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- 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.

$$g(a)=\frac{2}{2\pi} \int_{0}^{\infty} e^{-x} e^{-iax} dx$$,

$$g(a)=\frac{1}{\pi} \int_{0}^{\infty} e^{(-x)(1+a)} dx$$,

$$g(a)=\frac{1}{\pi} \left[\frac{e^{-ix(1+a)}}{-i(1+a)} \right]^{\infty}_{0}$$.

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.