Inverse Fourier Transform of e^{-|\omega|\alpha}

[PiRho31416]

[Solved] Inverse Fourier Transform

Homework Statement

If
$$F(\omega)=e^{-|\omega|\alpha}\,(\alpha>0)$$,
determine the inverse Fourier transform of $$F(\omega)$$. The answer is $$\frac{2\alpha}{x^{2}+\alpha^{2}}$$

Homework Equations

Inverse Fourier Transform is defined as:
$$f(x)=\int_{-\infty}^{\infty}F(\omega)e^{-i\omega x}\, d\omega$$

The Attempt at a Solution

So I broke up the equation into two different integrals.
$$F(\omega)=e^{-|\omega|\alpha}=\int_{-\infty}^{\infty}e^{-|\omega|\alpha}e^{-i\omega x}d\omega=\int_{-\infty}^{0}e^{\omega(\alpha-ix)}+\int_{0}^{\infty}e^{-\omega(\alpha+ix)}d\omega$$

$$\frac{e^{\omega(\alpha-ix)}}{\alpha-ix}\bigg|_{\omega=-\infty}^{\omega=0}+\frac{e^{-\omega(\alpha-ix)}}{\alpha-ix}\bigg|_{\omega=0}^{\omega=\infty}=\frac{1}{\alpha-ix}+\frac{1}{\alpha+ix}=\frac{\alpha+ix+\alpha-ix}{\alpha^{2}+x^{2}}=\frac{2\alpha}{\alpha^{2}+x^{2}}$$

 

[vela]

You got the sign of the second term wrong.

 

[PiRho31416]

I don't see where the sign is wrong.

I broke it up the absolute value so it reads like follows:

$$f(\omega)=\begin{cases} e^{\omega\alpha} & \mbox{if }\omega<0\\ e^{-\omega\alpha} & \mbox{if }\omega>0\end{cases}$$

 

[vela]

I'm referring to the second integral. Once you integrated and before you plugged in the limits, you should have a minus sign in front of the second term.

 

[PiRho31416]

Got it! Dang those minus signs Thanks!