oleifere45

2022-06-27

Prove that ${\int }_{0}^{1}\frac{1}{1+{\mathrm{ln}}^{2}x}\phantom{\rule{thinmathspace}{0ex}}dx={\int }_{1}^{\mathrm{\infty }}\frac{\mathrm{sin}\left(x-1\right)}{x}\phantom{\rule{thinmathspace}{0ex}}dx$

sleuteleni7

Expert

Hint. One may observe that
$\frac{1}{1+{\mathrm{ln}}^{2}x}=-\mathrm{\Im }\frac{1}{i-\mathrm{ln}x}=-\mathrm{\Im }{\int }_{0}^{\mathrm{\infty }}{e}^{-\left(i-\mathrm{ln}x\right)t}dt,\phantom{\rule{1em}{0ex}}x\in \left(0,1\right),$
gives
$\begin{array}{rl}{\int }_{0}^{1}\frac{1}{1+{\mathrm{ln}}^{2}x}\phantom{\rule{thinmathspace}{0ex}}dx& =-\mathrm{\Im }{\int }_{0}^{1}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{\int }_{0}^{\mathrm{\infty }}{e}^{-\left(i-\mathrm{ln}x\right)t}dt\phantom{\rule{mediummathspace}{0ex}}dx\\ \\ & =-\mathrm{\Im }{\int }_{0}^{\mathrm{\infty }}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\left({\int }_{0}^{1}{x}^{t}dx\right){e}^{-it}dt\\ \\ & =-\mathrm{\Im }{\int }_{0}^{\mathrm{\infty }}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\frac{1}{t+1}{e}^{-it}dt\\ \\ & ={\int }_{0}^{\mathrm{\infty }}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\frac{\mathrm{sin}t}{t+1}dt\\ \\ & ={\int }_{1}^{\mathrm{\infty }}\frac{\mathrm{sin}\left(x-1\right)}{x}\phantom{\rule{thinmathspace}{0ex}}dx\end{array}$
as announced.

Lucille Cummings

Expert

Define $I\left(a\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-\left(x+1\right)a}\frac{\mathrm{sin}\left(x\right)}{x+1}dx$. Then ${I}^{\prime }\left(a\right)=-{\int }_{0}^{\mathrm{\infty }}{e}^{-\left(x+1\right)a}\mathrm{sin}\left(x\right)dx=-\frac{{e}^{-a}}{{a}^{2}+1}$, and since $\underset{a\to \mathrm{\infty }}{lim}I\left(a\right)=0$, we have