计算 (0,+∞) 上的广义定积分 ∫(0,+∞)lnxsinx／x dx

elim

题：计算 \(\displaystyle\int_0^{\infty} \frac{\ln{x}}{x}\sin{x} dx\)

elim

网上找来的。应该有简化的余地。可能有些地方需要说明论证. 欢迎参与！

\(\small\begin{align}\int\limits_{0}^{+\infty }{\frac{\ln x\sin x}{x}dx}&=\int\limits_{0}^{+\infty }\big( \int\limits_{0}^{+\infty }{{{e}^{-xt}}\ln x\sin xdt} \big)dx\\
&=\frac{1}{2i}\int\limits_{0}^{+\infty }{\big( \int\limits_{0}^{+\infty }{\left( {{e}^{-x\left( t-i \right)}}-{{e}^{-x\left( t+i \right)}} \right)\ln xdx} \big)dt}\\
&=\frac{1}{2i}\int\limits_{0}^{+\infty }{\left( \frac{\gamma +\ln \left( t+i \right)}{t+i}-\frac{\gamma +\ln \left( t-i \right)}{t-i} \right)dt}\\
&=-\gamma \int\limits_{0}^{+\infty }{\frac{dt}{{{t}^{2}}+1}}+\frac{1}{2i}\int\limits_{0}^{+\infty }{\left( \frac{\ln \left( t+i \right)}{t+i}-\frac{\ln\left( t-i \right)}{t-i} \right)dt}\\
&=-\gamma \int\limits_{0}^{+\infty }{\frac{dt}{{{t}^{2}}+1}}+\frac{1}{2i}\int\limits_{0}^{+\infty }{ \frac{t\ln \left( \frac{t+i}{t-i} \right)-i\ln \left( {{t}^{2}}+1 \right)}{{{t}^{2}}+1}dt}\\
&=-\frac{\pi \gamma }{2}+\frac{1}{2i}\int\limits_{0}^{+\infty }{ \frac{t\ln \left( {{e}^{2i\arctan \frac{1}{t}}} \right)-i\ln \left( {{t}^{2}}+1 \right)}{{{t}^{2}}+1}dt}\\
&=-\frac{\pi \gamma }{2}-\frac{\pi }{2}\log 2+\int\limits_{0}^{+\infty }{\frac{t\arctan \frac{1}{t}}{{{t}^{2}}+1}dt}\\
&=-\frac{\pi \gamma }{2}-\frac{\pi }{2}\log 2+\int\limits_{0}^{+\infty }{\left( \frac{t}{{{t}^{2}}+1}\int\limits_{t}^{+\infty }{\frac{1}{{{u}^{2}}+1}du} \right)dt}\\
&=-\frac{\pi \gamma }{2}-\frac{\pi }{2}\log 2+\int\limits_{1}^{+\infty }{\left( \int\limits_{0}^{+\infty }{\frac{t}{{{t}^{2}}+1}\cdot \frac{t}{{{u}^{2}}{{t}^{2}}+1}dt} \right)du}\\
&=-\frac{\pi \gamma }{2}-\frac{\pi }{2}\log 2+\frac{\pi }{2}\int\limits_{1}^{+\infty }{\frac{du}{u\left( u+1 \right)}}\\
&=-\frac{\pi \gamma }{2}-\frac{\pi }{2}\log 2+\frac{\pi }{2}\log 2=-\frac{\pi \gamma }{2}\end{align}\)

luyuanhong

楼上 elim 的帖子很好！已收藏。