# Convergent of improper integral [closed]

Let $$f \in C^1[0,\infty)$$ be an increasing function with $$f(0)>0$$, suppose $$\int_0^\infty \frac{1}{f(x)+f'(x)} < \infty$$, prove that $$\int_0^\infty \frac{1}{f(x)} < \infty$$.

I find it weird since the behaviour of $$f'$$ is random, so I don't know how to control $$f$$ in terms of $$f+f'$$.

• You ask to prove it, what is the evidence that this is true? Commented Sep 12, 2020 at 11:56
• isn't it just a consequence of $f(x)+f'(x)>f(x)>0$ for all $x>0$? Commented Sep 12, 2020 at 12:13
• @CarloBeenakker no, it seems that it is not Commented Sep 12, 2020 at 12:16
• @FedorPetrov It is an exercise Commented Sep 12, 2020 at 12:23
• In general exercises are not welcome here (although I like this one) Commented Sep 12, 2020 at 12:58

## 1 Answer

For $$a\geqslant 0,b>0$$ we have $$\frac1b\leqslant \frac2{a+b}+\frac{a}{b^2}$$ (if $$a\leqslant b$$, then $$\frac1b\leqslant \frac2{a+b}$$; if $$a\geqslant b$$, then $$\frac1b\leqslant \frac{a}{b^2}$$). Apply this for $$b=f$$, $$a=f'$$ and use that $$\int f'/f^2=\int (-1/f)'$$ converges.