Solution

f (z) = \frac{e ^{i z}}{z ( z ^{2} - 1 )} .

Close the contour with a large semicircle in the upper half-plane, indenting above the three real pole

- 1, 0, 1

. The large-arc integral vanishes by Jordan's lemma, while the upper-half-plane indentation rule gives

PV \int_{- \infty}^{\infty} \frac{e ^{i x}}{x ( x ^{2} - 1 )} d x = iπ \sum_{a \in {- 1, 0, 1}} Res (f, a) .

The three residue are

Res (f, - 1) = \frac{e ^{- i}}{2}, Res (f, 0) = - 1, Res (f, 1) = \frac{e ^{i}}{2} .

Their sum is

cos 1 - 1

, so the residue theorem yields

PV \int_{- \infty}^{\infty} \frac{e ^{i x}}{x ( x ^{2} - 1 )} d x = iπ (cos 1 - 1) .

The real part of the integrand is an odd function, so its full-line principal value is zero. Its imaginary part,

\frac{s i n x}{x ( x ^{2} - 1 )},

is an even function. Taking imaginary parts and halving the full-line integral therefore gives

PV \int_{0}^{\infty} \frac{sin x}{x ( x ^{2} - 1 )} d x = \frac{π}{2} (cos 1 - 1) .

Solved by gpt-5.6-sol high.

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