汝坟
这个是超越积分,一般就是直接作为定理。下面写一个∫(-∞→∞) e^(-x²) dx的算法。(以下设的未知数跟你题目中未知数没关联。)解:积分域为 x ∈(-∞,+∞)令: F = (-∞,+∞)∫e^(-2x²)dx同样 F= (-∞,+∞)∫e^(-2y²)dy由于x,y是互不相关的的积分变量,因此:F² = (-∞,+∞)∫e^(-2x²)dx * (-∞,+∞)∫e^(-2y²)dy= [D]∫∫e^(-2x²)*dx * e^(-2y²)*dy= [D]∫∫e^[-2(x²+y²)]*dx *dy式中积分域D = {(x,y)|x ∈(-∞,+∞),y∈(-∞,+∞)}对x,y进行极坐标变换,则:x²+y² = ρ²;dxdy = ρ*dρ*dθF² = [D]∫∫e^[-2(x²+y²)]*dx *dy= [0,+∞)[0,2π]∫∫e^(-2ρ²) ρ*dρ*dθ= [0,2π]∫dθ *(0,+∞)∫e^(-2ρ²) ρ*dρ= 2π* 1/4*[0,+∞)*∫e^(-2ρ²) *d(2ρ²)= π/2因此 F = (-∞,+∞)∫e^(-2x²)dx = √(π/2) = (-∞,+∞)∫e^(-2y²)dy所以答案为(-1/2)√(π/2) + (-1/2)√(π/2)=-√(π/2)本回答被提问者和网友采纳