求积分0到无穷大e^-xsinxdx

Question

求积分 0到无穷大 e^-xsinxdx

Answer 1

求广义积分【0,+∞】∫[e^(-x)]sinxdx
先求不定积分：∫[e^(-x)]sinxdx=-∫e^(-x)dcosx=-[e^(-x)cosx-∫cosxde^(-x)]
=-[e^(-x)cosx+∫(cosx)e^(-x)dx]=-[e^(-x)cosx+∫e^(-x)d(sinx)]
=-[e^(-x)cosx+e^(-x)sinx-∫sinxde^(-x)]=-[e^(-x)cosx+e^(-x)sinx+∫sinxe^(-x)dx]
=-e^(-x)cosx-e^(-x)sinx-∫[e^(-x)]sinxdx
移项得2∫[e^(-x)]sinxdx=-e^(-x)(cosx+sinx)
故∫[e^(-x)]sinxdx=-(1/2)[e^(-x)(cosx+sinx)]
∴【0,+∞】∫[e^(-x)]sinxdx=-(1/2)[e^(-x)(cosx+sinx)]【0,+∞】
=x→+∞lim{-(1/2)[e^(-x)(cosx+sinx)]+1/2=x→+∞lim[-(cosx+sinx)/(2e^x)]+1/2=1/2
因为当x→+∞时,分子cosx+sinx是有解函数,而分母2e^x→+∞,故x→+∞lim[-(cosx+sinx)/(2e^x)]=0.