[Home Squat University Mathematics magazine] an integral inequality in the No. 409 period relating to the sine logarithm

Trial: $$\bex 0<\int_0^\infty \frac{\sin t}{\ln (1+x+t)}\rd t<\frac{2}{\ln (1+x)}. \eex$$

proof: $$\beex \bea \int_0^\infty \frac{\sin t}{\ln (1+x+t)}\rd T & =\sum_{k=0}^\infty\sez{\int_{2k\pi}^{2k\pi+\pi} \frac{\sin t}{\ln (1+x+t)}\rd t +\int_{2k\pi+\pi}^{2k\pi+2\pi} \frac {\sin T} {\ln (1+x+t)}\rd t}\\ &=\sum_{k=0}^\infty \sez{\int_0^\pi \frac{\sin s}{\ln (1+x+2k\pi +s)}\rd s-\int_0^\pi\frac{\ Sin s}{\ln (1+x+2k\pi+\pi+s)}\rd s}\\ &=\sum_{k=0}^\infty \int_0^\pi \sin s\sez{\frac{1}{\ln (1+x+2k\pi+s)}-\frac{ 1}{\ln (1+x+2k\pi+\pi+s)}}\rd s\\ &>0. \eea \eeex$$ on the other hand, $$\beex \bea \int_0^\infty \frac{\sin s}{\ln (1+x+s)}\rd s&=\int_0^\pi \sin s\sez{\frac{1}{\ln (1+x +2k\pi+s)}-\frac{1}{\ln (1+x+2k\pi+\pi+s)}}\rd s\\ &<\int_0^\pi \sin s\sez{\frac{1}{\ln (1+X+2K\PI)}-\frac{1} {\ln (1+X+2K\PI+\PI)}} \rd s\\ &\quad\sex{f (s) \equiv\frac{1}{\ln (1+x+2k\pi+s)}-\frac{1}{\ln (1+x+2k\pi+\pi+s)}\ \searrow}\\ &<\ Int_0^\pi \frac{\sin s}{\ln (1+x)}\rd s\\ &=\frac{2}{\ln (1+x)}. \eea \eeex$$

[Home Squat University Mathematics magazine] No. 409 issue an integral inequality related to sine logarithm