Discussion
The Cauchy principal value of the integral \(\int_{-\infty}^{\infty} dx \frac{e^{2ix}\sin x}{x}\) is :
Answer: C
The principal value is 0.
Write \(\sin x = (e^{ix} - e^{-ix})/2i\), so the integrand splits into two oscillating terms divided by \(x\):
\[ I = \frac{1}{2i}\mathrm{P}\int\frac{e^{3ix} - e^{ix}}{x}dx \]
For any \(k \gt 0\), closing the contour in the upper half plane and indenting around the pole at the origin gives \(\mathrm{P}\int e^{ikx}/x\,dx = i\pi\), independent of \(k\). The two terms are therefore equal and cancel:
\[ I = \frac{1}{2i}(i\pi - i\pi) = 0 \]
The options \(\pi i\) and \(2\pi i\) are what one term alone (or twice one term) would give, which forgets the subtraction; 1 has no source in the calculation.
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