Answer to Question #308238 in Calculus for Hwazen

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)} \$