= Solution
Consider the <meromorphic function>
$$
f(z)=\frac{e^{iz}}{z(z^2-1)}.
$$
Close the <contour integration>[contour] with a large semicircle in the upper half-plane, indenting above the three real <poles>[pole] $-1,0,1$. The large-arc integral vanishes by <Jordan lemma>[Jordan's lemma], while the <upper-half-plane indentation rule> gives
$$
\operatorname{PV}\int_{-\infty}^{\infty}
\frac{e^{ix}}{x(x^2-1)}\,dx
=i\pi\sum_{a\in\{-1,0,1\}}\operatorname{Res}(f,a).
$$
The three <residues>[residue] are
$$
\operatorname{Res}(f,-1)=\frac{e^{-i}}2,
\qquad
\operatorname{Res}(f,0)=-1,
\qquad
\operatorname{Res}(f,1)=\frac{e^i}2.
$$
Their sum is $\cos1-1$, so the <residue theorem> yields
$$
\operatorname{PV}\int_{-\infty}^{\infty}
\frac{e^{ix}}{x(x^2-1)}\,dx
=i\pi(\cos1-1).
$$
The real part of the integrand is an <odd function>, so its full-line principal value is zero. Its imaginary part,
$$
\frac{\sin x}{x(x^2-1)},
$$
is an <even function>. Taking imaginary parts and halving the full-line integral therefore gives
$$
\boxed{
\operatorname{PV}\int_0^\infty\frac{\sin x}{x(x^2-1)}\,dx
=\frac\pi2(\cos1-1)
}.
$$
Back to article page
