Note that
$ \int_0^\infty\int_0^\infty \sin (x^2+y^2)\,\mathrm dx\,\mathrm dy= \operatorname{Im}\left[\int_0^\infty\int_0^\infty \exp (i(x^2+y^2))\,\mathrm dx\,\mathrm dy\right] $
so, recalling that $\int f(x)\, \mathrm dx=\int f(y)\, \mathrm dy$
$ \int_0^\infty\int_0^\infty \exp (i(x^2+y^2))\,\mathrm dx\,\mathrm dy= \int_0^\infty \exp(ix^2) \,\mathrm dx\int_0^\infty \exp (iy^2)\,\mathrm dy= \left(\int_0^\infty \exp(ix^2) \,\mathrm dx \right)^2 $
Expanding the $\exp (ix^2)$ into $\cos x^2+i\sin x^2$, we find
$ \left(\int_0^\infty \exp(ix^2) \,\mathrm dx \right)^2= \left(\int_0^\infty \cos x^2+i\sin x^2 \,\mathrm dx \right)^2=\left((1+i)\frac{1}{2}\sqrt{\frac{\pi}{2}}\right)^2= 2i \cdot \frac{\pi}{8}= i\frac{\pi}{4} $
So, taking the imaginary part, we see the answer is $\frac{\pi}{4}$. As an added bonus, we take the real part of that answer ($0$) to determine that $\int_0^\infty\int_0^\infty \cos (x^2+y^2)\,\mathrm dx\,\mathrm dy=0$
Here is an alternate proof:
If we use polar coordinates, we see ($x=r\cos \theta$, $x=r\sin \theta$, $\mathrm dx\,\mathrm dy = r\mathrm dr\,\mathrm d\theta$)
$ \int_0^\infty\int_0^\infty \sin (x^2+y^2)\,\mathrm dx\,\mathrm dy= \int_{0}^{2\pi}\int_0^\infty r\sin (r^2)\,\mathrm dr\,\mathrm d\theta $
which diverges. So, we introduce a "dummy function," $\exp(-\delta (x^2+y^2))$ that equals $1$ when $\delta \to 0$. Then,
$ \int_0^\infty\int_0^\infty \exp(-\delta (x^2+y^2))\sin (x^2+y^2)\,\mathrm dx\,\mathrm dy= \int_{0}^{\pi/2}\int_0^\infty r\exp(-\delta r^2)\sin (r^2)\,\mathrm dr\,\mathrm d\theta=$ $=\frac{\pi}{2} \int_0^\infty r\exp(-\delta r^2)\sin (r^2)\,\mathrm dr $
Substituting $r^2=u$, $r \mathrm dr = \frac{1}{2}\mathrm du$, the integral becomes
$\frac{\pi}{4} \int_0^\infty \exp(-\delta u)\sin (u)\,\mathrm du= \frac{\pi}{4(\delta^2+1)}$
and we see that when $\delta \to 0$ the integral converges to $\frac{\pi}{4}$.