Question

计算如图的定积分，其中X是常数

Answer 1

因为∫(0,1)(x^b-x^a)/lnxdx=[x^y/lnx](a,b)=∫(a,b)x^ydy
原式=∫(0,1)dx∫(a,b)sin(ln1/x)x^ydy
=∫(a,b)dy∫(0,1)sin(ln1/x)x^ydx（二次积分交换次序）
而∫sin(ln1/x)x^ydx=sin(ln1/x)x^(y+1)/(y+1)-∫cos(ln1/x)(-1/x)x^(y+1)/(y+1)dx
=sin(ln1/x)x^(y+1)/(y+1)+∫cos(ln1/x)x^y/(y+1)dx
=sin(ln1/x)x^(y+1)/(y+1)+cos(ln1/x)x^(y+1)/(y+1)^2-∫(-sin(ln1/x))(-1/x)x^(y+1)/(y+1)^2dx
=sin(ln1/x)x^(y+1)/(y+1)+cos(ln1/x)x^(y+1)/(y+1)^2-∫sin(ln1/x)x^y/(y+1)^2dx
故∫sin(ln1/x)x^ydx=[sin(ln1/x)x^(y+1)/(y+1)+cos(ln1/x)x^(y+1)/(y+1)^2]/(1+1/(y+1)^2)+c
=[sin(ln1/x)x^(y+1)*(y+1)+cos(ln1/x)x^(y+1)]/(y^2+2y+1)+c（用不定积分计算原函数，主要是定积分不怎么会。。）
∫(0,1)sin(ln1/x)x^ydx=1/(y^2+2y+1)
所以原式=∫(a,b)dy/(y^2+2y+1)=[arctan(1+y)](a,b)=arctan(1+b)-arctan(1+a)
（以上括号外未标注符号的，统一是乘号）