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Mathematics
Calculus
Multiplying divergent integrals using Hardy fields approach
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[QUOTE="Mark44, post: 6627854, member: 147785"] Yes, but to go from this proper definite integral to the improper with an infinite limit requires taking the limit as x increases without bound. Doing so shows that each integral is unbounded and positive, so the product is also unbounded and positive. A small quibble is that you have used x both as a dummy variable in the integral as well as one of the integration limits. The usual practice is to use a different letter for one of these. Germs? I'm not familiar with that term. Is it a synonym for antiderivative? In any case, once you take the limit, the expression on the right has a "limit" of infinity. Every one of the integrals above diverges, so how is this useful? [/QUOTE]
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Mathematics
Calculus
Multiplying divergent integrals using Hardy fields approach
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