Suppose f is integrable on (0,1), then show lim
I tried to write (0,1) = \bigcup _{k=0}^{{a-1}} \left(\sqrt{\frac{k}{a}},\sqrt{\frac{k+1}{a}}\right), but cannot make the integral converge to 0.
Answer
Suggestion:
Split [0, 1]
into the intervals where
ax^2 = 2\pi n,
ax^2 = \pi(2 n+1),
ax^2 = \pi(2 n+2).
These are
I_{2n} =[\sqrt{\frac{2\pi n}{a}}, \sqrt{\frac{\pi(2 n+1)}{a}})
and
I_{2n+1} =[\sqrt{\frac{\pi(2 n+1)}{a}}, \sqrt{\frac{\pi(2 n+2)}{a}}) .
Since
\sin(ax^2) > 0
in
I_{2n}
and
\sin(ax^2) < 0
in
I_{2n+1},
show that
the integral over
I_{2n} \cup I_{2n+1}
goes to zero.
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