# Evaluate the integral using principal value and complex analysis

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.

• The function $z \rightarrow \frac{\sin^2 z}{z^2}$ is entire... Apr 29, 2014 at 11:44
• ^so? I'm sorry could you elaborate? Apr 29, 2014 at 12:06
• When you use residue theorem to compute an integral, what is interesting is the poles of your function. Here your function has no poles, so you have to split it some way. See here for a complete solution. Apr 29, 2014 at 12:13

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 $$I = \lim_{R\rightarrow +\infty, r\rightarrow 0^{+}} I_1(r,R) = \mbox{Re}\left( \int_{C_r}\frac{1-e^{iz}}{z^2}\mathrm{d}z - \int_{C_R}\frac{1-e^{iz}}{z^2}\mathrm{d}z \right) = \pi$$