# Help evaluating or upper-bounding integral $\int_{-\infty}^{\infty} \frac{ab\operatorname{sinc}^2(cx)}{a+b\operatorname{sinc}^2(cx)}\;dx$

I am trying to evaluate the following definite integral:

$$\int_{-\infty}^\infty \frac{ab\operatorname{sinc}^2(cx)}{a+b\operatorname{sinc}^2(cx)}\;dx$$

where $\operatorname{sinc}(x)=\dfrac{\sin x}{x}$ is the Sinc function and $a$, $b$, and $c$ are positive constants. I would also be happy with a reasonably tight upper bound. Does anyone have any ideas?

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As often happens, the word "solve" is being used here where "evaluate" would be appropriate. One solves problems; one solves equations; one evaluates expressions. –  Michael Hardy Mar 1 '12 at 22:51
Sorry, changed wording of the question accordingly... –  M.B.M. Mar 1 '12 at 22:55
Did you try complex contour integration? maybe some more info on a,b,c would be needed –  Dennis Gulko Mar 1 '12 at 23:27
@DennisGulko I've never used complex contour integration (and didn't even know that something like that exists before you mentioned it). Is there a good book/tutorial to get me started? Also, what kind of info would be needed about $a$, $b$ and $c$? –  M.B.M. Mar 2 '12 at 5:38
@Bullmoose: The lecture notes of William Chen are a good start: rutherglen.science.mq.edu.au/wchen/lnicafolder/lnica.html. What you really need is just chapter 11, but, probably, you'll need the other chapters for background. –  Dennis Gulko Mar 2 '12 at 6:07

I an assuming $a,b,c>0$. Let $\delta=\sqrt{b/a}$. Then $$\int_{-\infty}^\infty \frac{a\,b\,\operatorname{sinc}^2(c\,x)}{a+b\,\operatorname{sinc}^2(c\,x)}\,dx=\frac{b}{c}\int_{-\infty}^\infty\frac{\sin^2x}{x^2+\delta^2\sin^2x}\,dx=\frac{b}{c}\,I(\delta).$$ Integrating the inequalities $$\frac{\sin^2x}{x^2+\delta^2}\le\frac{\sin^2x}{x^2+\delta^2\sin^2x}\le\frac{\sin^2x}{x^2}$$ we get $$\frac{1-e^{-2\delta}}{2\,\delta}\,\pi\le I(\delta)\le\pi.$$ This bounds are good only for for small $\delta$. For Large $\delta$ try the following: \begin{align*} I(\delta)&=2\int_0^\delta\frac{\sin^2x}{x^2+\delta^2\sin^2x}\,dx+2\int_\delta^\infty\frac{\sin^2x}{x^2+\delta^2\sin^2x}\,dx\\ &\le\frac{2\,\delta}{1+\delta^2}+2\int_\delta^\infty\frac{\sin^2x}{x^2}\,dx\\ &=\frac{2\,\delta }{\delta ^2+1}+\frac{1-\cos2\,\delta}{\delta}+{\pi -2\,\operatorname{Si}(2\,\delta )}, \end{align*} where $\operatorname{Si}$ is the sine integral. in the graph, $I(\delta)$ is in red and the upper and lower bound in blue.
Thank you! I've got a question about the upper bound for large $\delta$: could you please elaborate on how $\int_0^\delta\frac{\sin^2 x}{x^2+\delta^2\sin^2x}dx\leq\frac{\delta}{1+\delta^2}$? –  M.B.M. Mar 2 '12 at 21:48
$x^2\ge\sin^2x$. –  Julián Aguirre Mar 2 '12 at 21:58