How to evaluate the integral $\iint_C \sin (x^2+y^2)\,\mathrm dx\,\mathrm dy$ I would like to evaluate
$$\iint\limits_C  \sin (x^2+y^2)\,\mathrm dx\,\mathrm dy$$
where $C$ is the first quadrant, i.e.
$$\int\limits_0^\infty\int\limits_0^\infty  \sin (x^2+y^2)\,\mathrm dx\,\mathrm dy$$
where
$$\int\limits_0^\infty \sin x^2\, dx=\int\limits_0^\infty \cos x^2\, dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}$$
 A: 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 $\frac{\pi}{4}$.
A: $$\begin{array}{c l}\int_0^\infty\int_0^\infty \sin(x^2+y^2)dxdy & = \int_0^\infty\int_0^\infty \sin(x^2)\cos(y^2)+\cos(x^2)\sin(y^2)dxdy \\[10pt] & =\int_0^\infty\sin(x^2)dx\int_0^\infty\cos(y^2)dy \\ & \qquad~~~+\int_0^\infty\cos(x^2)dx\int_0^\infty\sin(y^2)dy \\[10pt] & = 2\left(\frac{1}{2}\sqrt{\frac{\pi}{2}}\right)^2=\frac{\pi}{4}. \end{array}$$
