How to calculate this basic Fourier Transform?

Question

I am trying to calculate the Fourier Transform of $g(t)=e^{-\alpha|t|}$, where $\alpha > 0$.

Because there's an absolute value around $t$, that makes $g(t)$ an even function, correct? If that's the case, then I made the assumption that I can calculate FT integral from $0$ to $\infty$ and then multiply the result by $2$ to get the answer:

$G^*(f) = \int_{0}^{\infty} e^{-\alpha t} e^{-j\omega t}dt = G(f) = \int_{0}^{\infty} e^{-(j\omega + \alpha)t}dt = \frac{e^{-(j\omega + \alpha)t}}{-(j\omega + \alpha)}\bigg|_0^\infty = 0 - \frac{1}{-(j\omega + \alpha)} = \frac{1}{j\omega + \alpha}$

$G(f) = 2G^*(f) = \frac{2}{j\omega + a}$

This is not the correct answer. If wolfram alpha is to be believed, then the answer should be:

$\frac{\alpha}{\omega^2 + \alpha^2}$

What did I do wrong? What are the correct steps to calculate this FT?

Answer 1

$$\int_{-\infty}^{\infty}e^{-\alpha|t|}e^{-j\omega t}dt=\int_{-\infty}^{0}e^{\alpha t}e^{-j\omega t}dt+\int_{0}^{\infty}e^{-\alpha t}e^{-j\omega t}dt=\frac{1}{\alpha-j\omega}+\frac{1}{\alpha+j\omega}=\frac{2\alpha}{\alpha^2+\omega^2}$$

Answer 2

Since $f$ is even you have $$ \hat g(\omega)=2\,\int_0^\infty g(t)\,\cos(j\,\omega\,t)\,dt, $$ wich is the real part of $$ 2\,\int_0^\infty g(t)\,e^{j\,\omega\,t}\,dt. $$