點樣prove 到In(exp(-x^2/2))=sqrt(2*pi())?

你識唔識double integral先？識唔識jacobian？

[In(exp(-x^2/2))dx]^2
=[In(exp(-x^2/2))dx][In(exp(-x^2/2))dx]
=[In(exp(-x^2/2))dx][In(exp(-y^2/2))dy]
=InIn(exp(-x^2/2)exp(-y^2/2))dxdy
=InIn[exp(-(x^2+y^2)/2))]dxdy

Then use the change of variable x= r cos(s), y= r sin(s).
The jacobian of the change of variable is
(其實唔係dx/dr, 係 partial differentiation)
| (dx/dr)(dy/ds)-(dx/ds)(dy/dr) | = r

And the integration limit for r is from 0 to infinity and for s is 0 to 2*pi
because the integration domain is the whole R^2 plane.

Then
InIn[exp(-(x^2+y^2)/2))]dxdy
= InIn[exp(-r^2/2))]rdrds
= In{In[exp(-r^2/2))] d(r^2/2)}ds (again change of single variable)
= In{ -{ lim(r->infinity)[exp(-r^2/2)] - exp(-0^2/2)}} ds
= In(1)ds
= 2*pi

So [In(exp(-x^2/2))dx]^2=2*pi

Hence, In(exp(-x^2/2))=sqrt(2*pi)

做乜問兩次？