青青子衿

回复 6# dim

回复  战巡
    不是那里，是这步
dim 发表于 2016-5-29 03:25

\[g'(a)=\lim_{\Delta a\to 0}\frac{1}{\Delta a}\left\{\int_0^{a+\Delta a}\frac{\ln\big[1+(a+\Delta a)x\big]}{1+x^2}{\rm d}x-\int_0^a\frac{\ln(1+ax)}{1+x^2}{\rm d}x\right\}\]
\[=\lim_{\Delta a\to 0}\frac{1}{\Delta a}\left\{\int_0^{a+\Delta a}\frac{\ln\big[1+(a+\Delta a)x\big]}{1+x^2}{\rm d}x{\color{blue}{-\int_0^{a}\frac{\ln\big[1+(a+\Delta a)x\big]}{1+x^2}{\rm d}x}}{\color{red}{+\int_0^{a}\frac{\ln\big[1+(a+\Delta a)x\big]}{1+x^2}{\rm d}x}}-\int_0^a\frac{\ln(1+ax)}{1+x^2}{\rm d}x\right\}\]
\[=\lim_{\Delta a\to 0}\frac{1}{\Delta a}\left\{\color{blue}{\int_a^{a+\Delta a}\frac{\ln\big[1+(a+\Delta a)x\big]}{1+x^2}{\rm d}x}{\color{red}{+\int_0^{a}\frac{\ln\big[1+(a+\Delta a)x-\ln(1+ax)\big]}{1+x^2}{\rm d}x}}\right\}\]
利用积分中值定理，可得
\[=\lim_{\Delta a\to 0}\frac{1}{\Delta a}\left\{\color{blue}{\Delta a\cdot\frac{\ln\big[1+(a+\Delta a)\xi\big]}{1+\xi^2}}{\color{red}{+\int_0^{a}\frac{\ln\big[1+(a+\Delta a)x-\ln(1+ax)\big]}{1+x^2}{\rm d}x}}\right\}\]
\[=\color{blue}{\lim_{\Delta a\to 0}\frac{\ln\big[1+(a+\Delta a)\xi\big]}{1+\xi^2}}+{\color{red}{\int_0^{a}\lim_{\Delta a\to 0}\frac{\ln\big[1+(a+\Delta a)x-\ln(1+ax)\big]}{\Delta a}\frac{1}{1+x^2}{\rm d}x}}\]
\[=\color{blue}{\frac{\ln(1+a^2)}{1+a^2}}+{\color{red}{\int _0^a\frac{x}{(1+ax)(1+x^2)}{\rm d}x}}\]

青青子衿

回复  dim
\[g(a)=\int_0^a\frac{\ln(1+ax)}{1+x^2}dx\]
\[g'(a)=\dfrac{\partial g(a)}{\partial a}=\frac{\ln(1+a^2)}{1+a^2}+\int_0^a\frac{x}{(1+ax)(1+x^2)}dx\]
\[=\frac{\ln(1+a^2)}{1+a^2}+\frac{2a\arctan(a)-\ln(1+a^2)}{2(1+a^2)}=\frac{2a\arctan(a)+\ln(1+a^2)}{2(1+a^2)}\]
战巡 发表于 2016-5-28 13:09

\[{\color{red}{\begin{align*}
\int _0^a\frac{x}{(1+ax)(1+x^2)}{\rm d}x&=\int _0^a\frac{1}{1+a^2}\left(\dfrac{x+a}{1+x^2}-\dfrac{a}{1+ax}\right){\rm d}x\\
&=\frac{1}{1+a^2}\int _0^a\left(\dfrac{x}{1+x^2}+\dfrac{a}{1+x^2}-\dfrac{a}{1+ax}\right){\rm d}x\\
&=\frac{1}{1+a^2}\left.\left[\frac{1}{2}\ln(1+x^2)+a\arctan{x}-\ln(1+ax)\right]\right|_0^a\\
&=\frac{1}{1+a^2}\left[a\arctan{a}-\frac{1}{2}\ln(1+a^2)\right]\\
&=\frac{a\arctan{a}}{1+a^2}-\frac{\ln(1+a^2)}{2(1+a^2)}
\end{align*}}}\]
\[{\color{red}{\frac{x}{(1+ax)(1+x^2)}=\frac{{\rm A}x+{\rm B}}{1+x^2}+\frac{{\rm C}}{1+ax}}}\]
\[{\color{red}{\begin{align*}
x
&=({\rm A}x+{\rm B})(1+ax)+{\rm C}(1+x^2)\\
&=({\rm A}a+{\rm C})x^2+({\rm A}+{\rm B}a)x+({\rm B}+{\rm C})
\end{align*}} }\\
\,\\
\,\\
{\color{red}{\Rightarrow
\begin{cases}
{\rm A}a+{\rm C}=0\\
{\rm A}+{\rm B}a=1\\
{\rm B}+{\rm C}=0
\end{cases}\quad\Longrightarrow\quad
\begin{cases}
{\rm A}=\phantom{-}\dfrac{1}{1+a^2}\\
{\rm B}=\phantom{-}\dfrac{a}{1+a^2}\\
{\rm C}=-\dfrac{a}{1+a^2}
\end{cases}
} }\]
另外，可以参考梅加强《数学分析》 P587 例  16.1.4

青青子衿

$\displaystyle \int_0^a \frac{\arctan x}{1+ax} dx=\int_0^{\arctan a} \frac{u\sec^2 u\,du}{1+a\tan u}=\frac{1}{\sqrt{1+a^2}}\int_0^{u_0} \frac{u\sec u\,du}{\cos u_0 \cos u+\sin u_0 \sin u}$ ...
tommywong 发表于 2016-5-28 20:47

$\displaystyle \begin{align*}
\int_0^a \frac{\arctan x}{1+ax}{\rm d}x
\quad&\overset{x\,=\,\tan u}{\overline{\overline{\hspace{2cm}}}}\quad
\int_0^{\arctan a} \frac{u\sec^2 u\,{\rm d}u}{1+a\tan u}\\
&\overset{a\,=\,\tan u_0}{\overline{\overline{\hspace{2cm}}}}\quad\int_0^{u_0} \frac{u\sec^2 u\,{\rm d}u}{1+\dfrac{\sin u_0}{\cos u_0} \dfrac{\sin u}{\cos u}}=\int_0^{u_0} \frac{u\cos u_0\sec u\,{\rm d}u}{\cos u_0 \cos u+\sin u_0 \sin u}\\
&=\cos u_0\int_0^{u_0} \frac{u\sec u\,{\rm d}u}{\cos u_0 \cos u+\sin u_0 \sin u}= \cos u_0 \int_0^{u_0} \frac{u\sec u\,du}{\cos(u_0-u)}\\
&=\cos u_0 \int_0^{u_0} \frac{u\,{\rm d}u}{\cos(u_0-u)\cos u}
\end{align*}$

$\displaystyle \begin{align*}
\int_0^{u_0} \frac{u\,{\rm d}u}{\cos(u_0-u)\cos u}\quad&\overset{u\,=\,u_0-v}{\overline{\overline{\hspace{2cm}}}}\quad\int_{u_0}^0 \frac{(u_0-v)(-{\rm d}v)}{\cos(u_0-v)\cos v}\\
&=\int_0^{u_0} \frac{(u_0-v){\rm d}v}{\cos(u_0-v)\cos v}=\frac{u_0}{2}\int_0^{u_0} \frac{{\rm d}u}{\cos(u_0-u)\cos u}
\end{align*}$

$\displaystyle \begin{align*}
\frac{1}{\cos(u_0-u)\cos u}&=\frac{\sin(u_0-u+u)}{\sin u_0\cos(u_0-u)\cos u}=\frac{\sin(u_0-u)\cos u+\cos(u_0-u)\sin u}{\sin u_0\cos(u_0-u)\cos u}\\
&\\
&=\frac{\tan(u_0-u)+\tan u}{\sin u_0}
\end{align*}$

$\displaystyle \int_0^{u_0} \frac{{\rm d}u}{\cos(u_0-u)\cos u}=\int_0^{u_0} \frac{\tan(u_0-u)+\tan u}{\sin u_0}{\rm d}u=\frac{-2\ln\cos u_0}{\sin u_0}$

$\displaystyle \begin{align*}
\int_0^a \frac{\arctan x}{1+ax} dx&=\cos u_0\cdot\frac{u_0}{2}\cdot\int_0^{u_0} \frac{{\rm d}u}{\cos(u_0-u)\cos u}=\cos u_0 \cdot\frac{u_0}{2}\cdot\frac{-2\ln\cos u_0}{\sin u_0}\\
&=\frac{1}{\tan u_0 }\cdot\frac{u_0}{2}\cdot\ln\left(\dfrac{1}{\cos^2 u_0}\right)=\frac{1}{\tan u_0 }\cdot\frac{u_0}{2}\cdot\ln\left(1+\tan^2u_0\right)\\
&=\frac{1}{a}\cdot\frac{\arctan a}{2}\cdot\ln(1+a^2)
\end{align*}
$

青青子衿

回复 12# 青青子衿
补充几个
\begin{align*}
15.&\quad\int_0^{+\infty}\dfrac{\ln(1+x)}{(ax+b)^2}{\rm\,d}x=\dfrac{\ln\dfrac{a}{b}}{a(a-b)}\qquad\qquad[ab>0]\\
16.&\quad\int_0^1\dfrac{\ln(a+x)}{a+x^2}{\rm\,d}x=\dfrac{\operatorname{arccot}{\sqrt{a}}}{2\sqrt{a\,}}\ln[a(1+a)]\qquad\qquad[a>0]\\
17.&\quad\int_0^{+\infty}\dfrac{\ln(a+x)}{(b+x)^2}{\rm\,d}x=\dfrac{a\ln a-b\ln b}{b(a-b)}\qquad\qquad[a>0,b>0,a\ne b]\\
18.&\quad\int_0^a\dfrac{\ln(1+ax)}{1+x^2}{\rm\,d}x=\dfrac{\arctan a}{2}\ln\left(1+a^2\right)\qquad\qquad[a>0,b>0,a\ne b]\\
19.&\quad\int_0^1\dfrac{\ln(1+ax)}{1+ax^2}{\rm\,d}x=\dfrac{\arctan\sqrt{a}}{2\sqrt{a\,}}\ln\left(1+a\right)\qquad\qquad[a>0]\\
20.&\quad\int_0^1\dfrac{\ln(ax+b)}{(1+x)^2}{\rm\,d}x=\dfrac{\frac{a+b}{2}\ln(a+b)-b\ln b-a\ln2}{a-b}\qquad\qquad[a>0,b>0,a\ne b]\\
\end{align*}