∫∫_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．

∫∫_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∫₀^1-xdy∫₀^1-x-y [dz/(1+x+y+z)³] =-(1/2)∫₀¹dx∫₀^1-xdy[1/(1+x+y+z)²]∣₀^1-x-y =-(1/2)∫₀¹dx∫₀^1-x[1/4-1/(1+x+y+z)²]dy =-(1/2)∫₀¹dx•[(1/4)y+1/(1+x+y+z)]∣₀^1-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