Convergence of $\int_0^{\infty} \sin(t^2) dt$ in Rudin. How do the $\limsup$ and $\liminf$ help? Before answering, note that I know how to prove this integral converges by integrating by parts. This question is about proving convergence via a particular path presented below.
I'm doing this problem out of baby Rudin and I just don't quite see how to complete the path he leads me down. It's problem 6.13 concerning $\int_0^{\infty} \sin(t^2) dt $. He writes: 

Define $$ f(x) = \int_x^{x+1} \sin(t^2) dt $$ 
a. Prove $|f(x)| < 1/x$ if $x>0$. 
b. Prove that $2x f(x) = \cos(x^2) - \cos((x+1)^2) +r(x)$ where
  $|r(x)| <c/x$ for some $c$. 
c. Find the upper and lower limits of $xf(x)$ as $x \to \infty$. 
d. Does $\int_0^{\infty} \sin(t^2) dt $ converge?

I've done the first three parts. For c, I think the $\limsup_{x\to \infty} xf(x) = 1 $ and $\liminf_{x\to \infty} xf(x) = -1 $. I'm having trouble doing part d in the way the outline seems to lead. It seems as though I'm supposed to say:
$$\int_0^{n+1} \sin(t^2) dt = \sum_{k=0}^{n} f(k) $$
and use the info I got from the earlier parts to show that the sum on the right converges. However, I do not see how the info I've gotten earlier helps. For instance, its doesn't seem to help much to know the upper and lower limits, since whatever bound they might give me I already had from $|f(x)| <1/x$. 
What am I missing? I would like to solve this on my own, so more of a hint rather than a full solution would be appreciated. 
 A: Note that
$$\int_0^n \sin t^2 \, dt = \sum_{k=0}^{n-1} f(k) = f(0)+ \sum_{k=1}^{n-1} \frac{r(k)}{2k} + \frac{1}{2}\sum_{k=1}^{n-1} \left(\frac{\cos k^2}{k} - \frac{\cos (k+1)^2}{k}\right)$$
The first sum on the RHS is absolutely convergent as $n \to \infty$ since $\frac{|r(k)|}{k} < \frac{c}{k^2}$.
The second sum is
$$\sum_{k=1}^{n-1} \left(\frac{\cos k^2}{k} - \frac{\cos (k+1)^2}{k}\right) = \sum_{k=1}^{n-1}\left(\frac{\cos k^2}{k} - \frac{\cos (k+1)^2}{k+1}\right) - \sum_{k=1}^{n-1}\frac{\cos (k+1)^2}{k(k+1)} \\ = \cos (1)- \frac{\cos n^2}{n}- \sum_{k=1}^{n-1}\frac{\cos (k+1)^2}{k(k+1)}$$
The second term on the RHS converges to $0$ as $n \to \infty$ and the last sum is absolutely convergent since $\frac{|\cos (k+1)^2|}{k(k+1)} \leqslant \frac{1}{k^2}$.
Thus, $\lim_{n \to \infty}\int_0^n \sin t^2 \, dt$ exists.
To finish use integration by parts and show that $\lim_{x \to \infty}\int_{n}^x \sin t^2 \, dt= 0$ where $n = \lfloor x \rfloor$. Together these results imply that $\int_0^\infty \sin t^2 \, dt$ is convergent, since
$$\lim_{x \to \infty}\int_0^x \sin t^2 \, dt = \lim_{x \to \infty}\int_0^{\lfloor x \rfloor} \sin t^2 \, dt + \lim_{x \to \infty}\int_{\lfloor x \rfloor}^x \sin t^2 \, dt= \lim_{n \to \infty\\ n =\lfloor x \rfloor}\int_0^n \sin t^2 \, dt$$
