Walter Rudin Analysis Chapter 6 problem 13 I have been able to solve most of the question but i'm stuch at the last two parts.
the problem is the following:
Define $f(x) = \int sin(t^2) dt ~~from~~ t = x ~~to ~~t = (x+1) $
a) Prove that $|f(x)| < 1/x ~~if~~ x> 0$ by first showing that $f(x) = \frac{cos(x^2)}{2x} - \frac{cos[(x+1)^2]}{2(x+1)} - \int \frac{cos(u)}{4u^{3/2}} du$
the integral is from $x^2$ to $(x+1)^2$
b) Prove that $2xf(x) = cos(x^2) - cos[(x+1)^2] + r(x)$ where $|r(x)| < c/x $ and c is a constant
c) Find the upper and lower limits of $xf(x) $ as $x \rightarrow \infty$
d) Does $\int sin(t^2) ~~from ~~0 ~~to ~~\infty $ converge
As I said, I've already solved parts a and b, and this my attempt at part c:
$xf(x) = \frac{cos(x^2)}{2} - \frac{cos(x+1)^2}{2} + r(x)$
$r(x) \rightarrow 0$ and $\frac{cos(x^2) - cos(x+1)^2}{2} = sin(x^2 + x + 1/2)sin(1/2 + x)$
but then how to proceed? 
thanks
 A: Part (c): Start with 
Lemma 1: For any $\epsilon > 0$ there exists some $(k,q)\in \Bbb N$ such that 
$ \left| \pi-\frac{k^2}q \right| < \epsilon$.
Proof: choose any $q_0 > \frac{8\sqrt{\pi}+4}{\epsilon^2}$  and any rational approximation to $\pi \approx \frac{p}{q}$ with $q > q_0$ such that $|\pi-p/q| < \frac{\epsilon}{2}$.  Then a bit of algebra shows that for all $s \leq 2p+1$, 
$$
\left|\pi-\frac{p+s}{q}\right| < \epsilon
$$
But one of those $2p+1$ integers starting at $p$ must be a perfect square, thus proving lemma 1.
Similarly (lemma 1a), for any $\epsilon > 0$ there exists some $(k,n)\in \Bbb N$ such that 
$ \left| 4\pi-\frac{k^2}n \right| < \epsilon$.
Now for arbitrary $x$ we can use lemma 1a to show that for any $\epsilon > 0$ there will always be some $n$ such that $x_1 = n\pi > x$ such that 
$$
|\pi - (x+1)^2| < \epsilon
$$ 
because $(x+1)^2 = x^2 + 2\sqrt{\pi}\sqrt{n} + 1$.  So we know that we can make 
$ x^2$ and $(x+1)^2$ arbitrarily close to successive integer multiples of $\pi$, thus making  $\cos( x^2) - \cos(x+1)^2$ arbitrarily close to (our choice of $\pm 2$.  
So using the results of part (b), the upper and lower bounds on $xf(x)$ are respectively $1$ and $-1$.
For part (d), you could use the exact result $$\sqrt{\frac{\pi}{8}}$$ but I don't know how to prove that result for the integral.  The mere fact that pieces of the integral fall as $1/x$ does not quite prove convergence, and I don't see how to proceed on part $d$ using part (c). 
