Determining if $ I_n = \int_0^{\sqrt{\pi n}} \sin(t^2) dt $ is a Cauchy Sequence If I define the sequence $I_n$:
$$ I_n = \int_0^{\sqrt{\pi n}} \sin(t^2) dt$$
Is this a Cauchy sequence? It appears to be, but I'm not quite sure how to prove it. Here is my attempt so far:
Assume that  $\epsilon > 0$ and $m,n \geq N \;(N\in\mathbb{N})$
$$ \left| \int_0^{\sqrt{\pi m}} \sin(t^2)dt - \int_0^{\sqrt{\pi n}} \sin(t^2)dt \right| < \epsilon $$
$$ \left| \int_{\sqrt{\pi n}}^{\sqrt{\pi m}} \sin(t^2)dt \right| < \epsilon$$
and I'm stuck here. Where can I go from this point? In addition, is the same true for $\int_0^{\sqrt{\pi n}} \cos(t^2)dt$?
 A: Write $$I_n=\int_0^{n\pi}\frac{\sin(t)}{2\sqrt{t}}\,dt$$Then we have
$$\begin{align}
\left|I_n-I_m\right|&=\left|\int_{m\pi}^{n\pi}\frac{\sin(t)}{2\sqrt{t}}\,dt\,\right|\\\\
&=\frac12\,\left| \sum_{k=m}^{n-1}\int_{k\pi}^{(k+1)\pi}\frac{\sin(t)}{\sqrt{t}}\,dt\,\right|
\end{align}$$
Then, note that for $t\in[k\pi,(k+1)\pi]$, $\sin(t)$ is non-negative for even values of $k$ and non-positive for odd values of $k$.  
Heuristically, for large $k$, we have
$$\int_{k\pi}^{(k+1)\pi}\frac{\sin(t)}{\sqrt{t}}\,dt\approx \frac{1}{\sqrt{k\pi}}\int_{k\pi}^{(k+1)\pi}\sin(t)\,dt= (-1)^k\frac{2}{\sqrt{k\pi}}$$
Inasmuch as $\lim_{m,n\to \infty}\sum_{k=m}^{n-1}\frac{(-1)^k}{\sqrt{k\pi}}=0$ (Leibniz's Test for alternating series), the sequence $I_n$ is indeed Cauchy. 
A: Starting from Mark's suggestion,
$$ I_n = \int_{0}^{n\pi}\frac{\sin(t)}{2\sqrt{t}}\,dt $$
leads for sure to a Cauchy sequence, since $\frac{\sin(t)}{2\sqrt{t}}$ is improperly Riemann-integrable over $\mathbb{R}^+$ by Dirichlet's test: $\sin(x)$ is a function with a bounded primitive and $\frac{1}{\sqrt{t}}$ is decreasing to zero. As an alternative, you may just apply integration by parts:
$$ I_n = \int_{0}^{\pi n}\frac{1-\cos(t)}{4t\sqrt{t}}\,dt $$
where the new integrand function is non-negative, behaves like $\sqrt{t}$ in a right neighbourhood of the origin and decays like $\frac{1}{t\sqrt{t}}$ (that is an integrable function on $(1,+\infty)$) far from the origin.
Additionally, through the Laplace transform:
$$ \lim_{N\to +\infty}\int_{0}^{N}\frac{\sin(t)}{2\sqrt{t}}\,dt = \int_{0}^{+\infty}\frac{ds}{2\sqrt{\pi s}(1+s^2)}=\frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{dt}{1+t^4}=\color{red}{\frac{1}{2}\sqrt{\frac{\pi}{2}}}.$$
We have just computed a Fresnel integral. If $\sin$ is replaced by $\cos$, it is pretty much the same.
