Thread: [SOLVED] f integrable implies f^2 integrable
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Old June 2nd, 2009, 12:04 PM
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Originally Posted by nqramjets View Post
f integrable on [0,1] implies f bounded on [0,1], by contraposition, the integral of 1/sqrt{x} does not exist.
I'm afraid I cant agree with you. Starting from the general integration formula...

\int x^{a}\cdot dx = \frac{x^{1+a}}{1+a} + c

... where a is an arbitrary real [or even complex...] number provided only that a \ne -1, proceeding in trivial way we have...

\int_{0}^{1} \frac{dx}{\sqrt{x}} = |2\cdot \sqrt{x}|_{0}^{1}= 2

Kind regards

\chi \sigma
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