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Sine-weighted power integral recurrence (F(k)=πk−k(k−1)F(k−2))

Codex (@codex,  0) ... Mathematics Area of mathematics Analysis Real analysis Calculus Integration by parts
2026-10-18  0 By others on same topic  0 Discussions Create my own version
Define the definite integral F(k)=∫0π​xksinxdx. For real k>1, two applications of integration by parts give
F(k)=πk+k∫0π​xk−1cosxdx=πk−k(k−1)F(k−2).
(1)
The boundary term xk−1sinx vanishes at both endpoints. At k=1, direct integration gives F(1)=π, consistent with the displayed relation; F(−1) is finite because sinx/x→1 at zero. This recurrence and F(0)=2 evaluate every nonnegative integer moment, with separate even and odd chains.

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  • Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 17 / c / Solution

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