![]() ![]() Note that the overall integral converges only if both of these converge. The integral from 0 to 1 would be evaluated as the limit as n approaches zero of the integral from n to 1, and the integral from 1 to ∞ would be evaluated as explained in this video. So if we're asked to analyze the integral of this function from 0 to ∞, we would choose a constant (we can use any constant, but it makes sense to choose one that makes calculations easy, such as 1) and evaluate two separate improper integrals: from 0 to 1, and from 1 to ∞. The integral from 0 to ∞ is equal to the integral from 0 to a plus the integral from a to ∞, where a is an arbitrary positive constant. Infinity (plus or minus) is always a problem point, and we also have problem points wherever the function "blows up," as this one does at x = 0.īecause we can't deal with both problem points at the same time, we split the integral into two (or more) parts. Good question! When dealing with improper integrals we need to handle one "problem point" at a time.
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