Question

find fourier integral for the following f(x)= e-x x>0

Answer 1

Solution

Given that

$f(x)=e^{-x}$ here $x>0$

The Fourier expansion is defined as

f\left( x \right) = \sum\limits_0^\infty {\left( {A\left( \alpha \right)\cos \left( {\alpha x} \right) + B\left( \alpha \right)\sin \left( {\alpha x} \right)} \right)} \

Here,

A\left( \alpha \right) = \frac{1}{\pi }\int\limits_{ - \infty }^\infty {f\left( u \right)\,\cos \left( {\alpha u} \right)\,du} \

and

B\left( \alpha \right) = \frac{1}{\pi }\int\limits_{ - \infty }^\infty {f\left( u \right)\,\sin \left( {\alpha u} \right)\,du} \

Now, Integrating by parts, we get 

A\left( \alpha \right) = \frac{1}{{{\alpha ^2} + 1}}\ and B\left( \alpha \right) = \frac{\alpha}{{{\alpha ^2} + 1}}\

Hence

\begin{array}{l} f\left( x \right) = {e^{ - x}} = \sum\limits_0^\infty {\left( {A\left( \alpha \right)\cos \left( {\alpha x} \right) + B\left( \alpha \right)\sin \left( {\alpha x} \right)} \right)} \\ {e^{ - x}} = \sum\limits_0^\infty {\left( {\frac{1}{{{\alpha ^2} + 1}}\cos \left( {\alpha x} \right) + \frac{\alpha }{{{\alpha ^2} + 1}}\sin \left( {\alpha x} \right)} \right)} \\ {e^{ - x}} = \frac{1}{{{\alpha ^2} + 1}}\sum\limits_0^\infty {\left( {\cos \left( {\alpha x} \right) + \alpha \sin \left( {\alpha x} \right)} \right)} \end{array}\

Hence, the Fourier expansion of $f(x)=e^{-x}$ is

{e^{ - x}} = \frac{1}{{{\alpha ^2} + 1}}\sum\limits_0^\infty {\left( {\cos \left( {\alpha x} \right) + \alpha \sin \left( {\alpha x} \right)} \right)} \