|||
$\int^{a}_{0}\sqrt{a^2+x^2}dx=a^2\int^{1}_{0}\sqrt{1+y^2}dy$
(令$y=x/a$)
$a^2\int^{1}_{0}\sqrt{1+y^2}dy=a^2\int^{\frac{\pi}{4}}_{0}\sec^3\theta d\theta$
(令$y=\tan\theta$)
$=\frac{a^2}{2}\left[\sqrt{2}+\ln(\sqrt{2}+1)\right]$
177.
$\int^{\infty}_{0}\frac{x^{p-1}}{1+x^q}dx=\int^{\infty}_{0}\frac{x^{p-q}x^{q-1}}{\sqrt{1+x^q}}dx=\frac{1}{q}\int^{\infty}_{0}\frac{y^{\frac{p}{q}-1}}{\sqrt{1+y}}dy$
(令$y=x^q$)
$\frac{1}{q}\int^{\infty}_{0}\frac{y^{\frac{p}{q}-1}}{\sqrt{1+y}}dy=\frac{1}{q}\int^{\infty}_{1}\frac{(z-1)^{\frac{p}{q}-1}}{z^{\frac{1}{2}}}dz$
(令$z=1+y$)
$\frac{1}{q}\int^{\infty}_{1}\frac{(z-1)^{\frac{p}{q}-1}}{z^{\frac{1}{2}}}dz=\frac{1}{q}\int^{1}_{0}\frac{\left(\frac{1}{t}-1\right)^{\frac{p}{q}-1}}{\left(\frac{1}{t}\right)^{\frac{1}{2}}t^2}dt$
(令$t=1/z$)
$\frac{1}{q}\int^{1}_{0}\frac{\left(\frac{1}{t}-1\right)^{\frac{p}{q}-1}}{\left(\frac{1}{t}\right)^{\frac{1}{2}}t^2}dt=\frac{1}{q}\int^{1}_{0}t^{\frac{1}{2}-\frac{p}{q}-1}(1-t)^{\frac{p}{q}-1}dt=\frac{1}{q}B\left(\frac{p}{q},\frac{1}{2}-\frac{p}{q}\right)$
Archiver|手机版|科学网 ( 京ICP备07017567号-12 )
GMT+8, 2024-5-17 16:30
Powered by ScienceNet.cn
Copyright © 2007- 中国科学报社