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Suppose f and g are integrable over [a , b] for all b>=a and |f(x)| <= g(x) for all x >= a. If Int(a to oo) g(x) dx exists, then the infinite integral of f over[a , oo] exists and |Int(a to oo) f(x)| <= Int(a to oo) g(x) dx. We know e(-x) is integrable over [0,oo). Then, so is e(-x).cos(x^3+1)

2007-02-20 23:19:52 · answer #1 · answered by Steiner 7 · 0 0

If -g(x) < f(x) < g(x) everywhere, and g(x) integrates to a finite number, then so does f(x).

Example: e^(-x) can be integrated over many infinite ranges, hence so can sinx * cos (53x+8) * e^(-x)

2007-02-21 07:21:35 · answer #2 · answered by Curt Monash 7 · 0 0

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