principal value of $\int_{-\infty}^{\infty}\frac{\sin^2(x)}{x^2}\mathrm{d}x$

Question

I know the answer is $\pi$ there is a proof here. Now looking to my textbook (textbook image) the result should be $0$. Using the last equation on the right hand page we have: $$ i\pi(\sin^2(x))'|_{x=0} = 0 $$ and there are no other poles except $0$ so the result should be $0$.

What is the problem? Is the definition for principal value different in the two cases?

Answer 1

One may observe that, by the Taylor series expansion, as $x \to 0$, $$ \sin x=x+O(x^3) $$ giving $$ \frac{\sin^2 x}{x^2}=1+O(x^2), $$ on the other hand, as $x \to \infty$, $$ \left|\frac{\sin^2 x}{x^2}\right|\leq \frac1{x^2} $$ consequently the given integral is convergent: $$ 0<\int_{-\infty}^{+\infty}\frac{\sin^2 x}{x^2}\:dx<\infty. $$ No need to consider a principal value: there is no singularity of $\dfrac{\sin^2 x}{x^2}$ over $\mathbb{R}$.

We have

$$ \int_{-\infty}^{+\infty}\frac{\sin^2 x}{x^2}\:dx=2\int_0^{+\infty}\frac{\sin^2 x}{x^2}\:dx=2\cdot \frac{\pi}2=\pi, $$

a proof of the latter integral evaluation may be found here.

Answer 2

That kind of integrals are best managed through integration by parts. We have:

$$ \int_{-\infty}^{+\infty}\left(\frac{\sin x}{x}\right)^2\,dx = \int_{-\infty}^{+\infty}\frac{\frac{d}{dx}\sin^2 x}{x}\,dx = \int_{-\infty}^{+\infty}\frac{\sin(2x)}{x}\,dx = \int_{-\infty}^{+\infty}\frac{\sin z}{z}\,dz=\color{red}{\pi}.$$

Answer 3

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\begin{align} \color{#f00}{\int_{-\infty}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\,\dd x} & = \int_{-\infty}^{\infty}\ \overbrace{\bracks{\half\int_{-1}^{1}\expo{\ic kx}\,\dd k}} ^{\ds{{\sin\pars{x} \over x}}}\ \overbrace{\bracks{\half\int_{-1}^{1}\expo{-\ic qx}\,\dd q}\,\dd x} ^{\ds{{\sin\pars{x} \over x}}} \\[3mm] & = {\pi \over 2}\int_{-1}^{1}\int_{-1}^{1}\ \overbrace{% \int_{-\infty}^{\infty}\expo{\ic\pars{k - q}x}\,{\dd x \over 2\pi}} ^{\ds{\delta\pars{k - q}}}\ \,\dd k\,\dd q = {\pi \over 2}\int_{-1}^{1}\ \overbrace{\int_{-1}^{1}\delta\pars{k - q}\,\dd k}^{\ds{=\ 1}}\ \,\dd q \\[3mm] & = {\pi \over 2}\int_{-1}^{1}\dd q = \color{#f00}{\pi} \end{align}