Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

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


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

share|cite|improve this question
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
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 $\frac{\pi}{4}$.

share|cite|improve this answer
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

$$\begin{array}{c l}\int_0^\infty\int_0^\infty \sin(x^2+y^2)dxdy & = \int_0^\infty\int_0^\infty \sin(x^2)\cos(y^2)+\cos(x^2)\sin(y^2)dxdy \\[10pt] & =\int_0^\infty\sin(x^2)dx\int_0^\infty\cos(y^2)dy \\ & \qquad~~~+\int_0^\infty\cos(x^2)dx\int_0^\infty\sin(y^2)dy \\[10pt] & = 2\left(\frac{1}{2}\sqrt{\frac{\pi}{2}}\right)^2=\frac{\pi}{4}. \end{array}$$

share|cite|improve this answer
+1 Oh, so painfully simple and beautiful... – DonAntonio Aug 13 '12 at 3:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.