b＞0 时，求 (0,+∞) 上的定积分 ∫(0,+∞)xe^(-bx^2)cosh(x)sin(x)dx

anyon

RT，是不是需要用一些复变函数的知识啊，本人目前数分还没上完

yichang

\[\begin{aligned}
&\underbrace{\int_0^\infty}_{\frac12\int_{-\infty}^\infty} xe^{-bx^2}\quad\underbrace{\cosh{x}}_{\displaystyle\frac{e^x+e^{-x}}{2}}\quad\underbrace{\sin{x}}_{\displaystyle\operatorname{Im}(e^{ix})}\,\mathrm{d}x
=\frac14\operatorname{Im}\left[\int_{-\infty}^\infty xe^{(1+i)x-bx^2}\,\mathrm{d}x+\int_{-\infty}^\infty xe^{-(1-i)x-bx^2}\,\mathrm{d}x\right] \\
&=\frac14\operatorname{Im}\left[e^{(1+i)^2/(4b)}\int_{-\infty}^\infty xe^{-b(x-(1+i)/(2b))^2}\,\mathrm{d}x+e^{(1-i)^2/(4b)}\int_{-\infty}^\infty xe^{-b(x+(1-i)/(2b))^2}\,\mathrm{d}x\right] \\
&=\frac14\operatorname{Im}\left[e^{(1+i)^2/(4b)}\int_{-\infty}^\infty \left(t+\frac{1+i}{2b}\right)e^{-bt^2}\,\mathrm{d}t+e^{(1-i)^2/(4b)}\int_{-\infty}^\infty \left(t-\frac{1-i}{2b}\right)e^{-bt^2}\,\mathrm{d}y\right] \\
&=\frac14\sqrt{\frac{\pi}{b}}\operatorname{Im}\left[e^{(1+i)^2/(4b)}\frac{1+i}{2b}-e^{(1-i)^2/(4b)}\frac{1-i}{2b}\right] \\
&=\frac14\sqrt{\frac{\pi}{b^3}}\left(\cos{\frac{1}{2b}}+\sin{\frac{1}{2b}}\right)
\end{aligned}\]

anyon

yichang 发表于 2021-10-30 16:01
\[\begin{aligned}
&%underbrace{\int_0^\infty}_{\frac12\int_{-\infty}^\infty} xe^{-bx^2}\quad%underb ...

非常感谢！

luyuanhong

楼上 yichang 的解答很好！已收藏。