The Cauchy principal value of the integral \(\displaystyle\int_{-\infty}^{\infty} \dfrac{e^{2ix}\sin x}{x}\,dx\) is:
\(0\)
Write \(\sin x = \dfrac{e^{ix} - e^{-ix}}{2i}\), so \(\dfrac{e^{2ix}\sin x}{x} = \dfrac{1}{2i}\left(\dfrac{e^{3ix}}{x} - \dfrac{e^{ix}}{x}\right)\).
Using the standard principal value \(\text{P.V.}\displaystyle\int_{-\infty}^{\infty} \dfrac{e^{iax}}{x}\,dx = i\pi\,\operatorname{sgn}(a)\), both \(a = 3\) and \(a = 1\) are positive, giving \(i\pi\) each.
Therefore the integral equals \(\dfrac{1}{2i}(i\pi - i\pi) = 0\).
Hence, the Cauchy principal value is \(0\).
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