Improper integrals: does convergence imply monotonicity?

Question

If the integral from $1$ to $\infty$ for a positive, continuous function (on $[1, \infty)$) converges, does this imply the function is decreasing as well?

Answer 1

No. You can even have $f$ unbounded. For instance, define $f$ to be, on the interval $[k, k+1]$, a triangle of the height $k$ and base $2/k^3$. Then $$\int_1^\infty f(t) \, dt=\sum_k \frac 1{k^2} <\infty,$$ and $f$ is unbounded on any interval $[m, \infty]) $.

If you center the triangles on each $k$, you can get $f(k) =k$.

If you want $f(t)>0$ for all $t$, you can raise each triangle by $1/k^2$ and join the endpoint of one triangle with the starting point of the next one by a line segment.

Answer 2

Not necessarily, the function does not have to be monotonous, example:

$$f(x) = \frac{\sin^2{x}+1}{x^2}$$