Calculate the improper double integral: integral

Calculate the improper double integral: $$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}\cos(x^2+y^2)\,dx\,dy. $$

Step 1 · Convert to polar coordinates

With $x=r\cos\theta$, $y=r\sin\theta$, $x^2+y^2=r^2$, $dx\,dy=r\,dr\,d\theta$:

$$ I = \int_0^{2\pi}\int_0^\infty r\,e^{-r^2}\cos(r^2)\,dr\,d\theta = 2\pi\int_0^\infty r\,e^{-r^2}\cos(r^2)\,dr. $$

Step 2 · Substitute $u=r^2$

With $du=2r\,dr$:

$$ \int_0^\infty r\,e^{-r^2}\cos(r^2)\,dr = \frac12\int_0^\infty e^{-u}\cos u\,du. $$

Step 3 · Evaluate $\displaystyle\int_0^\infty e^{-u}\cos u\,du$

Integrating by parts twice (or invoking the standard Laplace-transform formula $\int_0^\infty e^{-au}\cos(bu)\,du=\frac{a}{a^2+b^2}$ with $a=b=1$):

$$ \int_0^\infty e^{-u}\cos u\,du = \frac1{1^2+1^2} = \frac12. $$

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 $\pi/2$.
• The same computation with $\cos$ replaced by $\sin$ gives an identical value $\pi/2$ (Problem 110), since $\int_0^\infty e^{-u}\sin u\,du=\int_0^\infty e^{-u}\cos u\,du=\frac12$ as well — both single-variable Laplace transforms happen to coincide at $a=b=1$.
• A slicker route uses $\cos(r^2)=\operatorname{Re}\big(e^{ir^2}\big)$: then $\int_0^\infty re^{-r^2}\cos(r^2)\,dr=\operatorname{Re}\!\left(\int_0^\infty re^{-(1-i)r^2}\,dr\right)=\operatorname{Re}\!\left(\dfrac1{2(1-i)}\right)=\dfrac14$, matching Step 2–3 combined.