solve integral of $\frac{\sin (ax)}{ \sin(x)}$

Question

I would like to find the area under the curve of $\frac{\sin(ax/2)}{\sin(x/2)}$, namely between the first zero crossing on the left and right:

$$ \int_{-\frac{2\pi}{a}}^{\frac{2\pi}{a}} \frac{\sin(\frac{ax}{2})}{\sin(\frac{x}{2})} $$

I realized from Wolfram Alpha that this was not a simple solution, so I was wondering if there is a good approximation for the area. For my application, $a$ is usually over $10000$.

Thank you for any advice and suggestions.

Answer 1

Using symmetry and the change of variable $x=2\,t$ we have $$ \int_{-\frac{2\pi}{a}}^{\frac{2\pi}{a}} \frac{\sin(\frac{ax}{2})}{\sin(\frac{x}{2})}\,dx= 4\int_0^{\pi/a} \frac{\sin(a\,x)}{\sin x}\,dx. $$ If $a>2$ then $$ \frac{a}{\pi}\,\Bigl(\sin\frac{\pi}{a}\Bigr)\,x\le\sin x\le x\quad 0\le x\le\frac{\pi}{a} $$ and $$ \int_0^{\pi/a}\frac{\sin(a\,x)}{x}\,dx\le\int_0^{\pi/a}\frac{\sin(a\,x)}{\sin x}\,dx\le\frac{\pi}{a}\,\csc\frac{\pi}{a}\int_0^{\pi/a}\frac{\sin(a\,x)}{x}\,dx. $$ Finally we get $$ \int_0^{\pi}\frac{\sin x}{x}\,dx\le\int_0^{\pi/a}\frac{\sin(a\,x)}{\sin x}\,dx\le\frac{\pi}{a}\,\csc\frac{\pi}{a}\int_0^{\pi}\frac{\sin x}{x}\,dx. $$ For $a$ large $$ 1\le\frac{\pi}{a}\,\csc\frac{\pi}{a}\le1+\frac16\,\frac{\pi^2}{a^2}. $$ Thus $$ 0\le\int_0^{\pi/a}\frac{\sin(a\,x)}{\sin x}\,dx-\int_0^{\pi}\frac{\sin x}{x}\,dx\le\frac16\,\frac{\pi^2}{a^2}\int_0^{\pi}\frac{\sin x}{x}\,dx. $$

Answer 2

If $a$ is so large, then the bounds of the integral are not far from 0, so you can approximate both numerator and denominator with Maclaurin series expansion.