pi (1-2i \omega+2  n)}$$

Notice something particularly interesting about the above sum $f(\omega)$. It has evenly spaced poles along the imaginary axes at for every integer n. However the $(-1)^n$ term means the polarity of these poles alternates. That being said we can use this observation to remind us of a fairly useful and convenient identity for cotangent (see proof here) , namely, 

$$\pi \cot(\pi \omega) = \sum_{n=-\infty}^{\infty} \frac{1}{\omega+n} $$ 

Our goal is to convert this cotangent identity into an expression for  our Fourier transform $f(\omega)$ defined above. And we will see that this is precisely, $\sech(\pi \omega)$. We will begin by manipulating the cotangent sum. 

$$\sum_{n=-\infty}^{\infty} \frac{2}{\omega+2n} =\sum_{n=-\infty}^{\infty} \frac{1}{\frac{\omega}{2}+n} = \pi \cot(\frac{\pi \omega}{2})  $$

Likewise

$$\sum_{n=-\infty}^{\infty} \frac{-2}{\omega+2n+1} =\sum_{n=-\infty}^{\infty} \frac{-1}{\frac{\omega}{2}+(n+\frac{1}{2})} = \pi \tan(\frac{\pi \omega}{2})  $$

Notice here that tangent's poles are offset by half an integer and have flipped polarity. This is because $\tan(\omega) = -\cot(\omega+\frac{\pi}{2})$. Next we can use the half angle formula to show

$$ \pi \cot(\frac{\pi \omega}{2})  = \pi (\csc(\pi \omega)+\cot(\pi \omega))$$

and 

$$ \pi \tan(\frac{\pi \omega}{2})  = \pi (\csc(\pi \omega)-\cot(\pi \omega))$$

Note that we can simply add these two expressions together to get 

$$ 2\pi \csc(\pi \omega)= \pi \cot(\frac{\pi \omega}{2})+\pi \tan(\frac{\pi \omega}{2})=\sum_{n=-\infty}^{\infty} \frac{2}{\omega+2n}+\sum_{n=-\infty}^{\infty} \frac{-2}{\omega+2n+1}$$

Dividing by $2$, and cleaning up we get

$$ \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{\omega+n}=\pi \csc(\pi \omega)$$

Recall now that $\pi \csc(\pi (\omega+\frac{1}{2})) = \pi \sec(\pi \omega)$. Using this identity we get

$$ \pi \sec(\pi \omega)= \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{\omega+n+\frac{1}{2}}$$

We can finesse this equation into a more fitting form with a little algebra to get

$$\pi \sec(\pi \omega)= \sum_{n=0}^{\infty} \frac{(-1)^n}{\omega+n+\frac{1}{2}} -\sum_{n=0}^{\infty} \frac{(-1)^n}{\omega-n-\frac{1}{2}}$$

Now lets remove the $\frac{1}{2}$ from the expression. This yields, 

$$\pi \sec(\pi \omega)= 2\sum_{n=0}^{\infty} \frac{(-1)^n}{2\omega+2n+1} -2\sum_{n=0}^{\infty} \frac{(-1)^n}{2\omega-2n-1}$$

Now we can use Euler's famous identity to note that $\cosh(\omega) = \cos( i \omega)$ which implies that $\sech(\omega)= \sec(i \omega)$ we can rewrite this sum as, 

$$\pi \sech(\pi \omega)=\pi \sec(i\pi \omega)= 2\sum_{n=0}^{\infty} \frac{(-1)^n}{2i\omega+2n+1} -2\sum_{n=0}^{\infty} \frac{(-1)^n}{2i\omega-2n-1}$$

divide out the $\pi$ to get

$$\sech(\pi \omega)= 2\sum_{n=0}^{\infty} \frac{(-1)^n}{\pi (2i\omega+2n+1)} -2\sum_{n=0}^{\infty} \frac{(-1)^n}{\pi(2i\omega-2n-1)}$$

Notice that this expression is precisely the definition we had derived above for $f(\omega)$, thus completing the proof that the Fourier transform of  $\sech(\pi t)$ is $\sech(\pi \omega)$.

I hope this proof was enlightening. It shows several useful identities that are extremely useful in Fourier analysis. Specifically, it shows how important manipulating trig identities and sums are in evaluating Fourier transforms. Furthermore, several of the identities such as the cotangent identity come up often in defining the phase of a signal (which is useful in analyzing local field potentials, EGGs and synchronization). Moreover, we showed how Taylor series are a key tool in evaluating rational functions with exponents. All in all, these calculus and trigonometry techniques are very important to get a handle on. I hope this worked example gives you an idea of some less talked about tools a computational neuroscientist can use to manipulate and simplify signals.

Author: Alexander White