试证: $$ex 0<int_0^infty frac{sin t}{ln(1+x+t)} d t<frac{2}{ln(1+x)}. eex$$
证明: $$eex ea int_0^infty frac{sin t}{ln(1+x+t)} d t &=sum_{k=0}^inftysez{ int_{2kpi}^{2kpi+pi} frac{sin t}{ln(1+x+t)} d t +int_{2kpi+pi}^{2kpi+2pi} frac{sin t}{ln(1+x+t)} d t}\ &=sum_{k=0}^infty sez{int_0^pi frac{sin s}{ln(1+x+2kpi +s)} d s -int_0^pifrac{sin s}{ln(1+x+2kpi+pi+s)} d s}\ &=sum_{k=0}^infty int_0^pi sin ssez{ frac{1}{ln(1+x+2kpi+s)}-frac{1}{ln(1+x+2kpi+pi+s)}} d s\ &>0. eea eeex$$ 另一方面, $$eex ea int_0^infty frac{sin s}{ln(1+x+s)} d s&=int_0^pi sin ssez{ frac{1}{ln(1+x+2kpi+s)} -frac{1}{ln (1+x+2kpi+pi+s)}} d s\ &<int_0^pi sin ssez{ frac{1}{ln(1+x+2kpi)} -frac{1}{ln (1+x+2kpi+pi)}} d s\ &quadsex{f(s)equivfrac{1}{ln(1+x+2kpi+s)} -frac{1}{ln (1+x+2kpi+pi+s)} searrow}\ &<int_0^pi frac{sin s}{ln(1+x)} d s\ &=frac{2}{ln(1+x)}. eea eeex$$