e的负t的平方积分等于根下派的那个是怎么证明的

I=∫(-∞ ->+∞)e^-t^2 dt
与下面两式等效
I=∫(-∞ ->+∞)e^-x^2 dx
I=∫(-∞ ->+∞)e^-y^2 dy
∴I^2=∫(-∞ ->+∞)∫(-∞ ->+∞)e^(-x^2-y^2)dxdy
令x=pcosa y=psina
∴上式化为
I^2=∫(0->2π)∫(0->+∞) e^-p^2 pdpda
=∫(0->2π)da∫(0->+∞) e^-p^2 pdp
=-1/2∫(0->2π)da∫(0->+∞) e^-p^2d(-p^2)
=-1/2∫(0->2π)da∫(0->+∞) de^-p^2
=-1/2∫(0->2π)da e^-p^2 | (0->+∞)
=-1/2 *(2π)*(0-1)
=π
∴ I=√π