Question

find fourier transform of f(x)=e^(-ax)

Answer 1

We are finding the Fourier transform for the function e(-ax) * u(x),  where  u(x) = unit step function and
  
   u(x) = 0 for  x < 0  and 
         = 1  for  x >= 0

We find the transform for the signal given, for  x >= 0 part only.

Fourier transform of the signal  f(x) is defined as:

$F(f)= \int\limits^{\infty}_{0} {f(x)\ e^{-(2 \pi\ f\ x)i}} \, dx \\\\=\int\limits^{\infty}_{0} {e^{(-ax)}\ e^{-(2 \pi\ f\ x)i}} \, dx \\\\=\int\limits^{\infty}_{0} {e^{-(a+2 \pi f\ i)x}} \, dx \\\\=\frac{-1}{a+2 \pi f i}\ [e^{-(a+2 \pi f\ i)x}]^\infty_{0} \\\\=\frac{-1}{a+2 \pi f i}\ [-1 + 0] \\\\\frac{1}{a+2 \pi f i}\\\\=\frac{a-2 \pi\ f\ i}{\sqrt{a^2+4 \pi^2\ f^2}}$

The given function grows to infinity as  x tends to -∞.  So its Fourier transform is not possible to determine if the lower limit is -∞ for the span of x.  So we take the real time  for x >= 0 only.  We can take x >= -A  for a real value of A and find the transform.