一道简单高数积分题,∫(上限∞,下限0)(t+b)^1/2·e（-st次幂）dt

一道简单高数积分题,∫(上限∞,下限0)(t+b)^1/2·e（-st次幂）dt

∫√(t+b)e^(-st)dt
=∫2(t+b)e^(-st)d√(t+b)
√(t+b)=u
=∫2u^2e^-s(u^2-b)du
=(2e^sb)∫u^2e^(-su^2)du
=(-e^2b /s)∫ue^(-su^2)d(-su^2)
=(-e^2b)/s *∫ude^(-su^2)
=(-e^2b)/s*[e^(-su^2)u-∫e^(-su^2)du]
=(-e^2b)/s *e^(-su^2)*u +(e^2b)/(2s)*∫de^(-su^2)/u
=-e^(2b-su^2)/s +e^(2b-su^2)/2su -e^(2b)/(2s) *∫e^(-su^2)d(1/u)
=-e^(2b-su^2)/s+ (e^2b-su^2)/2su - e^(2b-su^2)/[(4s^2)*u^3] +e^2b/(4s^2)∫e^(-su^2)d(1/u^3)
=-e^(2b-su^2)/s+e^2b-su^2)/2su-e^(2b-su^2)/[(4s^2)u^3]+e^(2b-su^2)/[(8s^3)u^5]+...
+ (-1)^n e^(2b-su^2)/[(2s)^(n-1) * u^(2n-3)]
∫(0,+∞)(t+b)^(1/2)*e^(-st)dt ≈-e^(2b-st+sb)/s|(0,∞]=-e^(2b-sb) /s
再问： 感谢drug2009的热心回复，但是还是有点不太明白 第8行到第9行∫e^(-su^2)du积分得1/2∫de^(-su^2)/u这一步是怎么来的？正确吗？ 谢谢!
再答： ∫e^(-su^2)du积分得1/2∫de^(-su^2)/u,有失误 ∫e^(-su^2)du=(1/2s)∫e^(-su^2)2sudu/su=(-1/2s)∫e^(-su^2)d-su^2/su =(-1/2s)∫de^(-su^2)/u 因此后面应该是 =(-e^2b)/s*[e^(-su^2)u-∫e^(-su^2)du] =(-e^2b)/s *e^(-su^2)*u +(e^2b)/(2s^2)*∫de^(-su^2)/u =-e^(2b-su^2)/s +e^(2b-su^2)/2s^2u -e^(2b)/(2s^2) *∫e^(-su^2)d(1/u) =-e^(2b-su^2)/s+ (e^2b-su^2)/2s^2u - e^(2b-su^2)/[(4s^3)*u^3] +e^2b/(4s^2)∫e^(-su^2)d(1/u^3) =-e^(2b-su^2)/s+e^2b-su^2)/2s^2u-e^(2b-su^2)/[(4s^3)u^3]+e^(2b-su^2)/[(8s^4)u^5]+... + (-1)^n e^(2b-su^2)/[(2s)^(n) * u^(2n-3)] ∫(0,+∞)(t+b)^(1/2)*e^(-st)dt ≈-e^(2b-st+sb)/s|(0,∞]=-e^(2b-sb) /s