Convergence of $ I=\int_0^\infty \sin x\sin(x^2)\mathrm{d}x$
We shall use only substitution and integration by parts to show that the integral of interest, $\int_1^L \sin(x)\sin(x^2)\,dx$, is convergent.
First, enforcing the substitution $x\to \sqrt{x}$ reveals
$$\int_1^L \sin(x)\sin(x^2)\,dx=\frac12\int_1^L \frac{\sin(\sqrt{x})\sin(x)}{\sqrt{x}}\,dx \tag 1$$
Second, integrating by parts the integral on the right-hand side of $(1)$ with $u=\frac{\sin(\sqrt{x})}{\sqrt{x}}$ and $v=-\cos(x)$ yields
$$\begin{align} \int_1^L \sin(x)\sin(x^2)\,dx&=\frac12\left.\left(-\frac{\sin(\sqrt{x})\cos(x)}{\sqrt{x}}\right)\right|_{x=1}^{x=L}\\\\ &+\frac14 \int_1^L \left(\frac{\cos(\sqrt{x})\cos(x)}{x}-\frac{\sin(\sqrt{x})\cos(x)}{x^{3/2}}\right)\,dx\tag2 \end{align}$$
Third, integrating by parts the first term in the integral on the right-hand side of $(2)$ with $u=\frac{\cos(\sqrt{x})}{x}$ and $v=\sin(x)$, we obtain
$$\begin{align}\int_1^L \frac{\cos(\sqrt{x})\cos(x)}{x}\,&=\left.\left(\frac{\cos(\sqrt{x})\sin(x)}{x}\right)\right|_{x=1}^{x=L}\\\\ &- \int_1^L \left(\frac{\sin(\sqrt{x})\sin(x)}{2x^{3/2}}+\frac{\cos(\sqrt{x})\sin(x)}{x^2}\right)\,dx\tag 3 \end{align}$$
Substituting $(3)$ into $(2)$ shows that all integrals involved are of the forms
$$I_1=\int_1^L \frac{\sin(\sqrt{x})\cos(x)}{x^{3/2}}\,dx$$
and
$$I_2=\int_1^L \frac{\cos(\sqrt{x})\sin(x)}{x^2}\,dx$$
Both $I_1$ and $I_2$ are absolutely convergent as $L\to\infty$ since $\int_1^\infty \frac{1}{x^{3/2}}\,dx<\infty$ and $\int_1^L \frac{1}{x^2}\,dx<\infty$.
Therefore, the integral of interest converges as was to be shown!.
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With $\ds{\Lambda > 0}$:
\begin{align} \int_{0}^{\Lambda}\sin\pars{x}\sin\pars{x^{2}}\,\dd x & = {1 \over 2}\int_{0}^{\Lambda}\cos\pars{x^{2} - x}\,\dd x - {1 \over 2}\int_{0}^{\Lambda}\cos\pars{x^{2} + x}\,\dd x \\[5mm] & = {1 \over 2}\int_{1/2}^{\Lambda + 1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x - {1 \over 2}\int_{-1/2}^{\Lambda - 1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x \\[1cm] & = {1 \over 2}\int_{1/2}^{\Lambda - 1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x + {1 \over 2}\int_{\Lambda - 1/2}^{\Lambda + 1/2} \cos\pars{x^{2} - {1 \over 4}}\,\dd x \\[2mm] & - {1 \over 2}\int_{-1/2}^{1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x - {1 \over 2}\int_{1/2}^{\Lambda - 1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x \\[1cm] & = {1 \over 2}\int_{-1/2}^{1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x \\[2mm] & + \bracks{% {1 \over 2}\int_{1/2}^{\Lambda + 1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x - {1 \over 2}\int_{1/2}^{\Lambda - 1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x} \end{align}
As $\ds{\Lambda \to \infty}$, the last two integrals converge since they can be reduced to convergent Fresnel Integrals.
$$ \int_{0}^{\infty}\sin\pars{x}\sin\pars{x^{2}}\,\dd x = \bbx{\int_{0}^{1/2}\cos\pars{x^{2} - {1 \over 4}}\,\dd x} \quad\mbox{which is clearly}\ convergent. $$