I think that since the function gets closer and closer to 0, it closes in on an value and thus it does converge. I also know that for $[1,\infty)$, functions $\frac{1}{x^p}$, converge if $p > 1$. How can I put this proof in math language? I have a hard time doing it.
Prove/disprove that if $f$ is cont, decreasing on $[1, \infty)$, and $\lim_{x \to \infty} f(x) = 0$, then $\int^{\infty}_{1}f(x)dx$ converges.
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calculus
integration
convergence
definite-integrals
indefinite-integrals
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2How about $f(x)=1/x$ it is decreasing,$\lim$ goes to $0$ though the integral doesn't converge. – 2017-02-16
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0@kingW3 Well, didn't see your comment there. I added a similar answer. – 2017-02-16
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0I think the idea, if anything, would have been the following : dominate any such given function by a power of $x$ larger than $1$, like $\frac{1}{x^{1.01}}$ etc. and these have convergent integrals. As it turns out, a stronger condition is required for convergence of the integral. How about if $f$ were "uniformly" continuous? – 2017-02-16
1 Answers
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Well, what about $f(x)=\frac{1}{\sqrt{x}}$? It is continuous, decreasing, and $f(x) \to 0$ as $x \to \infty$. Yet the integral does not converge as $$\lim_{x \to \infty}\int_{1}^{x}f(t)dt=\lim_{x \to \infty}(2 \sqrt{x} -2)$$
Note that the same goes for $\frac{1}{x^{p}}$ when $0
Your conjecture is false.