Fourier transform of piecewise function

Question

I am trying to calculate the Fourier transform of the following function: $$f(x)=\begin{cases}e^{-x}&|x|<1\\0&\text{otherwise}\end{cases}$$ What I have so far is: $$\hat{f}(\omega)=\int_{-\infty}^\infty f(x)e^{-j\omega x}dx=\int_{-1}^1e^{-x}e^{-j\omega x}dx=\int_{-1}^1e^{-(1+j\omega)x}dx$$ which I got as: $$(1-j\omega)\frac{e^{(1+j\omega)}-e^{-(1+j\omega)}}{1+\omega^2}$$ which I have tried to simplify to the following: $$2(1-j\omega)\frac{\sinh(1)\cos(\omega)+j\cosh(1)\sin(\omega)}{1+\omega^2}$$ which if I separate the real and imaginary parts left me with: $$\frac{2\left[\sinh1\cos\omega+\cosh1\,\omega\sin\omega\right]+2j\left[\cosh1\sin\omega-\sinh1\,\omega\cos\omega\right]}{1+\omega^2}$$ My question is am I missing an identity that makes this a lot easier or is this as simple as I can get it? Thanks

Answer 1

$$\hat{f}(\omega)=\int_{-\infty}^\infty f(x)e^{-j\omega x}dx=\int_{-1}^1e^{-x}e^{-j\omega x}dx=\int_{-1}^1e^{-(1+j\omega)x}dx$$

Your above definition of Fourier Transform is valid if you are assuming non unitary angular frequecy $\omega$ but there are two other types of frequency, check their definitions also. But i will proceed with your definition and using $j=\sqrt{-1}$. Solving the exponential integral easily we get $$=-\left[\frac{e^{-x(1+j\omega)}}{1+j\omega}\right]_{-1}^{1}$$ $$=-\frac{1}{1+j\omega}\left[e^{-(1+j\omega)}-e^{(1+j\omega)}\right]$$ Using the property $\sinh x=\frac{e^x-e^{-x}}{2}$ we get $$=-\frac{1}{1+j\omega}\left[-2\sinh(1+j\omega)\right]$$ $$=\boxed{\color{blue}{\frac{2\sinh(1+j\omega)}{1+j\omega}}}$$

Edit : This is the best closed form in which we can express the answer, but if you wish to separate real and imaginary parts then you need to express in those long expressions as you did. Unfortunately there can't be simple closed forms for real and imaginary parts expressions here.

Answer 2

Here's one way to calculate the Fourier transform:

The distributional derivative of $f$ satisfies the equation $$f'(x)=-f(x)+e^{1}\delta(x+1)-e^{-1}\delta(x-1).$$

Taking the Fourier transform of both sides gives $$ j\omega\hat f(\omega) = -\hat f(\omega)+e^{1}e^{j\omega}-e^{-1}e^{-j\omega} $$ i.e. $$ \hat f(\omega) = \frac{e^{1+j\omega}-e^{-(1+j\omega)}}{1+j\omega}. $$

I don't think that the splitting into real and imaginary parts can be done in a simple way.