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Let $f:[0,\infty)\rightarrow[0,\infty)$ be a real differentiable function such that $f(1)=1$.
If $f'$ is continuous and $\,\forall x\in [1,\infty):\,f'(x)\leqslant-\frac{2}{x^3}$, then the improper integral $\int^{\infty}_0f(x)\,\mathrm dx$ converges.

My idea is to use the FTC and imply a similar inequality with $\frac{1}{x^2}$ and $f$.
Currently I'm trying to think if $|f|$ would satisfy the same inequality, so I can use the comparison test. Assumably, I will have to use the assumption $f(1)=1$ but I struggle to see how this can be implemented.

I would like to have some ideas\guidance. Thanks!

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  • $\begingroup$ The more I read your commentary, the less sense it makes. Was "Use the FTC to imply a similar inequality with $\frac1{x^2}$ and $f$" a hint given to you by someone else that you're just passing off as your idea? It sticks out like a sore thumb amid your other comments, because you would use $f(1)=1$ in the FTC that gives you that inequality for $f$, and as the function is non-negative, $|f|=f$. $\endgroup$ Commented Aug 26, 2023 at 7:11
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    $\begingroup$ Hint: if two functions $f$ and $g$ are such that $f\leq g$, then $\int f(x) dx \leq \int g(x)dx$. You can see that using the area-under-the-curve interpretation of integration for example. Also, note that $|f(x)| = f(x)$ since $f$ is defined as a non-negative function. And yes, the FTC is exactly what you need here :) $\endgroup$ Commented Aug 26, 2023 at 7:14
  • $\begingroup$ @BrianMoehring I literally don't understand why you're being angry. Also that's not kind to assume someone is lying based on nothing. I'm happy it doesn't make sense to you because it tells me I was right when I looked for help here. $\endgroup$ Commented Aug 26, 2023 at 8:10

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Since $\,f’$ is continuous and $\,f’(x)\leqslant-\dfrac2{x^3}\,,\;\forall\,x\in[1,+\infty)\,,\,$ by applying the Fundamental Theorem of Calculus, we get :

$f(x)-f(1)=\displaystyle\int_1^x\!f’(t)\,\mathrm dt\leqslant\int_1^x\!\!-\dfrac2{t^3}\,\mathrm dt=\left[\dfrac1{t^2}\right]_1^x=\dfrac1{x^2}-1$

for any $\,x\in(1,+\infty)\,.$

Consequently, given that $\,f(1)=1\,,\,$ it follows that

$f(x)\leqslant\dfrac1{x^2}\;$ for any $\,x\in[1,+\infty)\,.$

On the other hand, the improper integral

$\displaystyle\int_0^{+\infty}\!\!\!\!\!f(x)\,\mathrm dx=\lim\limits_{b\to+\infty}\int_0^b\!f(x)\,\mathrm dx=I$

exists and $\,I\in\Bbb R^+\cup\{+\infty\}\,$ because the function $\,f\,$ is continuous and non-negative on the interval $\,[0,+\infty)\,$ and $\,f(1)=1\in\Bbb R^+.$

Moreover ,

$\displaystyle I=\int_0^{+\infty}\!\!\!\!\!f(x)\,\mathrm dx=\int_0^1\!f(x)\,\mathrm dx+\int_1^{+\infty}\!\!\!\!\!f(x)\,\mathrm dx\leqslant$

$\displaystyle\leqslant\int_0^1\!f(x)\,\mathrm dx+\int_1^{+\infty}\!\!\dfrac1{x^2}\,\mathrm dx\leqslant$

$\displaystyle\leqslant\int_0^1\!f(x)\,\mathrm dx+\left[-\dfrac1x\right]_1^{+\infty}=\int_0^1\!f(x)\,\mathrm dx+1\in\Bbb R^+.$

So $\,I\in\Bbb R^+$ and it implies that the integral $\,I=\!\displaystyle\int_0^{+\infty}\!\!\!\!\!f(x)\,\mathrm dx\,$ is convergent.

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  • $\begingroup$ About the last part - if we would reason that $\int_{0}^{1} f(x)\space dx$ exists (as a real number) since $f$ is continues at $[a, b]$ hence by limit's arithmetics $I$ is convergent, would that still be correct? Thanks! $\endgroup$ Commented Aug 26, 2023 at 10:21
  • $\begingroup$ @X4J, it is necessary to prove that $I<+\infty$ in order to prove that the integral is convergent. For that reason I proved that $I\leqslant\int_0^1f(x)\mathrm dx+1$ which is a positive real number. Say that $\int_0^1f(x)\mathrm dx$ exists as a real number is not enough to prove that $I$ is convergent, in fact it is necessary to prove that $I<+\infty$. $\endgroup$ Commented Aug 26, 2023 at 10:31
  • $\begingroup$ Then I struggle to see where we actually use the assumption that $f'$ is continues. E.g. if $f: [0, \infty) \rightarrow [0, \infty)$ is an anti-derivative of $f'$ at $[a, b] \subset (0, \infty)$ for each $a < b$, and $f'$ is integrable at $[a, b]$, then we seemingly assume less right? $\endgroup$ Commented Aug 26, 2023 at 10:38
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    $\begingroup$ @X4J, it is not necessary that $f’$ is continuous. If $f’$ is Riemann integrable on any interval $[1,b]$ for any $b>1$, you can still use the Fundamental Theorem of Calculus. See here $\endgroup$ Commented Aug 26, 2023 at 11:01

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