# Math Help - Integral of sin(x)/x

1. ## Integral of sin(x)/x

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

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

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

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

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

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)?

2. just put $t=\bigg(N+\frac12\bigg)x$ on the RHS and you're done!

3. Thanks! What about the limit point being $b=\pi$?

Do you mean do that in the very first equation in my post? The 'show that...' part?

Or in the last line of my post.

4. Ok, now that I've looked at it... Is it (using your way just working with the LHS though), set $x = (N + \frac{1}{2})t$

Then $dx = (N + \frac{1}{2})dt$

So $\int \frac{\sin(x)}{x} dx = \int \frac{\sin((N + \frac{1}{2})t)}{(N + \frac{1}{2})t)} \cdot (N+\frac{1}{2}) dt = \int \frac{\sin((N + \frac{1}{2})t}{t}dt$

I've left out integral limits as I'm not too sure how to deal with them...

5. You flipped the variables, look at my substitution and look at yours.

What I'm saying, it's easier to work with the RHS instead of LHS.

6. Ok that parts easy enough (I just thought I'd do it the other way round to see if it worked).

But I'm not sure how to change the range of integration from $\pi$ to b. Any how to justify the limit changes (from N to b).

Is it that we have $b = (N + \frac{1}{2})\pi$ so as N goes to infinity so must b.