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If I have the following expression:

$\displaystyle\int_{0}^{\infty} f(x) dx = \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^3}$

Can I then deduce that $\int_{0}^{\infty} f(x) dx$ converges, because right hand side does?

2 Answers 2

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Well...yes, of course! What you wrote says the integral equals a number , so yes: this means the integral converges.

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    Oops Sorry. I see :) I'm just a little co$n$fused.2012-05-28
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If I were to say, for example, for a finite $C$

$\int^{\infty}_{0}f(x)\, dx=C < \infty \implies \int^{\infty}_{0}f(x)\, dx < \infty$

Thus

$\displaystyle\int_{0}^{\infty} f(x) \,dx=\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^3}=\zeta(3)$

So the integral converges! You even know what the integral converges to.