An improper integral I = a (x) dx is called absolutely convergent if
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An improper integral I = ∫∞a ƒ(x) dx is called absolutely convergent if ∫∞a |ƒ(x)| dx converges. It can be shown that if I is absolutely convergent, then it is convergent.
Show that ∫∞1 sin x/x2 dx is absolutely convergent.
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Because t Int for t 2 Thus Fx o as x o Moreover ...View the full answer
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