1.已知∫(+∞,-∞) e^(-x^2)dx=√π,则∫(+∞,0 )e^(-ax^2)dx=
发布网友
发布时间：2024-10-09 08:43
共1个回答
热心网友
时间：2024-11-24 15:27
则∫（+∞,0 ）e^（-ax^2）dx
=1/√a∫（+∞,0 ）e^（-ax^2）d(√ax)
=√π/√a
lim(x→0) ∫(x^2,0)f(t)dt/[x^2∫(x,0)f(t)dt]
=lim(x→0) 2xf(x^2)/[2x∫(x,0)f(t)dt+x^2f(x)]
=lim(x→0) 2f(x^2)/[2∫(x,0)f(t)dt+xf(x)]
=lim(x→0) 4xf'(x^2)/[2f(x)+f(x)+xf'(x)]
=lim(x→0) 4f'(x^2)/[3f(x)/x+f'(x)]
=4f'(x^2)*lim(x→0) 1/[lim(x→0) 3f(x)/x+f'(x)]
=lim(x→0) 4f'(x^2)/(4f'(x))
=1
