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Possible Duplicate:
Why is $\int^\infty_{-\infty} \frac{x}{x^2+1} dx$ not zero?

I've always learned that the improper integral $\int\limits_{-\infty}^{\infty}f(x)dx$ only exists when the integrals $\int\limits_{-\infty}^af(x)dx$ and $\int\limits_a^{\infty}f(x)dx$ exists. Why is that?

If we have for example the function $f(x)=x^3$ then the integral $\int\limits_{-a}^{a}f(x)dx$ is defined for any a. However, $\int\limits_{-\infty}^{\infty}f(x)dx$ isn't defined. Why is that?

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    See the discussion in [this answer](http://math.stackexchange.com/questions/80262/why-is-int-infty-infty-fracxx21-dx-not-zero/80274#80274). The main point is that the value of the integral should *not* depend on how we choose to evaluate, so long as the usual rules are obeyed.2012-01-26
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    In principle, one could try to define a principle value (like the Cauchy principal value) which indicates that the limit should be taken symmetrically.2012-01-26
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    @Arturo Thanks, that was exactly what I was looking for!2012-01-26
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    @Tiddo: So... you think it would be okay to mark this as a duplicate of that question, even though the precise query is not the same?2012-01-26
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    @Arturo - Sure! If I'd found that question before, I didn't need to ask mine.2012-01-26

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