e∧xsinx积分上下限正负无穷。求证是否发散

Question

如题所述

Answer 1

∫(e^x)sinxdx=∫sinxd(e^x)=(e^x)sinx-∫(e^x)cosxdx=(e^x)sinx-∫cosxd(e^x)
=(e^x)sinx-(e^x)cosx-∫(e^x)sinxdx
移项2∫(e^x)sinxdx=(sinx-cosx)e^x
故∫(e^x)sinxdx=(1/2)(sinx-cosx)e^x=(√2/2)[sin(x-π/4)]e^x.
广义积【-∞+∞】∫(e^x)sinxdx=【-∞0】∫(e^x)sinxdx+【0+∞】∫(e^x)sinxdx
=x→-∞lim{(√2/2)[sin(x-π/4)]e^x}+ x→+∞lim{(√2/2)[sin(x-π/4)]e^x}
= x→+∞lim{(√2/2)[sin(x-π/4)]e^x}=存(趋于任何极限).
原积发散