Evaluate using contour the integral $\displaystyle\int_\mathbb{R}\left(\dfrac{\sin(x)}{x}\right)^2dx\;$

I am using the the boundaries of a large semicircle with radius $R$ and a small semicircle with radius $\varepsilon$ for the contour and the integration over the contour breaks into four parts. I am stuck at evaluating at the piece along the small circle, namely: $$\int_\gamma\left(\frac{\sin(z)}{z}\right)^2 dz = - \int_0^\pi \frac{e^{2i\varepsilon e^{i\theta}}}{\varepsilon^2 e^{2i\theta}}i\varepsilon e^{i\theta} d\theta$$ It appears to me that the integral on the right hand side is zero due to the extra power of $\varepsilon$ on the denominator as $\varepsilon \rightarrow 0$.

I wonder how we deal with this problem?


1 Answer 1


First of all, it is never a good idea to use trigonometric functions in such exercises. It is very hard to bound them on semicircle curves. So always replace them with exponential. For example, here we have $\frac{\sin^2x}{x^2}=\frac{1}{2}\frac{1-\cos(2x)}{x^2}$. So the complex function you should define is $f(z)=\frac{1}{2}\frac{1-e^{2iz}}{z^2}$. Your choice of the contour is correct. The integral over the large semicircle clearly tends to $0$ as $R\to\infty$, the integrals over the straight lines should not present any problems.

Now let $\gamma$ be the small semicircle. Note that $f$ has a simple pole at $z=0$, so its Laurent series has the form $\sum\limits_{n=-1}^\infty a_nz^n$. So in a small neighborhood of $0$ we can write $f(z)=\frac{a_{-1}}{z}+h(z)$ where $h$ is a holomorphic function. So now we have:


Since $h$ is holomorphic it is bounded near $0$, and so the second integral tends to $0$ when $\epsilon\to 0$. And the first integral is just $-a_{-1}\pi i$, this is a direct computation. Finally, $a_{-1}$ is the residue of $f$ at $z=0$. I believe you know how to compute residues.


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