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.