# Exponential Fourier transform

## Homework Statement:

Find the exponential Fourier transform of
##f(x)=e^{-|x|}## and write the inverse transform. You should find:
$$\int_{0}^{\infty} \frac{\cos{ax}}{a^2+1} da = \frac {\pi}{2} e^{-|x|}$$

## Relevant Equations:

Fourier transform:
$$g(a)=\frac{1}{2\pi} \int_{-\infty}^{\infty} f(x) e^{-iax} dx$$
Inverse Transform:
$$f(x)=\int_{-\infty}^{\infty} g(a) e^{iax} da$$
From the sketch, I know that this function is an even function. So, I simplify the Fourier transform in the limit of the integration (but still in exponential form). Then, I try to find the exponential FOurier transform. Here what I get:
$$g(a)=\frac{2}{2\pi} \int_{0}^{\infty} e^{-x} e^{-iax} dx$$,
$$g(a)=\frac{1}{\pi} \int_{0}^{\infty} e^{(-x)(1+a)} dx$$,
$$g(a)=\frac{1}{\pi} \left[\frac{e^{-ix(1+a)}}{-i(1+a)} \right]^{\infty}_{0}$$.
As x approaching infinite ##e^{-ix(1+a)}## approaching zero. So,
$$g(a)=\frac{1-ia}{\pi(1+a^2)}$$.

Knowing this transform, I did the inverse transformation.
$$f(x)=\int_{-\infty}^{\infty} \frac{1-ia}{\pi(1+a^2)} e^{iax} da$$, where ##e^{iax}=\cos {(ax)} + i \sin {(ax)}##
So,
$$f(x)=\int_{-\infty}^{\infty} \frac{(1-ia)\left(\cos{ax} + i \sin {ax}\right)}{\pi(1+a^2)} da$$.

I observe that ##\frac{\sin{ax}}{1+a^2}##; ##\frac{(-a)\cos{ax}}{1+a^2}## are odd functions. But, ##\frac{\cos{ax}}{1+a^2}##; ##\frac{(a)\sin{ax}}{1+a^2}## are even functions. So,
$$f(x)=\frac{2}{\pi}\int_{0}^{\infty} \frac{\cos {ax} + a \sin {ax}}{(1+a^2)} da$$.

The sin term of the answer shouldn't be there. I have double-checked my work and still haven't find the mistake. Could you please explain how I get the answer term, in the problem statement? Thanks.

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PeroK
Homework Helper
Gold Member
Then, I try to find the exponential FOurier transform. Here what I get:
$$g(a)=\frac{2}{2\pi} \int_{0}^{\infty} e^{-x} e^{-iax} dx$$,
How did you get that?

I change ##e^{-|x|}## from ##-\infty## to ##\infty## becomes ##e^{-x}## from 0 to ##\infty##.

PeroK
Homework Helper
Gold Member
I change ##e^{-|x|}## from ##-\infty## to ##\infty## becomes ##e^{-x}## from 0 to ##\infty##.

It should be the same, isn't it? Because of the general formula?

PeroK
Homework Helper
Gold Member
It should be the same, isn't it? Because of the general formula?
It's not an even function:
$$e^{-iax} = \cos(ax) - i\sin(ax)$$
And ##\sin(ax)## is an odd function.

Oh Gee. I forgot that ##f(x)## should be multiplied by ##e^{-iax}##. I will try to fix this.

PeroK
I mean, an odd or even function is multiplied function; not f(x) itself.

Yeah, finally I can show the solution. Thanks, Pero K for the correction!