Fourier transform of exponent

82 Views Asked by At

I need to count the Fourier transform of the following function but it does not seem so obvious for me.

$f(x)=(e^{-ab})-1$ for $x\ge0$ and $f(x)=0$ for $x<0$

where: $a=1$ and $b=-1$

I don't know whether I should use the standard method or is there any other efficient way?

1

There are 1 best solutions below

5
On BEST ANSWER

The original post has $f$ defined such that

$$f(x)=(e^{-ab}-1)H(x)$$

where $H$ is the Heaviside function defined as

$$H(x) = \begin{cases} 1, & \text{if $x\ge 0$} \\ 0, & \text{if $x<0$} \end{cases} $$

The Fourier-Transform $\hat H(k)$ of the Heaviside function is

$$\hat H(k)\equiv \int_{-\infty}^{\infty}H(x)e^{ikx}dx=\pi \delta(k)+\text{PV}\left(\frac{i}{k}\right)$$

where PV denotes the Cauchy Principal Value and where $\delta$ is the Dirac delta and is a Generalized function or Distribution defined as

$$\int_{-\infty}^{\infty}\delta(x)f(x)dx=f(0)$$

for all test functions $f$.

Thus, we have for $a=1$ and $b=-1$ (as in the original post)

$$\hat f(k)=(e-1)\left(\pi \delta(k)+\text{PV}\left(\frac{i}{k}\right)\right)$$