sin(x^2)/x^2的无穷积分,

Question

sin(x^2)/x^2的无穷积分,

Answer 1

0,+∞)∫(sinx/x)^2dx=(1/2)*(0,+∞)∫(1-cos2x)/x^2dx
=(1/2)*(0,+∞)∫1/x^2dx-(1/2)*(0,+∞)∫cos2x/x^2dx
=(1/2)*(0,+∞)∫1/x^2dx-(1/2)*(0,+∞)∫cos2x/d(1/x)
=(1/2)*(0,+∞)(-1/x)-(1/2)*(0,+∞)cos2x/x+(1/2)*(0,+∞)∫sin2x/xdx
=(1/2)*(0,+∞)(cos2x-1)/x+(0,+∞)∫sint/tdt
=(0,+∞)∫sinx/xdx
下面是高手做的.
I=∫∫ e^(-xy) ·sinxdxdy
D
其中D = [0,+∞)×[0,+∞),
今按两种不同的次序进行积分得
I=∫sinxdx ∫e^(-xy)dy
0 +∞ 0 +∞
= ∫sinx·(1/x)dx
0 +∞
另一方面,交换积分顺序有：
I=∫∫ e^(-xy) ·sinxdxdy
D
=∫dy ∫e^(-xy)·sinxdx
0 +∞ 0 +∞
=∫dy/(1+y^2）=arc tan+∞-arc tan0
0 +∞
= π/2
所以：
∫sinx·(1/x)dx=π/2
0 +∞