I'm having some problems solving this integral:
$$ I = \mathcal{P} \int_{-\infty}^{+\infty} \frac{1-e^{2ix}}{x^2} \ dx$$
where $\mathcal{P}$ is the Cauchy principal value. The exercise suggests to use the fact that:
$$I_* = \frac{1}{2} \operatorname{Re} \left[I\right]=\mathcal{P} \int_{-\infty}^{+\infty} \frac{\sin^2 x}{x^2} \ dx$$
since $\sin^2 x = \frac{1}{2} \left(1- \cos(2x)\right)$.
My solution. I went on and tried to solve $I_*$ as follows: I used the fact that the analytic extension of the integrand has no poles, which makes the integral equals to $0$ by using residue theory:
$$\lim_{R\to + \infty}\oint_{\Gamma_R} \frac{\sin^2z}{z^2} \ dz = \mathcal{P} \int_{-\infty}^{+\infty} \frac{\sin^2 x}{x^2} \ dx = 0$$
where the second equality is true since
$$\oint_{\Gamma_R}\frac{\sin^2z}{z^2} \ dz =\left(\int_{-R}^{+R} + \int_{C_R}\right) \frac{\sin^2z}{z^2} \ dz$$
where $C_R = \{z = r e^{i \theta}\in \mathbb{C} : 0\le r \le R\}$ and
$$\left\lvert \int_{C_R} \frac{\sin^2 z}{z^2} \ dz \right\rvert \le \int_{C_R} \frac{1}{|z^2|} \ dz \le \int_{C_R} \frac{1}{|R^2|} \ dz \to 0, \ R\to +\infty$$
$$\lim_{R\to+\infty} \int_{-R}^{+R} \frac{\sin^2z}{z^2} \ dz = \lim_{R\to+\infty} \frac{\sin^2 x}{x^2} \ dx \equiv \mathcal{P} \int_{-\infty}^{+\infty} \frac{\sin^2 x}{x^2} \ dx$$
Since the residues of this function are all $0$, this means that also
$$\mathcal{P} \int_{-\infty}^{+\infty} \frac{\sin^2 x}{x^2} \ dx =0 $$
Ok, now, since this implies that $\operatorname{Re}I = 0$, I thought that $I$ must have just an imaginary part; for this reason, I then tried to calculate the following:
$$\operatorname{Im} [I] = \mathcal{P} \int_{-\infty}^{+\infty} \frac{\sin(2x)}{x^2} \ dx \equiv \mathcal{P}\int_{-\infty}^{+\infty} h(x) \ dx$$
I extended $h(x)\to h(z)$, which has a first order pole in $z=0$:
$$\operatorname{Res}\left[h(z) , z=0\right]=\lim_{z\to 0 } \left(z \frac{\sin 2z}{z^2}\right) = 2$$
Then I integrated $h(z)$ as follows:
$$\oint_{\Gamma_{r,R} } \frac{\sin 2z}{z^2} \ dz = \left(\int_{-R} ^{-r} + \int_{C_r^-} +\int_{r}^{R} + \int_{C_R} \right) \frac{\sin 2z}{z^2} \ dz $$
where:
$$\lim_{r\to 0} \int_{C_r^-} \frac{\sin 2z}{z^2} \ dz \to -i\pi\operatorname{Res}\left[h(z), z=0\right] = -2i\pi $$
$$\left\lvert \int_{C_R} \frac{\sin 2z}{z^2} \ dz \right\rvert \le \int_{C_R} \frac{1}{R^2}\to 0, \ R\to+\infty $$
$$\lim_{r \to 0, \ R\to +\infty} \left(\int_{-R}^{-r} + \int_{r} ^{R} \right) h(z) \ dz \equiv \mathcal{P}\int_{-\infty} ^{+\infty} h(x) \ dx $$
Putting all of this together, we get:
$$\mathcal{P}\int_{-\infty} ^{+\infty} \frac{\sin 2x}{x^2 } \ dx = -2i\pi$$
What bothers me the most and that makes me think I did something wrong is that the result of this real valued integral is an imaginary number. Moreover, this would mean $\operatorname{Im} (I) = -2i\pi\Rightarrow I\stackrel{?}{=} 2\pi$ or $I \stackrel{?}{=} -2i\pi$. Did I make some errors? Can you help me getting to the correct solution? I really need help with this because I feel like I'm missing something very important. Thanks a lot in advance for the help!!
EDIT 1: This is the solution by calculating directly $I$ with the residues. So we define $F(z) = \frac{1-e^{2iz} }{z^2}$, which is the analytical extension of the integrand of $I$. This clearly has a first order pole in $z=0$, which residue is obtained by:
$$\operatorname{Res}\left[F(z), z=0\right] = \lim_{z \to 0} \frac{d}{dz}\left(z^2 \frac{1-e^{2iz} }{z^2}\right) = -2i $$
The complex integral we need to calculate would be:
$$\lim_{r\to 0, \ R \to +\infty} \oint_{\Gamma_{r,R} } \frac{1-e^{2iz} }{z^2}\ dz = \lim_{r\to 0, R\to+\infty} \left(\int_{-R}^{-r} + \int_{C_r^-} +\int_{r} ^R + \int_{C_R} \right) \frac{1-e^{2iz} }{z^2}\ dz =0 $$
since of the first order pole in $z=0$ (and it is equal to $0$ because there are no poles inside the contour taken), it is needed to create a small arc of circumference $C_r$ to avoid calculating the function in $z=0$. After making sure that the integral on $C_R$ goes to $0$, which is done by using Jordan's Lemma and the fact that $\frac{1}{|z^2|}=\frac{1}{R^2}\to 0$ as $R\to+\infty$, we can calculate the integral on $C_r^-$ (where the minus sign is because it is “walked'' clockwise); this should be integral that gives the residue in $z=0$ of $F(z)$ when $r\to 0$ because of the theorem that states that:
$$\lim_{r \to 0} \int_{C_{r_\alpha } } F(z) \ dz = i \alpha \operatorname{Res}\left[F(z), z=0\right] $$
(which is true just for first order poles). This theorem implies automatically that
$$\lim_{r \to 0} \int_{C_r^-} \frac{1-e^{2iz} }{z^2} \ dz = -i \pi (-2i) = 2\pi =-2\pi$$
Since the other two integrals are such that:
$$\lim_{r \to 0, \ R\to+\infty} \left(\int_{-R} ^{-r} + \int_{r} ^R \right) \frac{1-e^{2iz} }{z^2} \ dz \equiv \lim_{r \to 0, \ R\to+\infty} \left(\int_{-R} ^{-r} + \int_{r} ^R \right) \frac{1-e^{2ix} }{x^2} \ dx \equiv \mathcal{P}\int_{-\infty} ^{+\infty} \frac{1-e^{2ix} }{x^2} \ dx $$
it means that:
$$\mathcal{P}\int_{-\infty} ^{+\infty} \frac{1-e^{2ix} }{x^2} \ dx = 2\pi$$
As pointed out in the comments, this is the most concise and easy solution for the problem; nonetheless why does the exercise, which is directly taken from a past exam that my professor made public, suggests to use the integral $I_*$?
EDIT 2: I'll contact my professor to ask him why he gave that hint in the exercise. I'll update this post afterwards to let you know what he replied to me.
EDIT 3: My professor told me that that was not an hint, rather it was possibile to evaluate that in integral ($I_*$) once obtained the main one, since $I_*$ is not obtainable by standard integration.
Best Answer
This inequality is incorrect: $$ \left| \int_{C_R} \frac{\sin^2(z)}{z^2} \, dz \right| \leq \int_{C_R} \frac{1}{|z^2|} \, dz $$ As pointed out by Gribouillis, your approach can be directly applied to the original integral.
Consider the complex function: $$ f(z)=\frac{1-e^{2iz}}{z^2} $$ which possesses a pole at $z=0$. We define the contour $C$ as an Indented semicircle.
By Cauchy's theorem, we obtain: $$ \oint_C f(z) \, dz = \int_{C_R} f(z) \, dz + \int_{-R}^{-\varepsilon} f(x) \, dx +\int_{\varepsilon}^{R} f(x) \, dx +\int_{C_{\varepsilon}} f(z) \, dz = 0 $$ where $C_R$ denotes the semicircle with radius $R$, and $C_{\varepsilon}$ represents the semicircle with radius $\varepsilon$.
For $C_R$, we have: \begin{align*} \left| \int_{C_R} f(z) \right| &= \left| \int_0^{\pi}\frac{1-e^{2(iR\cos \theta -R\sin \theta)}}{R^2 e^{2i\theta}} Re^{i\theta}\, d\theta \right| \\ &\leq \int_0^{\pi} \frac{1+e^{-2R\sin \theta}}{R} \, d\theta \to 0 \quad \text{as } R \to \infty \end{align*}
As for $C_{\varepsilon}$, by expressing $ f(z) $ as $ f(z)=\frac{-2iz}{z^2}+R(z) $, where $ R(z) $ is bounded as $ z \to 0 $, we get: \begin{align*} \int_{C_{\varepsilon}} f(z) &\to \int_0^{\pi} \frac{-2i}{z} dz= -2\pi \quad \text{as } \varepsilon \to 0 \end{align*}
Thus, we achieve the desired result: $$ \int_{-\infty}^{\infty} \frac{1-e^{2iz}}{z^2} dz=2\pi$$