∫∫Ω[1/(1+x+y+x)3]dυ，其中Ω是由x=0，y=0，z=0和x+y+z=1所围空间立体．
因Ω={(x，y，z)∣0≤x≤1,0≤y≤1-x,0≤z≤1-x-y} 所以 ∫∫∫Ω[1/(1+x+y+z)3]dΩ =∫01dx∫01-xdy∫01-x-y [dz/(1+x+y+z)3] =-(1/2)∫01dx∫01-xdy[1/(1+x+y+z)2]∣01-x-y =-(1/2)∫01dx∫01-x[1/4-1/(1+x+y+z)2]dy =-(1/2)∫01dx•[(1/4)y+1/(1+x+y+z)]∣01-x =-(1/2)∫01[1/4(1-x)+1/2-1/(1+x)]dx =-(3/8)∫01dx+1/8∫01xdx+1/2∫01[1/(1+x)]dx =-(3/8)+(1/16)x2∣01+(1/2)ln(1+x)∣01 =(1/2)ln2-5/16
