Show $\int_{0}^{\infty}f(x^2)dx$ exists if $f(x)$ is continuous and $\forall a\gt 0, |\int_{0}^{a}f(x)dx| \leq M$ 
Given a continuous function $f(x)$ and
  $\forall a\gt 0, |\int_{0}^{a}f(x)dx| \leq M$, show
  $\int_{0}^{\infty}f(x^2)dx$ exists.

I tried substituting $t=x^2$ which gave me $\displaystyle \int_{0}^{\infty}f(x^2)dx=\frac{1}{2}\int_{0}^{\infty}\frac{f(t)}{\sqrt t}dt$ but I can't see how to use the fact that $f(x)$ has a bounded indefinite integral in $[0, \infty)$. Simply doing $\frac{1}{2}|\int_{0}^{\infty}\frac{f(t)}{\sqrt t}dt|\leq \frac{M}{2}|\int_{0}^{\infty}\frac{1}{\sqrt t}dt|$ doesn't really help because $\int \frac{1}{\sqrt t}$ doesn't converge in $[0, \infty)$
After substituting I tried integration by parts (using $u = \frac{1}{\sqrt t}, v' = f(t) \implies v = F(t)$) but that seems to complicate things further.

If $f$ is only assumed continuous on $(0,\infty)$, then $\int_0^\infty  {f(x^2 )dx}$ might diverge (consider $f$ such that $f(x)=1/\sqrt{x}$ near $0+$). So, assume that $f$ is continuous on $[0,\infty)$. Then, obviously $\int_0^1 {f(x^2 )dx}$ exists (since $f$ is uniformly bounded on $[0,1]$). So, we only have to prove that $\int_1^\infty  {f(x^2 )dx} $ exists. Now, after substituting and performing integration by parts as you did, it suffices to show, using your notation, that $F(t)/\sqrt{t} \to 0$ as $t \to \infty$, and that $\int_1^\infty  {\frac{{F(t)}}{{t^{3/2} }}} dt$ exists. This is clearly true... 
