一道反三角函数瑕积分

青青子衿

\begin{align*}
{\large\int}_0^1\frac{\arctan\left(x\right)\arcsin\left(x\right)}{x^2}\mathrm{d}x \\
{\large\int}_0^1\frac{\arctan\left(\frac{x}{2}\right)\arcsin\left(x\right)}{x^2}\mathrm{d}x
\end{align*}

Czhang271828

试了试推广：求积分$\displaystyle{\int_0^1\dfrac{\arctan(kx)\arcsin x}{x^2}\mathrm dx}$
记积分为$I(k)$，则
\begin{align*}
I'(k)=&\int_0^1\dfrac{\arcsin x}{x(1+k^2x^2)}\mathrm dx\\
=&\int_0^1\dfrac{\arcsin x}{x}\mathrm dx-\int_0^1\dfrac{k^2x\arcsin x}{(1+k^2x^2)}\mathrm dx\\
=&\dfrac{\pi\log 2}{2}-\int_0^1\dfrac{k^2x\arcsin x}{(1+k^2x^2)}\mathrm dx\\
=&\dfrac{\pi\log 2}{2}-\dfrac{1}{2}\arcsin x\log(1+k^2x^2)|_0^1+\dfrac{1}{2}\int_0^1\dfrac{\log(1+k^2x^2)}{\sqrt{1-x^2}}\mathrm dx\\
=&\dfrac{\pi\log 2}{2}-\dfrac{\pi\log(1+k^2)}{4}+\dfrac{1}{2}\int_0^1\dfrac{\log(1+k^2x^2)}{\sqrt{1-x^2}}\mathrm dx
\end{align*}
记积分$\displaystyle{J(k)=\dfrac{1}{2}\int_0^1\dfrac{\log(1+k^2x^2)}{\sqrt{1-x^2}}\mathrm dx}$，求导得
\begin{align*}
J'(k)=&\int_0^1\dfrac{x^2k}{(1+k^2x^2)\sqrt{1-x^2}}\mathrm dx\\
=&\int_0^{\pi/2}\dfrac{k\sin^2 t}{1+k^2\sin^2t}\mathrm dt\\
=&\dfrac{\pi}{2k}-\dfrac{1}{k}\int_0^{\pi/2}\dfrac{\mathrm dt}{1+k^2\sin^2t}\\
=&\dfrac{\pi}{2k}-\dfrac{1}{k\sqrt{1+k^2}}\int_0^{\pi/2}\dfrac{\mathrm d\sqrt{1+k^2}\tan{t}}{1+(1+k^2)\tan^2 t}\\
=&\dfrac{\pi}{2}\left(\dfrac{1}{k}-\dfrac{1}{k\sqrt{1+k^2}}\right)
\end{align*}
因此$\int J'(k)=\dfrac{\pi}{2}(\log k+\coth^{-1}\sqrt{1+k^2})+C=\dfrac{\pi}{2}\log(\sqrt{k^2+1}+1)+C$。
再由$J(0)=0$得$J(k)=\dfrac{\pi}{2}(\log(\sqrt{k^2+1}+1)-\log 2)$。因此求得$I'(k)=\dfrac{\pi}{2}\log\dfrac{\sqrt{k^2+1}+1}{\sqrt{k^2+1}}$，故
\[
I(k)=\dfrac{\pi}{2}\left(k\log\dfrac{\sqrt{k^2+1}+1}{\sqrt{k^2+1}}-\arctan(k)+\sinh^{-1}(k)\right)
\]

***
计算得
\begin{align*}
&\int_0^1\dfrac{\arctan x\arcsin x}{x^2}\mathrm dx=-\dfrac{\pi}{8}\left(\pi-4\log\left(2+\dfrac{3}{\sqrt 2}\right)\right)\\
&\int_0^1\dfrac{\arctan (x/2)\arcsin x}{x^2}\mathrm dx=\cdots\\
\end{align*}