Computation of $\int_{-\infty}^{\infty} \left(\frac{\sin x}{x}\right)^2e^{itx}dx$ for real $t$ [duplicate]

We need to compute the integral $~\displaystyle \int_{-\infty}^{\infty} \left(\frac{\sin x}{x}\right)^2e^{itx}dx~$ for real $t$.
Now notice that, for real $t$ $$\int_{-\infty}^{\infty} \left(\frac{\sin x}{x}\right)^2e^{itx}dx=\lim_{A \to \infty} \int_{-A}^{A} \left(\frac{\sin x}{x}\right)^2e^{itx}dx.$$ I know that, for $t=0$ we have $~\displaystyle \int_{-\infty}^{\infty} \left(\frac{\sin x}{x}\right)^2e^{itx}dx=\pi,~~$ and I can compute that. But for any $t\neq 0,~$ I am not able to find.
For simplicity, let us consider a complex function $f$ as $\displaystyle f(z)=\frac{\sin^2 z} {z^2}e^{itz}$. Then $f$ is entire for all $t\in \mathbb R.$ Now to find the integral we can take a positively oriented semicircle center at origin and radius of $A$ in upper half plane, then by the Cauchy's theorem we have $$\int_{-A}^{A} \left(\frac{\sin x}{x}\right)^2e^{itx}dx+\int_{\gamma} \left(\frac{\sin z}{z}\right)^2e^{itz}dz=0$$ where $\gamma $ is the upper semicircular part with length $\pi A$. Now I am not able to proceed from here. Please help me to solve this.

Here is a solution using contour integral.

Let $I(t)$ denote the integral. Then by using $\sin x = \frac{e^{ix} - e^{-ix}}{2i}$, we get

\begin{align*} I(t) &= \int_{-\infty}^{\infty} \left(\frac{\sin x}{x}\right)^2 e^{itx} \, \mathrm{d}x \\ &= \int_{-\infty}^{\infty} \frac{2e^{itx} - e^{i(t+2)x} - e^{i(t-2)x}}{4x^2} \, \mathrm{d}x \\ &= J(t) - J(t+2) - J(t-2), \end{align*}

where $J(\cdot)$ is defined by

$$ J(t) = \operatorname{PV}\!\! \int_{-\infty}^{\infty} \frac{e^{itx} - 1}{4x^2} \, \mathrm{d}x = \lim_{\varepsilon \to 0^+} \int_{|x|>\varepsilon} \frac{e^{itx} - 1}{4x^2} \, \mathrm{d}x. $$

So we turn to computing $J(t)$. We first consider the case $t \geq 0$. Then by using the contour

we find that

$$ \int_{\varepsilon<|x|<R} \frac{e^{itx} - 1}{4x^2} = \int_{\gamma_{\varepsilon}} \frac{e^{itz} - 1}{4z^2} \, \mathrm{d}z - \int_{\gamma_{R}} \frac{e^{itz} - 1}{4z^2} \, \mathrm{d}z. $$

Then by noting that $|e^{itz}| \leq 1$ whenever $t \geq 0$ and $\operatorname{Im}(z) \geq 0$, as $\varepsilon \to 0^+$ and $R \to \infty$, we get

$$ J(t) = \pi i \, \underset{z=0}{\mathrm{Res}} \, \frac{e^{itz} - 1}{4z^2} = -\frac{\pi t}{4}. $$

The case $t \leq 0$ can be tackled either by using the contour integrals along the lower semicircular contours or by using the identity $\overline{J(t)} = J(-t)$. Altogether, we get

$$ J(t) = -\frac{\pi |t|}{4}. $$

Therefore, plugging this into $I(t)$, we get

$$ I(t) = \frac{\pi}{4}(|t + 2| + |t - 2| - 2|t|) = \frac{\pi}{2} \max\{0, 2 - |t|\}. $$

Hint

Without complex analysis, contour integration or residues.

The antiderivative does not make any problem if you write $$\left(\frac{\sin x}{x}\right)^2e^{itx}=\frac{\left(1-e^{2 i x}\right)^2 e^{i (t-2) x}}{4 x^2}$$ which makes that you will face a bunch of exponential integral functions.

I am not sure that if we do not have $-2 < t < 2$, the result would be else than $0$. If we are in this range, the value should not depend on the sign of $t$.

For the definite integral, the result should be $$\int_{-\infty}^{\infty} \left(\frac{\sin x}{x}\right)^2e^{itx}\,dx=\pi \left(1+\frac{1}{2}|t|\right)$$