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Is it true to say that $\int_\mathbb{R}|f(x)|dx<\infty\Rightarrow\int_\mathbb{R}f(x)=0$?

SAMEER
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  • If no, then under what conditions above statement holds – SAMEER Jul 04 '14 at 07:51
  • R to L implication basically say: if a function is integrable, then it integrate to $0$. Of course it won't hold... And L to R implication basically say: if a function integrate to $0$, then it's integrable. Of course, that's trivially by definition. – Gina Jul 04 '14 at 07:54
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    Whoa, not at all. Any odd, integrable function has that property. Like $f(x)={\sin x\over e^{x^2}}$ – Adam Hughes Jul 04 '14 at 07:55
  • The condition you want is

    $$\int_\mathbb{R}f^+=\int_\mathbb{R}f^-$$ since

    $$\int_\mathbb{R} f = \int_\mathbb{R}f^+-\int_\mathbb{R}f^-$$

     – Adam Hughes Jul 04 '14 at 07:58
  • The answer to your new question is ALSO no. There are tons of integrable functions which have integral not $0$, take $e^{-x^2}$ for starters, or ${1\over 1+x^2}$. – Adam Hughes Jul 04 '14 at 08:00
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    @SAMEER: Maybe you meant $\int |f(x)|dx < \infty $, then $\int f(x)dx < \infty$ ? – user99680 Jul 04 '14 at 08:02
  • To anyone voting this question down: I don't think it is a bad question. In that sense, it does not deserve voting down. It does, if OP will continnue his no-response policy, deserve to be closed, but not voted down if you ask me. – 5xum Jul 04 '14 at 08:19

1 Answers1

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Your statement is far from correct. For example, it fails for simple functions like $f(x) = e^{-|x|}$, for which $\int f(x) dx \neq 0$ despite the fact that $\int |f(x)| dx < \infty$. The implication you wrote does not hold.

5xum
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