2⁢n⁢∫0x(1-y)α⁢(1+y)β⁢Pn(α,β)⁡(y)⁢ⅆ⁡y=Pn-1(α+1,β+1)⁡(0)-(1-x)α+1⁢(1+x)β+1⁢Pn-1(α+1,β+1)⁡(x).