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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.

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

where

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

  • 0
    First quadrant, sorry.2012-08-13

2 Answers 2

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

  • 0
    In case anyone else is wondering about the underlying theory here, I posted a question about it you might find interesting: http://math.stackexchange.com/questions/181971/abel-summability-of-integrals2012-08-13
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$\begin{array}{c l}\int_0^\infty\int_0^\infty \sin(x^2+y^2)dxdy & = \int_0^\infty\int_0^\infty \sin(x^2)\cos(y^2)+\cos(x^2)\sin(y^2)dxdy \\[10pt] & =\int_0^\infty\sin(x^2)dx\int_0^\infty\cos(y^2)dy \\ & \qquad~~~+\int_0^\infty\cos(x^2)dx\int_0^\infty\sin(y^2)dy \\[10pt] & = 2\left(\frac{1}{2}\sqrt{\frac{\pi}{2}}\right)^2=\frac{\pi}{4}. \end{array}$