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I need to find the value of the integral:

$\int_{-\infty}^{\infty} \frac{sin^2x}{x^2}dx $

Right now progress:

Because the value of $\frac{sin^2x}{x^2}$ is convergent, the integral will be equal to its principal value. So $\int_{-\infty}^{\infty} \frac{sin^2z}{z^2}dz = P\int_{-\infty}^{\infty}\frac{sin^2x}{x^2}dx $

We can write $sin^2 x = \frac{(e^{ix} - e^{-ix})^2}{(2i)^2x^2}= \frac{e^{2ix} - 2 e^{ix} e^{-ix} + e^{-2ix}}{(2i)^2x^2}$ however, now $e^{2ix} - 2+ e^{-2ix} = cosh(2ix)/2 -2$
For cosh, we know that the right contour to use is rectangular contour however, I'm kind of stuck and would really appreciate some insight. Thanks.

phoenix
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1 Answers1

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Let $I = \int_{-\infty}^{\infty} \frac{\sin^2 x}{x^2}\mathrm{d}x = \int_{-\infty}^{\infty} \frac{1 - \cos 2x}{2 x^2}\mathrm{d}x = \int_{-\infty}^{\infty} \frac{1-\cos x}{x^2}\mathrm{d}x$.

Now let $I_1(r,R) = (\int_{-R}^{-r} + \int_r^R)\frac{1 - \cos x}{x^2}\,\mathrm{d}x$, $I_2(r,R) = (\int_{-R}^{-r} + \int_r^R)\frac{\sin x}{x^2}\,\mathrm{d}x$, $R > r > 0$.(I am doing this because $\int_0^R \frac{\sin x}{x^2}\,\mathrm{d}x$ and $\int_{-R}^0 \frac{\sin x}{x^2}\,\mathrm{d}x$ diverge). Put $C_r = \{z: |z| = r, \mbox{Im}z \geq 0\}$, $C_R = \{z: |z| = R, \mbox{Im} z \geq 0\}$.

By Cauchy integral theorem, we have $$ I_1(r,R) - iI_2(r,R) + \int_{C_r^{-}}\frac{1 - e^{iz}}{z^2}\mathrm{d}z + \int_{C_R}\frac{1 - e^{iz}}{z^2}\mathrm{d}z = 0 $$ Jordan's lemma implies that $$ \lim_{R\rightarrow +\infty} \int_{C_R} \frac{1 - e^{iz}}{z^2} \mathrm{d}z = 0 $$ On the other hand, $$ \lim_{z\rightarrow 0} z \cdot \frac{1 - e^{iz}}{z^2} = -i $$ So $\lim_{r\rightarrow 0^{+}} \int_{C_r} \frac{1 - e^{iz}}{z^2} \mathrm{d}z = \pi$. Hence $$ I = \lim_{R\rightarrow +\infty, r\rightarrow 0^{+}} I_1(r,R) = \mbox{Re}\left( \int_{C_r}\frac{1-e^{iz}}{z^2}\mathrm{d}z - \int_{C_R}\frac{1-e^{iz}}{z^2}\mathrm{d}z \right) = \pi $$

zeno tsang
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