How to see this: ∫∫[0-∞] e^(-(x+y)^2) dx dy = 1/2?

Question

∫∫[0->∞] e^(-(x+y)^2) dx dy = 1/2.

Is there a way to see this without doing the full integral?

? · Accepted Answer

Change the variables

u = x + y,   v = x - y.

The Jacobian of this transformation is 1/2. The integral in terms of u and v takes the form

(1/2) ∫ ∫ exp( - u^2) du  dv.

Now, you need to determine the limits of the integrals. Since both x and y are non-negative, u varies from 0 to infinity. If u is fixed, then v has maximum at the point x=u, y=0, and v has minimum at the point x=0, y=u. So, v varies from -u to u and integration over v gives

∫_{-u}^{u}  dv = 2 u.

Then

(1/2) ∫ ∫ exp( - u^2) du  dv =  ∫_0^∞  exp( - u^2) u du = 1/2.

Anonymous · Answer

convert to polar coordinates --->∫e^-r^2 rdr ∫ d(0)

How to see this: ∫∫[0->∞] e^(-(x+y)^2) dx dy = 1/2?

2 Answers