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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.

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Please modify your post so that the question is also in the post text. – Jonas Teuwen Dec 7 '10 at 19:58
You used the M inequality backwards. You pull the sup of one function out (where you had an "M") and leave the other function in to use the M. You might also want to use that f(x^2) is integrable on [0,1] since it is continuous. That gives you sup(1/sqrt(t)) = 1, which is super. This sort of two-step thing is a reasonably common technique in (PDE or) analysis: a separate inequality for each type/location of singularity. – Jack Schmidt Dec 7 '10 at 20:15
I'm not sure I fully understand what you mean, could you explain this somehow differently? – daniel.jackson Dec 7 '10 at 21:01
up vote 3 down vote accepted

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...

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Yes... It somehow eluded me that I can split the integral to 2 to fix the problem near 0. Thanks, I owe you 20 points of my assignment for 2 questions you helped me with already :). – daniel.jackson Dec 7 '10 at 21:00

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