# Math Help - sin(cLt)*e^(-iLx)

1. ## sin(cLt)*e^(-iLx)

I am unsure if this belongs in calculus, differential equation, advance applied math, etc.

So I would like to figure out the trick to
$
g(x) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}\! \sin(c \omega t) e^{-i\omega x} \, d\omega$
.
Yes, I know this is an Inverse Fourier Transform, and I know there are tables. What I am looking for is a good explanation of how
$
g(x) = \frac{i}{2}\left[\delta(x+ct) - \delta(x-ct)\right]$
.
Also, I figure this might help me understand the Dirac-delta function a bit better, because I am unsure why
$f(x) \ast g(x) = f(x + ct) - f(x - ct)$.

Thank you in advance.

2. Originally Posted by lvleph
I am unsure if this belongs in calculus, differential equation, advance applied math, etc.

So I would like to figure out the trick to
$
g(x) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}\! \sin(c \omega t) e^{-i\omega x} \, d\omega$
.
Probably the simplest way to do this is to use the fact that $cos(x)= \frac{e^{ix}+ e^{-ix}}{2}$.

With that, the integral becomes $g(x)= \frac{1}{2\pi}\int_{-\infty}^\infty \frac{e^{i\omega(ct- x)}+ e^{-i\omega(ct+ x}}{2}d\omega$

Yes, I know this is an Inverse Fourier Transform, and I know there are tables. What I am looking for is a good explanation of how
$
g(x) = \frac{i}{2}\left[\delta(x+ct) - \delta(x-ct)\right]$
.
Also, I figure this might help me understand the Dirac-delta function a bit better, because I am unsure why
$f(x) \ast g(x) = f(x + ct) - f(x - ct)$.

Thank you in advance.

3. Yeah, I tried this and it was messing me up.
$
g(x)= \frac{1}{2\pi}\int_{-\infty}^\infty \frac{e^{i\omega(ct- x)} - e^{-i\omega(ct+ x)}}{2i}d\omega
$

$
g(x)= \frac{1}{4\pi i}\left[\frac{-i}{ct - x}e^{i\omega(ct- x)} - \frac{i}{ct + x} e^{-i\omega(ct+ x)} \right|_{-\infty}^{+\infty}
$
.
This is where I get messed up, and probably where the dirac-delta comes in. If you take these limits they are
$g(x) = \frac{-1}{4\pi}\left[\frac{+\infty }{ct - x} + \frac{+\infty}{ct + x}\right]$