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Lets suppose that $f:[a,\infty) \to \mathbb{R}$ is a Riemann integrable function and lets suppose that $$\lim_{x \to \infty} f(x) = l > 0$$ I am trying to prove that then $$\int_a^\infty f(x) \text{d}x=\infty$$ By hypothesis $f$ has limit $l > 0$ as $x \to \infty$, so we know that for all $\varepsilon>0$ exists $K_{\varepsilon} > 0$ such that for all $x \geq K_{\varepsilon}$ it is $f(x)>l-\varepsilon$.

Since by hypothesis $l > 0$ and for the arbitrarity of $\varepsilon>0$ we can choose $\varepsilon=l/2$; so we have the estime $f(x)>l/2$.

So it is

$$\int_a^\infty f(x) \text{d}x =\int_a^{K_\varepsilon} f(x) \text{d}x+\int_{K_\varepsilon}^\infty f(x) \text{d}x > \int_a^\infty \frac{l}{2} \text{d}x =\infty$$ Since the first integral on the right hand side is finite it does not influence the convergence, so we concentrate on the second integral; by the limit estimation we have that $$\int_{K_\varepsilon}^\infty f(x) \text{d}x > \int_{K_\varepsilon}^\infty \frac{l}{2} \text{d}x =\infty$$

So the integral is divergent.

Some questions:

1) is the context correct? I've assumed that "$f:[a,\infty)$ Riemann integrable" means that the only point we have to study is when $x \to \infty$ because of the unboundedness;

2) when I split the integral in two integrals I suppose that $a<K_\varepsilon$, can I do this? If yes, why?

3) is the proof correct in general? If not, where are the mistakes? If yes, how can I improve it?

Thanks.

ZaWarudo
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2 Answers2

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Let $ a\in\mathbb{R} \cdot $

By the definition there exists some $ A >\max\left(a,0\right) $ such that $ \left(\forall x\geq A\right),\ \left|f\left(x\right)-\ell\right|<\frac{\ell}{2} \cdot $

And from that we get that $ \left(\forall x\geq A\right),\ f\left(x\right)\geq\frac{\ell}{2} \cdot $

Let $ x\geq A $, then : \begin{aligned}\int_{a}^{x}{f\left(t\right)\mathrm{d}t}&=\int_{a}^{A}{f\left(t\right)\mathrm{d}t}+\int_{A}^{x}{f\left(t\right)\mathrm{d}t}\\ &\geq\int_{a}^{A}{f\left(t\right)\mathrm{d}t}+\frac{\ell}{2}\int_{A}^{x}{\mathrm{d}t}\\ &\geq\int_{a}^{A}{f\left(t\right)\mathrm{d}t}+\frac{\ell}{2}\left(x-A\right)\underset{n\to +\infty}{\longrightarrow}+\infty\end{aligned}

Hence, $$ \int_{a}^{+\infty}{f\left(x\right)\mathrm{d}x}=+\infty $$

CHAMSI
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1.: Yes. The Improper Riemann integral is a limit of Riemann integrals, so $f$ must be Riemann integrable on $[a,b]$ for all $b>a$.

2.: Yes, you can suppose that $K_\varepsilon > a$ because you don't need to pick the smallest $K_\varepsilon$. And if $K_\varepsilon$ is good, then so is $42a+42K_\varepsilon$.

3.: The idea is good, but you can improve it. I would not integrate it from $a$ to $+\infty$, just from $a$ to some $b$, and examine the behaviour as $b \to +\infty$. Technically, you did the same, but I think it's more formal.

Also, the following line is not correct: $$\int_a^\infty f(x) \text{d}x =\int_a^{K_\varepsilon} f(x) \text{d}x+\int_{K_\varepsilon}^\infty f(x) \text{d}x > \int_a^\infty \frac{l}{2} \text{d}x =\infty$$ I think you wanted to write $$\int_a^\infty f(x) \text{d}x =\int_a^{K_\varepsilon} f(x) \text{d}x+\int_{K_\varepsilon}^\infty f(x) \text{d}x > \int_{\color{red}{K_\varepsilon}}^\infty \frac{l}{2} \text{d}x =\infty$$ But it is still not correct, because $$\int_a^{K_\varepsilon} f(x) \text{d}x$$ can be negative.

Botond
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    Yes, I made a typo with the lower bound of integration! I've been sloppy there, since I know that the integral from $a$ to $K_\varepsilon$ is finite adding it to a divergent integral it doesn't affect the divergence, so I shoul've written $\int_{a}^{K_\varepsilon} f(x) \text{d}x + \int_{K_\varepsilon}^b f(x) \text{d}x > \int_{a}^{K_\varepsilon} f(x) \text{d}x + \int_{K_\varepsilon}^b \frac{l}{2} \text{d}x \to \infty$ when $b \to \infty$. Is this better? Thanks for solving my doubts! – ZaWarudo Apr 26 '20 at 18:27
  • @Dunkelheit Yes, it's better. I know that it does not matter here, because of the divergence, but it's not always the case (for example when you try to give a lower bound for a convergent integral), so it's good to not avoid it. – Botond Apr 26 '20 at 18:29
  • Just one last thing to confirm if I've understood well the question (2): so if $K_{\varepsilon}>a$ we have no problem, if $K_{\varepsilon}<a$ we can define a $K_{\varepsilon}'>a$ and since $a>K_{\varepsilon}$ we have $K_{\varepsilon}'>a>K_{\varepsilon}$ and so we have that the definition of limit works with $K_{\varepsilon}'$. Something like this? I have never seen this argument explicitly, so I suppose that implicitly in general they take $K_{\varepsilon}'=\max{a,K_{\varepsilon}}$ and with abuse of notation they call again $K_{\varepsilon}$ the new $K_{\varepsilon}'$, am I wrong? – ZaWarudo Apr 26 '20 at 19:26
  • @ZaWarudo $\lim_{x \to +\infty} f(x)=l$ guarantees that $\forall \varepsilon > 0 \exists K_\varepsilon \in \mathbb{R}$ so that $f(x) > l-\varepsilon \forall x > K_\varepsilon$. letting $\varepsilon=l/2$ we get that $\exists K_{l/2} \in \mathbb{R}$ so that $\forall x > K_{l/2}$ we have that $f(x) > l/2$. But this guarantees that $f(x) > l/2$ for all $x > K_{l/2} + 42$ too. So you don't just reach into a sack of $K_\varepsilon$s and get one of them blindly, but you can look into a sack and pick one that is good for you... – Botond Apr 26 '20 at 19:43
  • @ZaWarudo ... So you can say something like this: Let $K_\varepsilon > a$ be a real number so that $f(x) > l/2$ for all $x > K_\varepsilon$ (and the existence will be guaranteed by the fact that $\lim_{x \to +\infty} = l$), or just simply, without the loss of generality, assume that $K_\varepsilon > a$. – Botond Apr 26 '20 at 19:45
  • @ZaWarudo But yes, if you don't have time to search for the perfect $K_\varepsilon$ in that sack, you can pick one blindly and do what you did. – Botond Apr 26 '20 at 19:48