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If the semi-circular contour D of radius 2 is as shown in the figure, then the value of the integral ∫_D 1/(s² - 1) ds is

1. jπ
2. -jπ
3. -π
4. π

Correct answer: jπ

Solution

The integral evaluates to jπ because the integrand has simple poles at s = ±1, and the contour encloses the pole at s = 1, contributing to the integral's value through the residue theorem.

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