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$f: [a. \infty ) \rightarrow \mathbb R$ is differentiable and $\int_a^\infty f(t)dt $ and $\int_a^\infty f^{\prime}(t)dt $ are convergent then $f(t) \rightarrow 0$ as $t \rightarrow \infty$

my answer :

  1. $\int_a^x f^{\prime}(t)dt =f(x) - f(a)$

  2. $\int_a^\infty f^{\prime}(t)dt $ is convergent $\implies$ $\lim_{x \rightarrow \infty} f(x) = l $(where $l$ is finite ).to show $l=0$.

if $l \neq 0 \implies \exists <x_n> \rightarrow \infty $ such that $f(x_n) \geq \epsilon$ or $f(x_n) \leq \epsilon$ if i suppose $f(x_n) \geq \epsilon$ then as $f(x)$ is continuous then $f(x) \geq \frac{\epsilon}{2} \forall x \in(x_n-\delta,x_n+\delta)$ $\implies $ $\int_{x_n-\delta}^{x_n +\delta} f(t)dt \geq \epsilon\delta$ which will contradict the convergence of $\int_a^\infty f(t)dt $

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The last part of your argument needs more rigor (what is $\epsilon$? What is $\delta$?) and is a bit off. You could argue more simply here: If $l\ne0$, then there exists an $\epsilon>0$ and an $x_0$ such that either $f(x)>\epsilon$ for all $x>x_0$ or $f(x)<-\epsilon$ for all $x>x_0$. Then use this to show that $f$ is not improperly integrable on $[a,\infty)$. –  David Mitra Dec 16 '12 at 14:34
    
I should also point out that, in the last part of your proof, you don't need the fact that $f$ is continuous; all you need is the already proven fact that $\lim\limits_{x\rightarrow\infty}f(x)=l$. –  David Mitra Dec 16 '12 at 14:43

1 Answer 1

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You already know that $$ l:=\lim_{x \to \infty}f(x) $$ is finite because $$ f(x)=f(a)+\int_a^xf'(t)\,dt\ \text{ and }\ \int_a^\infty f'(t)\,dt \ \text{ converges }. $$ In particular you can see that $$ l=f(a)+\int_a^\infty f'(t)\,dt. $$ If you want to prove by contradiction that $l \ne 0$, you may w.l.o.g. assume that $l>0$.

Then there is an $r >0$ such that $$ |f(t)-l| \le \frac12l \quad \forall t \ge r, $$ i.e. $$ \frac12l \le f(t) \le \frac32l \quad \forall\ t \ge r, $$ and so we have $$ \int_a^\infty f(t)\,dt=\int_a^rf(t)\,dt+\int_r^\infty f(t)\,dt=\infty, $$ which is a contradiction.

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