广义积分提问1知道∫（Oto+∞）sinx/xdx=π/2,试证∫(oto+∞)sinxcosx/xdx=π/4和∫（oto+∞）（sinx）^2/x^2dx=π/22∫(oto+∞)1/x(lnx)^adx=?（oto+∞）就是区间

Question

广义积分提问
1 知道∫（O to +∞）sinx/xdx=π/2,试证∫(o to +∞)sinxcosx/xdx=π/4和∫（o to +∞）（sinx）^2/x^2dx=π/2
2∫(o to +∞)1/x(lnx)^adx=?
（o to +∞）就是区间

Answer 1

我会第一题,就帮你证一下吧,
∫(o to +∞)sinxcosx/xdx=∫(o to +∞)(1/2)sin2x/xdx
=∫(o to +∞)sin2x/2xdx=1/2∫(o to +∞)sin2x/2xd2x
接下来用换元法,令t=2x,则
1/2∫(o to +∞)sin2x/2xd2x=1/2∫(o to +∞)sint/tdt=1/2∫（O to +∞）sinx/xdx=π/4
证明∫（o to +∞）（sinx）^2/x^2dx=π/2 要用分部积分法
∫（o to +∞）（sinx）^2/x^2dx=[-(sinx)^2/x]（O to +∞）+∫(o to +∞)sin2x/xdx=∫(o to +∞)sin2x/2xd2x
还是用换元法,令t=2x,则
∫(o to +∞)sin2x/2xd2x=∫(o to +∞)sint/tdt=∫（O to +∞）sinx/xdx=π/2,即∫（o to +∞）（sinx）^2/x^2dx=π/2
注：[-(sinx)^2/x]（O to +∞）是指-(sinx)^2/x它以0和+∞为上下限的值,可以用极限的方法算一下,为0