If $f>0$ when does $\int_{a}^{\infty} f(x)dx$ converge/diverge?

Question

If $f>0$ when does $I=\int_{a}^{\infty} f(x)dx$ converge/diverge? In particular:

(1) If $f$ is not decreasing, does that imply that $I$ diverges?

(2) If $f$ does not approach $0$ as $x\to\infty$ does that imply $I$ diverges?

EDIT: For (1) seems answer is no. I have a follow up question:

Now suppose also that $f$ is periodic and continuous.

(3) If the local maxima are not decreasing as $x\to\infty$, does that imply $I$ diverges? E.g. the local maxima for $\sin^2(x)$ is always $1$ and diverges.

Answer 1

(1) yes if $f$ is non-decreasing : since $f(x)\geq f(a)>0$ in particular, so $$\int_a^{\infty}f(x) dx \geq \int_a^{\infty}f(a) dx=f(a)\int_a^{\infty} dx=+\infty $$

But you can't know if $f$ is just not decreasing : for exemple $f(x)= 1_{x\geq1}\frac1{x^2}(2+\cos(x))$ is not decreasing but the integral converge...

(2) no : you can imagine $f(x)=\exp(-x^2)$ for all $x\in \mathbb{R}\setminus \mathbb{N}^*$ and $f(x)=x$ for all $x\in \mathbb{N}^*$ : then $$ \int_a^{\infty}f(x) dx = \int_a^{\infty}\exp(-x^2) dx <\infty $$ but $f$ does not tend to $0$....

(3) : no, you can use the same example I gave in (2)

Answer 2

(1) Yes. With $c:=f(a)>0$, you have $f(x)\ge c$ for all $x$, hence $\int_A^B f(x)\,\mathrm dx\ge c(B-A)$ for all $B>A$, whereas convergence would require this to be $<\epsilon$ if only $A,B\gg 0$.

(2) The integral may still converge. $f$ might have a sequence of "bumps" of fixed positive height, but rapidly decreasing width.

Answer 3

Hint on 1):

Note that the function $1_{\mathbb N}$ is not decreasing.

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edit:

If $f$ is continuous and $f(x_0)>0$ for some $x_0$ then for $\epsilon>0$ small enough we have $x\in(x-\epsilon,x+\epsilon)\implies f(x)\geq\frac12 f(x_0)$ so that $\int_{x_0-\epsilon}^{x_0+\epsilon}f(x)\geq\int_{x_0-\epsilon}^{x_0+\epsilon}\frac12f(x_0)\geq\epsilon f(x_0)$.

If moreover it has a period $p$ then we get $\int_{x_0+np-\epsilon}^{x_0+np+\epsilon}f(x)\geq\epsilon f(x_0)$ for every $n\in\mathbb N$.

This leads to $\int_a^{\infty}f(x)dx=+\infty$.