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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.

$$\int\limits_0^\infty\int\limits_0^\infty \sin (x^2+y^2)\,\mathrm dx\,\mathrm dy$$


$$\int\limits_0^\infty \sin x^2\, dx=\int\limits_0^\infty \cos x^2\, dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}$$

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The double integral you wrote is not over the entire plane but only over its first quadrant, so which one is it? –  DonAntonio Aug 13 '12 at 2:17
First quadrant, sorry. –  qwerty Aug 13 '12 at 2:18
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2 Answers

up vote 7 down vote accepted

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 $\frac{\pi}{4}$.

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Thanks for the help! Both the methods are very interesting! –  qwerty Aug 13 '12 at 2:29
@qwerty No problem! –  Argon Aug 13 '12 at 2:30
@Potato: the integral converges, but not absolutely. Just like rearranging a conditionally convergent series can give you a series that diverges or converges to any value, change of variables in a conditionally converging integral could give you different results. –  Robert Israel Aug 13 '12 at 3:19
Let $R_n$ be the quarter-annulus $\{(r,\theta): \sqrt{\pi (n-1)} < r < \sqrt{\pi n}, 0 \le \theta \le \pi/2\}$. Then $J_n = \int_{R_n} \sin(x^2+y^2)\ dx dy = (\pi/2) \int_{\sqrt{\pi(n-1)}}^\sqrt{\pi n} r \sin(r^2)\ dr = (-1)^{n+1} \pi/2$, and essentially you're looking at $\sum_{n=1}^\infty J_n$. Introducing $\exp(-\delta r^2)$ is essentially Abel's summability method, while using $\int_0^R \int_0^R dx \ dy$ is rather similar to Cesaro summability. –  Robert Israel Aug 13 '12 at 3:39
By this I mean, instead of a straight partial sum of the $J_n$ you have basically a weighted partial sum where the weights (essentially the fraction of the annulus included in the square) go gradually to $0$ at the upper end. –  Robert Israel Aug 13 '12 at 3:51
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$$\begin{array}{c l}\int_0^\infty\int_0^\infty \sin(x^2+y^2)dxdy & = \int_0^\infty\int_0^\infty \sin(x^2)\cos(y^2)+\cos(x^2)\sin(y^2)dxdy \\[10pt] & =\int_0^\infty\sin(x^2)dx\int_0^\infty\cos(y^2)dy \\ & \qquad~~~+\int_0^\infty\cos(x^2)dx\int_0^\infty\sin(y^2)dy \\[10pt] & = 2\left(\frac{1}{2}\sqrt{\frac{\pi}{2}}\right)^2=\frac{\pi}{4}. \end{array}$$

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+1 Oh, so painfully simple and beautiful... –  DonAntonio Aug 13 '12 at 3:32
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