Evaluation of the principal value of $\int\limits_{-\infty}^\infty \frac{\sin 2x}{x^3} \, dx$ I'm trying to evaluate an integral $\int\limits_{-\infty}^\infty \frac{\sin 2x}{x^3}\,dx$ using Cauchy's theorem. Considering an integral from $-R$ to $-\epsilon$, then a semicircular indentation around $x=0$, then $\epsilon$ to $R$, then a semicircular contour from $R$ to $-R$. Around the pole at $x=0$, the semicircular contribution gives $$\int\limits_\pi^0 \, dz\frac{e^{2iz}}{z^3}=\int_\pi^0 (\epsilon e^{i\theta})(i \, d\theta) \frac{e^{\epsilon e^{i\theta}}}{(\epsilon e^{i\theta})^3}$$ What I need is the limiting value of this integral as $\epsilon\rightarrow 0$. But it seems to diverge.
 A: We have that $\;z=0\;$ is clearly a triple pole of
$$f(z):=\frac{e^{i2z}}{z^3} ,\;\;\text{and in this case it is probable easier to use power series for the residue:}$$
$$\frac{e^{2iz}}{z^3}=\frac1{z^3}\left(1+2iz-\frac{4z^2}{2!}-\ldots\right)=\frac1{z^3}+\frac{2i}{z^2}-\frac2z-\ldots\implies\text{Res}_{z=0}(f)=-2$$
so taking the usual contour with a "bump" around zero, we get
$$0=\lim_{R\to\infty,\,\epsilon\to0}\oint_{\Gamma_R}f(z)=\int_{-\infty}^\infty\frac{e^{2ix}}{x^3}dx-\int_{\gamma_\epsilon}f(z)dz= \int_{-\infty}^\infty\frac{e^{2ix}}{x^3}dx+2\pi i\implies$$
$$-2\pi i=\int_{-\infty}^\infty\frac{e^{2ix}}{x^3}dx=\int_{-\infty}^\infty\frac{\cos2x+i\sin2x}{x^3}dx\implies \int_{-\infty}^\infty\frac{\sin2x}{x^3}dx=-2\pi$$
the last equality following from comparing real and imaginary parts in both sides.
The above though is just CPV (Cauchy's Principal Value) of the integral, since it doesn't converge in the usual sense of the word.
A: $$
\frac{\sin (2x)}{x^3} = \left( \vphantom{\frac2{x^2}} \right.\underbrace{\frac{\sin(2x)}{2x}}_{\begin{smallmatrix} \text{This} \\ \text{approaches} \\[2pt] \text{$1$ as $x\to0$} \\[2pt] {} \end{smallmatrix}} \cdot \left. \frac 2 {x^2} \right)
$$
For $x$ close enough to $0$ you have $\dfrac{\sin(2x)}{2x}> 0.9$ and then $$ \int_{-a}^a \frac{\sin(2x)}{2x} \cdot \frac 2 {x^2} \, dx \ge 0.9 \int_{-a}^a \frac{2\, dx}{x^2} = +\infty. $$
And $$ \int_a^\infty \left|\frac{\sin(2x)}{x^3}\right| \, dx \le \int_a^\infty \frac{dx}{x^3} <\infty $$ so here we have absolute convergence.  (And $x\mapsto \dfrac{\sin(2x)}{x^3}$ is an even function, so that $\text{takes care of } \displaystyle \int_{-\infty}^{-a} \frac{\sin(2x)}{x^3} \, dx$.)
So you have $+\infty$ plus something that converges absolutely, and you conclude that the whole thing diverges to $+\infty$.
A: It is not difficult to show (through the Laplace transform, for instance) that
$$ \int_{-\infty}^{+\infty}\frac{\sin(2z)-2z}{z^3}\,dz = -2\pi \tag{1}$$
so the principal value of your integral is given by $-2\pi$ plus twice the principal value of
$$ \int_{-\infty}^{+\infty}\frac{dz}{z^2} \tag{2} $$
that is just $+\infty$, since the integrand function is even and it has a double pole at the origin.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
We can 'jump' directly into the $\mathrm{P.V.}$ definition:
\begin{align}
&\color{#f00}{%
\mathrm{P.V.}\int_{-\infty}^{\infty}{\sin\pars{2x} \over x^{3}}\,\dd x}\
\stackrel{\mbox{def.}}{=}\
\lim_{\epsilon \to 0^{+}}\bracks{%
\int_{-\infty}^{-\epsilon}{\sin\pars{2x} \over x^{3}}\,\dd x +
\int_{\epsilon}^{\infty}{\sin\pars{2x} \over x^{3}}\,\dd x}
\\[3mm] = &\
2\lim_{\epsilon \to 0^{+}}
\int_{\epsilon}^{\infty}{\sin\pars{2x} \over x^{3}}\,\dd x =
2\lim_{\epsilon \to 0^{+}}\bracks{%
{\sin\pars{2\epsilon} \over 2\epsilon^{2}}
+ \int_{\epsilon}^{\infty}{\cos\pars{2x} \over x^{2}}\,\dd x}
\\[3mm] & =
2\lim_{\epsilon \to 0^{+}}\bracks{%
{\sin\pars{2\epsilon} \over 2\epsilon^{2}} + {\cos\pars{2\epsilon} \over \epsilon}
- 2\int_{\epsilon}^{\infty}{\sin\pars{2x} \over x}\,\dd x} =
\color{#f00}{+\infty}
\end{align}
