# Thread: Definite Integral, infinite limits, complex analysis

1. ## Definite Integral, infinite limits, complex analysis

I need to show that:

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

It seems obvious that i need to use

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

but I can't get it to work.

I have tried thinking it of as a complex curve integral

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

with

$\displaystyle \gamma(t)=t, 0<t<\infty$

and trying to use a different parametrization for $\displaystyle \gamma(t)$

2. 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.

3. Yes, the -1 is definitely there. It is straight out of the book.

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

Edit: Your idea does seem to work since $\displaystyle \frac{\cos{x}-1}{x}$ 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!

4. Glad you got it to work. Have a good one!