∫e^(-x^2)dx=∫e^(-y^2)dy
而∫e^(-x^2)dx*∫e^(-y^2)dy
=∫∫e^(-y^2)*e^(-x^2)dxdy
=∫∫e^(-x^2-y^2)dxdy
然后是用極坐標(biāo)換元,x=rcosa,y=rsina r屬于[0,無(wú)窮大),a屬于[0,2π]
=∫∫re^(-r^2)drda (r屬于[0,無(wú)窮大),a屬于[0,2π])
=∫(0,2π)da*∫re^(-r^2)dr r屬于[0,無(wú)窮大),
=2π* 1/2*∫e^(-r^2)dr^2 r屬于[0,無(wú)窮大),
=π* ∫-de^(-r^2) r屬于[0,無(wú)窮大),
=π*[e^(-0^2)-lime^(-r^2)] r→無(wú)窮大
=π*(1-0)
=π
∫e^(-x^2)dx*∫e^(-y^2)dy=π=[∫e^(-x^2)dx]^2
易知∫e^(-x^2)dx>0
所以∫e^(-x^2)dx=√π