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!
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.