# Definite Integral, infinite limits, complex analysis

• Sep 30th 2010, 04:22 PM
lusheslewis
Definite Integral, infinite limits, complex analysis
I need to show that:

$\displaystyle{\int_{0}^{\infty}\frac{\sin{x}}{x}\, dt=\frac{1}{2i}\int_{-\infty}^{\infty}\frac{e^{ix}-1}{x}}$

It seems obvious that i need to use

$\sin{x}=\frac{e^{ix}-e^{-ix}}{2i}$

but I can't get it to work.

I have tried thinking it of as a complex curve integral

$\displaystyle{\int_{\gamma}^{}f(z)\,dz \ \,\ \ f(z)=\frac{\sin{z}}{z}$

with

$\gamma(t)=t, 0

and trying to use a different parametrization for $\gamma(t)$
• Sep 30th 2010, 04:30 PM
Ackbeet
I'd be willing to bet that you could take the RHS, substitute exp(ix) = cos(x) + i sin(x), and then use properties of even and odd functions to get the desired result. Are you sure that that -1 is in the numerator of the RHS's integrand? Because that's not such a nice feature of the problem.
• Sep 30th 2010, 04:34 PM
lusheslewis
Yes, the -1 is definitely there. It is straight out of the book.

Maybe it has something to do with the limit of $\frac{\cos{x}-1}{x}$ as x -> 0

Edit: Your idea does seem to work since $\frac{\cos{x}-1}{x}$ is odd and should integrate to zero. I was trying to go from the left and show it equals the right though which was obviously the wrong way. Cheers!
• Sep 30th 2010, 05:17 PM
Ackbeet
Glad you got it to work. Have a good one!