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fresh_42

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Why shouldn't it converge? ##\int_a^b \dfrac{1}{x}\dfrac{1}{x}\,dx = -\dfrac{1}{b}+\dfrac{1}{a}\,.##

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Sorry, I meant to replace ##a## with ##-1## and ##b## with ##1##Why shouldn't it converge? ##\int_a^b \dfrac{1}{x}\dfrac{1}{x}\,dx = -\dfrac{1}{b}+\dfrac{1}{a}\,.##

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fresh_42

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But the function ##x \mapsto \dfrac{1}{x}## you mentioned isn't defined on ##[-1,1]##. You also need functions which are at least integrable, usually Lebesgue integrable, or continuous. Real valued alone is too weak, because at least the inner product must be defined!Sorry, I meant to replace ##a## with ##-1## and ##b## with ##1##

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So is what Wolfram said incorrect?But the function ##x \mapsto \dfrac{1}{x}## you mentioned isn't defined on ##[-1,1]##. You also need functions which are at least integrable, usually Lebesgue integrable, or continuous. Real valued alone is too weak, because at least the inner product must be defined!

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fresh_42

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Do you have a link?So is what Wolfram said incorrect?

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martinbn

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http://mathworld.wolfram.com/InnerProduct.html

They are being sloppy. They don't mean all functions.

They are being sloppy. They don't mean all functions.

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