Show that $\displaystyle \lim_{b \rightarrow \infty} \int^b_0 \frac{\sin(x)}{x} dx = \lim_{N \rightarrow \infty} \int^{\pi}_0 \frac{\sin((N + \frac{1}{2})x)}{x} dx$

Thoughts...

First of, $\displaystyle \sin(y) = y$ for $\displaystyle y \rightarrow 0$. (I think)

So let $\displaystyle y = \frac{x}{2n}$, then $\displaystyle 2n\sin \Big{(}\frac{x}{2n}\Big{)} = x$.

Also, substitute $\displaystyle x = nt$, $\displaystyle dx = ndt$ then we get.

$\displaystyle \lim_{b \rightarrow \infty} \int^b_0 \frac{\sin(x)}{x} dx = \frac{1}{2} \lim_{n\to\infty} \int\limits_0^b \frac{\sin(nt)}{\sin(\frac{t}{2})} dt $$\displaystyle = \frac{1}{2} \lim_{N\to\infty} \int\limits_0^b \frac{\sin((N+\frac{1}{2})t)}{\sin(\frac{t}{2})} dt$

I do this as I seem to remember doing something similar in a course a year or two ago. Don't really know where to go next though and pretty sure I've messed up the integrals limits... Maybe the b should become nt (in the integral)?