∫(0→π) e^[-(y+1)t]sint dt
=-∫(0→π) e^[-(y+1)t] d(cost)
=-e^[-(y+1)t]cost - (y+1)∫(0→π) e^[-(y+1)t]cost dt
=-e^[-(y+1)t]cost - (y+1)∫(0→π) e^[-(y+1)t] d(sint)
=-e^[-(y+1)t]cost - (y+1)e^[-(y+1)t]sint - (y+1)²∫(0→π) e^[-(y+1)t]sint dt
将- (y+1)²∫(0→π) e^[-(y+1)t]sint dt 移到等式左边与左边合并,得：
[1+(y+1)²]∫(0→π) e^[-(y+1)t]sint dt=-e^[-(y+1)t]cost - (y+1)e^[-(y+1)t]sint |(0→π)
=e^[-(y+1)π]+1
因此：∫(0→π) e^[-(y+1)t]sint dt=(e^[-(y+1)π]+1)/[1+(y+1)²]