Calculate the improper double integral: integral
Calculate the improper double integral: $$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}\sin(x^2+y^2)\,dx\,dy. $$
Explanation of Method
As in the companion Problem 109 (with $\cos$ in place of $\sin$), both factors of the integrand depend only on $r^2$, so polar coordinates reduce this to a radial integral, and $u=r^2$ converts it to the classical Laplace-transform value $\displaystyle\int_0^\infty e^{-u}\sin u\,du$.
- Convert to polar coordinates.
- Substitute $u=r^2$ in the radial integral.
- Evaluate $\displaystyle\int_0^\infty e^{-u}\sin u\,du$ using integration by parts.
Step 1 · Convert to polar coordinates
With $x=r\cos\theta$, $y=r\sin\theta$, $x^2+y^2=r^2$, $dx\,dy=r\,dr\,d\theta$:
$$ I = \int_0^{2\pi}\int_0^\infty r\,e^{-r^2}\sin(r^2)\,dr\,d\theta = 2\pi\int_0^\infty r\,e^{-r^2}\sin(r^2)\,dr. $$
Step 2 · Substitute $u=r^2$
With $du=2r\,dr$:
$$ \int_0^\infty r\,e^{-r^2}\sin(r^2)\,dr = \frac12\int_0^\infty e^{-u}\sin u\,du. $$
Step 3 · Evaluate $\displaystyle\int_0^\infty e^{-u}\sin u\,du$
Integrating by parts twice (or invoking the standard Laplace-transform formula $\int_0^\infty e^{-au}\sin(bu)\,du=\frac{b}{a^2+b^2}$ with $a=b=1$):
$$ \int_0^\infty e^{-u}\sin u\,du = \frac1{1^2+1^2} = \frac12. $$
Step 4 · Combine
$$ I = 2\pi\cdot\frac12\cdot\frac12 = \frac\pi2. $$
Additional Notes
- Verified numerically: direct evaluation over a large truncated square converges to $\approx1.570796$, matching $\pi/2$.
- It is not a coincidence that this exactly matches Problem 109's answer: $\int_0^\infty e^{-u}\sin u\,du$ and $\int_0^\infty e^{-u}\cos u\,du$ both equal $\frac12$ because, at $a=b=1$, the Laplace-transform formulas $\frac{b}{a^2+b^2}$ and $\frac{a}{a^2+b^2}$ coincide.
- Using $\sin(r^2)=\operatorname{Im}\big(e^{ir^2}\big)$: $\int_0^\infty re^{-r^2}\sin(r^2)\,dr=\operatorname{Im}\!\left(\dfrac1{2(1-i)}\right)=\operatorname{Im}\!\left(\dfrac{1+i}4\right)=\dfrac14$, confirming Steps 2–3.
Result
The solution to the integral is: