Under which terms would $ \intop_{1}^{\infty}f\left(x^{2}\right)dx $ converge

Question

let $ f:[1,\infty) \to \mathbb{R} $ be continious function.

Originally I had to tell which demands on $ f $ would make sure that the integral $ \intop_{1}^{\infty}f\left(x^{2}\right)dx $ would converge.

There's just one demand that I cant decide yet.

So, I'll arrange the question:

Assume $ f:[1,\infty) \to \mathbb{R} $ continious function,

and assume that exists $ 0<T\in\mathbb{R} $ such that for any $ 1\leq x $ it follows that $ f\left(x+T\right)=f\left(x\right) $.

In addition assume that $ \intop_{1}^{1+T}f\left(x\right)dx=0 $.

Under those terms, is it true that $ \intop_{1}^{\infty}f\left(x^{2}\right)dx $ converges?

I'm stuck on this one, and dont really have a good idea. So any help would be great.

Thanks in advance.

Answer 1

Let $F(y) := \int_1^y f(s)\, ds$, $y\geq 1$. From the assumptions on $f$ it is easy to verify that $F$ is bounded, i.e. there exists a constant $C>0$ such that $|F(y)| \leq C$ for every $y\geq 1$.

Using the change of variable $y=x^2$ and integrating by parts we have that, for every $b > 1$, $$ \int_1^b f(x^2)\, dx = \int_1^{b^2} \frac{f(y)}{2 \sqrt{y}}\, dy = -\left.\frac{F( y)}{2\sqrt{y}}\right|_1^{b^2} +\int_1^{b^2}\frac{F(y)}{y^{3/2}}\, dy. $$ Since $F$ is bounded, the function $F(y) / y^{3/2}$ is absolutely integrable in $[1, +\infty)$, hence the r.h.s. converges for $b\to +\infty$.