Question

求如下含有三角函数的定积分

Answer 1

令f(r)=∫(0,2π) ln(r^2-2rcosθ+1)dθ
再令u=θ-π，则θ=u+π，dθ=du
则f(r)=∫(-π,π) ln(r^2+2rcosu+1)du
=2∫(0,π) ln(r^2+2rcosu+1)du
df/dr=(d/dr)*2∫(0,π) ln(r^2+2rcosu+1)du
=2∫(0,π) d[ln(r^2+2rcosu+1)/dx]du
=4∫(0,π) (r+cosu)/(r^2+2rcosu+1)du
令t=tan(u/2)，在u=2arctant，du=2/(1+t^2)dt，cosu=(1-t^2)/(1+t^2)
df/dr=4*∫(0,+∞) [r+(1-t^2)/(1+t^2)]/[r^2+2r(1-t^2)/(1+t^2)+1]*2/(1+t^2)dt
=8*∫(0,+∞) [(1+t^2)r+1-t^2]/[(1+t^2)r^2+2r-2rt^2+1+t^2](1+t^2)dt
=8*∫(0,+∞) [(r-1)t^2+r+1]/[(r-1)^2*t^2+(r+1)^2](1+t^2)dt
=(4/r)*∫(0,+∞) {(r^2-1)/[(r-1)^2*t^2+(r+1)^2]+1/(t^2+1)}dt
=(4/r)*[(r+1)/(r-1)]*∫(0,+∞) 1/{t^2+[(r+1)/(r-1)]^2}dt+(4/r)*arctant|(0,+∞)
=(4/r)*arctan[(r-1)t/(r+1)]|(0,+∞)+(4/r)*(π/2)
（1）当r<=-1或r>=1时，df/dr=(4/r)*(π/2)+(4/r)*(π/2)=4π/r
f(r)=4πln|r|+C，其中C是任意常数
因为f(1)=2*∫(0,π) ln(1+cosu)du=-2πln2
所以C=-2πln2
即f(r)=2π*[ln(r^2)-ln2]
（2）当-1<r<1时，df/dr=(4/r)*(-π/2)+(4/r)*(π/2)=0
f(r)=C，其中C是任意常数
因为f(0)=∫(0,2π) ln1dθ=0，所以C=0
即f(r)=0