Considering the function $f(x) = e^{-x}, x>0$ and $f(-x) = f(x)$. I am trying to find the Fourier integral representation of $f(x)$. Below are my workings but I am not sure if I proceeded correctly or how to use $f(-x)=x$. Also since the function is for $x>0$ should I use a low bound of $0$ and upper bound of $\infty$ for the integrals.
My attempt:
Using the formula for the Fourier integral representation,
$$f(x) = \int_0^\infty \left( A(\alpha)\cos\alpha x +B(\alpha) \sin\alpha x\right) d\alpha$$
Calculating $A(\alpha),$
$$A(\alpha) = \frac{1}{\pi}\int_{-\infty}^{\infty} f(u)\cos\alpha u du = \frac{1}{\pi} \int_0^\infty e^{-u}\cos \alpha u du$$
Notice here how I used $0$ and $\infty$ as my bounds, is this correct?
Below is the calculation of this integral, I will save myself sometime and just post the answer I got.
$$A(\alpha) = \frac{1}{\alpha ^2 +1}$$
Calculting $B(\alpha),$
$$B(\alpha) = \frac{1}{\pi}\int_{-\infty}^{\infty} f(u)\sin\alpha u du = \frac{1}{\pi} \int_0^\infty e^{-u}\sin \alpha u du$$
Notice again here how I used $0$ and $\infty$ as my bounds, is this correct?
Below is the calculation of this integral, I will save myself sometime and just post the answer I got.
$$B(\alpha) = \frac{\alpha}{\alpha^2 +1}$$
$$F(\omega)=\int_{-\infty}^{\infty}f(x)e^{-j\omega x}dx$$
$$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{j\omega x}d\omega$$
Expand the first formula yielding
$F(\omega)=A(\omega)+jB(\omega)$
where, $A(\omega)=\int_{-\infty}^{\infty}f(x)cos(\omega x)dx$, $B(\omega)=-\int_{-\infty}^{\infty}f(x)sin(\omega x)dx$
Since $f(x)$ is an even function, $B(\omega)=0$. All you have to do is caculate $A(\omega)$
$$A(\omega)=\int_{-\infty}^{\infty}f(x)cos(\omega x)dx=2\int_{0}^{\infty}f(x)cos(\omega x)dx=\frac{2}{1+\omega ^2}$$
Hence, $$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}{\frac{2}{1+\omega ^2}e^{j\omega x}\,d\omega}$$