Show that $\int_0^\infty \sin\left(x^2\right)dx$ converges, but that $\int_0^\infty \sqrt{\sin^2\left(x^2\right)}dx$ does not. Show that $\int_0^\infty \sin\left(x^2\right)dx$ converges, but that $\int_0^\infty \sqrt{\sin^2\left(x^2\right)}dx$ does not.
The first part I think I proved using triangles, but I could not prove the second part.
 A: Your second integral is
$$
\int_0^\infty|\sin x^2|\,dx.
$$
Note that $\sin t\geq1/2$ if $t\in[\frac\pi6+2k\pi,\frac{5\pi}6+2k\pi]$. So $\sin x^2\geq1/2$ if $x\in[\sqrt{\frac\pi6+2k\pi},\sqrt{\frac{5\pi}6+2k\pi}]$, $k\in\mathbb N$.
So
$$
\begin{eqnarray}
\int_0^\infty|\sin x^2|\,dx&\geq&\sum_{k=0}^\infty\int_{\sqrt{\frac\pi6+2k\pi}}^{\sqrt{\frac{5\pi}6+2k\pi}}|\sin x^2|\,dx\\
&\geq&\frac12\,\sum_{k=0}^\infty\left(\sqrt{\frac{5\pi}6+2k\pi}-\sqrt{\frac{\pi}6+2k\pi}\right)\\
&\geq&\frac12\,\sum_{k=0}^\infty\frac{\frac{4\pi}6}{\sqrt{\frac{5\pi}6+2k\pi}+\sqrt{\frac{\pi}6+2k\pi}}\\
&\geq&\frac12\,\sum_{k=0}^\infty\frac{\frac{4\pi}6}{2\sqrt{\frac{5\pi}6+2k\pi}}\\
&=&\frac\pi8\,\sum_{k=0}^\infty\frac{1}{\sqrt{\frac{5\pi}6+2k\pi}}=\infty
\end{eqnarray}
$$
(the last series clearly diverges as its terms grow as $k^{-1/2}$).
A: Here's a proof of the first part, using the general philosophy that if an integrand oscillates, you can often get a better sense of the size of the integral using integration by parts. Ignoring the perfectly tame integral from 0 to 1,
$$
\int_1^\infty \sin(x^2)\,dx = \int_1^\infty \frac1{-2x}(-2x\sin(x^2))\,dx = \frac{\cos(x^2)}{-2x}\bigg|_1^\infty - \int_1^\infty \frac{\cos(x^2)}{2x^2}\,dx.
$$
The boundary terms are finite, and the remaining integral converges absolutely by comparison to $\int x^{-2}dx$.
A: Note that, by change of variables we have
$$\int_0^\infty \sqrt{\sin^2\left(x^2\right)}dx = \int_0^\infty |\sin\left(x^2\right)|dx =\int_0^\infty \frac{|\sin\left(x\right)|}{2 x^{1/2}}dx  $$
More generally we study the following integrals
$$\varphi_1(\alpha) =\int_0^\infty \frac{\sin t}{t^\alpha}\,dt\tag{I}$$

Lemma$ \frac{\sin t}{t^\alpha} $ converges if and only if $0<\alpha<2$ and converges absolutely if and only if $1<\alpha <2$.
Therefore, $\int_0^\infty \sqrt{\sin^2\left(x^2\right)}dx = \int_0^\infty |\sin\left(x^2\right)|dx =\int_0^\infty \frac{|\sin\left(x\right)|}{2 x^{1/2}}dx  $ diverges

Proof of the Lemma
case $\alpha\gt 0$ 
Near $t=0$, $\sin t\approx t.$ Which yields,  $\frac{\sin t}{t^{\alpha}}\approx \frac{1}{t^{\alpha -1}}$ and the  convergence of the integral in (I)  holds nearby $t=0$ if and only if $\alpha<2 $. 
Now let take into play the case where $t $ is large.
case $\alpha\leq 0$ 
Employing integration by part, 
 \begin{eqnarray*}
\Big| \int_{\frac{\pi}{2}}^\infty \frac{\sin t}{t^\alpha}\,dt\Big|  &= & \Big| -\alpha \int_{\frac{\pi}{2}}^\infty \frac{\cos t}{t^{\alpha+1}}\,dt\Big|\\
%
&\leq &    \alpha \int_{\frac{\pi}{2}}^\infty \frac{ 1 }{t^{\alpha+1}}\,dt< \infty \qquad\text{since} \qquad \alpha +1>1~~\text{with} ~~\alpha >0.
 \end{eqnarray*}
 Thus for $\alpha>0 $, $\varphi_1(\alpha)$ exists if and only if $0<\alpha<2$.
We will later these are the only values of $\alpha$ which guarantee the existence of $\varphi_1$. For now let have  a look on the integrability of functions under (I). In other to see that, one can quickly check the following
$$ \mathbb{R}_+ =  \bigcup_{n\in\mathbb{N}} [n\pi, (n+1)\pi).$$
Then, 
$$\int_0^\infty \frac{|\sin t|}{t^\alpha}\,dt = \int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt+ \sum_{n=1}^{\infty}  \int_{n\pi}^{(n+1)\pi} \frac{|\sin t|}{t^\alpha}\,dt \\:= \int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt+\sum_{n=1}^{\infty} a_n$$
With suitable change of variable ($u = t-n\pi$) we get
\begin{eqnarray*}
a_n &=& \int_{0}^{\pi} \frac{\sin t}{{(t+n\pi)}^\alpha} \,dt\qquad\text{since } \sin(t+n\pi)= (-1)^n\sin t  
\end{eqnarray*} 
 On the oder hand, it is also easy to check
\begin{eqnarray}
 \frac{2}{(n+1\pi)^\alpha} \leq  a_n \leq \frac{2}{(n\pi)^\alpha}.
 %
 \end{eqnarray}
 These inequality together with the Riemann sums show that the series of general terms $(a_n)_n$ and $(b_n)_n$ converge if and only if $\alpha>1.$ Moreover we have seen from the foregoing that
$$\int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt$$ converges only for $\alpha <2$ 
Taking profite of the tricks above, we get the result for the case $\alpha \leq 0$ as follows
$$\int_0^\infty \frac{\sin t}{t^\alpha}\,dt = \int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt+ \sum_{n=1}^{\infty}  \int_{n\pi}^{(n+1)\pi} \frac{\sin t}{t^\alpha}\,dt \\:= \int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt+\sum_{n=1}^{\infty} a'_n $$
With
\begin{eqnarray*}
|a'_n| &=&\left|\int_{n\pi}^{(n+1)\pi} \frac{\sin t}{{(t+n\pi)}^\alpha} \,dt\right|= \left|\int_{0}^{\pi} \frac{\sin t}{{(t+n\pi)}^\alpha} \,dt\right| \geq \frac{2}{(\pi+n\pi)^\alpha}  \qquad\qquad\text{since } \sin(t+n\pi) = (-1)^n\sin t .
\end{eqnarray*}
and the equalities hold in both cases when $\alpha = 0.$ Therefore,
$$\lim |a'_n|= \begin{cases}
2 &~~if ~~\alpha = 0 \nonumber\\
\infty & ~~if ~~\alpha <0. \nonumber
\end{cases}$$
What prove that the divergence of the series $\sum\limits_{n=0}^{\infty} a'_n$ since $a_n'\not\to 0$. Consequently the left hand side of the previous relations always diverge since $\int_{0}^{\pi} \frac{\sin t}{{t}^\alpha} \,dt $ converges for $\alpha\leq 0.$

Conclusion$ \frac{\sin t}{t^\alpha} $ converges for $0<\alpha<2$ and converges absolutely for $1<\alpha <2$.

