Polar coordinates: e^−(x²+y²) integral

Short answer

First quadrant, polar: 0≤r<∞, 0≤θ≤π/2. e^{−(x²+y²)}=e^{−r²}, dx dy=r dr dθ. ∫₀^∞ e^{−r²} r dr = 1/2, θ-integral π/2, giving π/4.

Evaluate: ∫₀^∞ ∫₀^∞ e^−(x²+y²) dy dx by changing to polar coordinates.

Region. x and y both run from 0 to ∞: the whole first quadrant. In polar coordinates: 0 ≤ r < ∞, 0 ≤ θ ≤ π/2.

x² + y² = r², dy dx = r dr dθ.

I = ∫₀^{π/2} ∫₀^∞ e^(−r²) r dr dθ

Put t = r², dt = 2r dr:

∫₀^∞ e^(−r²) r dr = (1/2) ∫₀^∞ e^(−t) dt = (1/2)[ −e^(−t) ]₀^∞ = 1/2

I = (1/2) ∫₀^{π/2} dθ = (1/2)(π/2) = π/4

Answer: I = π/4.

Consequence. The integral separates as (∫₀^∞ e^(−x²) dx)² = π/4, so ∫₀^∞ e^(−x²) dx = √π/2.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions