\[\Large\int_{0}^{1}\frac{\arctan x}{\sqrt{1-x^{2}}}\mathrm{d}x \]
\(\Large\mathbf{Solution:}\)
首先第一種做法，含參積分.不多說直接上圖
第二種方法則是利用級數，易知
\[\begin{align*} \int_{0}^{1}\frac{\arctan x}{\sqrt{1-x^{2}}}\mathrm{d}x&=\int_0^{\pi/2}\arctan(\sin(x))\,\mathrm{d}x\\&=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\int_0^{\pi/2}\sin^{2k+1}(x)\,\mathrm{d}x\\ &=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\frac{2^k\,k!}{(2k+1)!!}\\ &=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}\frac{4^k}{\displaystyle\binom{2k}{k}} \end{align*}\]
下面來解決最后一個級數，利用Beta函數我們可以得到以下等式
\[\frac1{\displaystyle\binom{2n}{n}}=(2n+1)\int_0^1t^n(1-t)^n\mathrm{d}t \]
所以
\[\begin{align*} \sum_{n=0}^\infty\frac{(-4)^nx^{2n}}{(2n+1)\displaystyle\binom{2n}{n}} &=\int_0^1\frac1{1+4x^2t(1-t)}\mathrm{d}t\\ &=\int_0^1\frac1{1+x^2-x^2(2t-1)^2}\mathrm{d}t\\ &=\frac1{1+x^2}\int_0^1\frac1{1-\dfrac{x^2}{1+x^2}(2t-1)^2}\mathrm{d}t\\ &=\frac1{1+x^2}\int_{-1}^1\frac1{1-\dfrac{x^2}{1+x^2}t^2}\frac12\mathrm{d}t\\ &=\frac1{2x\sqrt{1+x^2}}\int_{-x/\sqrt{1+x^2}}^{x/\sqrt{1+x^2}}\frac1{1-t^2}\mathrm{d}t\\ &=\frac1{x\sqrt{1+x^2}}\mathrm{arctanh}\left(\frac{x}{\sqrt{1+x^2}}\right)\\ &=\frac1{x\sqrt{1+x^2}}\mathrm{arcsinh}(x) \end{align*}\]
兩邊積分可以得到
\[\begin{align*} \sum_{n=0}^\infty\frac{(-4)^n}{(2n+1)^2\displaystyle \binom{2n}{n}} &=\int_0^1\frac1{x\sqrt{1+x^2}}\mathrm{arcsinh}(x)\,\mathrm{d}x\\ &=-\int_0^1\mathrm{arcsinh}(x)\frac1{\sqrt{\vphantom{\big|}1+1/x^2}}\mathrm{d}\frac1x\\ &=-\int_0^1\mathrm{arcsinh}(x)\,\mathrm{d}\,\mathrm{arcsinh}\left(\frac1x\right)\\ &=-\,\mathrm{arcsinh}^2(1)+\int_0^1\mathrm{arcsinh}\left(\frac1x\right)\,\mathrm{d}\,\mathrm{arcsinh}(x)\\ &=-\,\mathrm{arcsinh}^2(1)-\int_1^\infty\mathrm{arcsinh}(x)\,\mathrm{d}\,\mathrm{arcsinh}\left(\frac1x\right)\\ &=-\,\mathrm{arcsinh}^2(1)+\int_1^\infty\frac1{x\sqrt{1+x^2}}\mathrm{arcsinh}(x)\,\mathrm{d}x\\ &=-\frac12\,\mathrm{arcsinh}^2(1)+\frac12\int_0^\infty\frac1{x\sqrt{1+x^2}}\mathrm{arcsinh}(x)\,\mathrm{d}x\\ &=-\frac12\,\mathrm{arcsinh}^2(1)+\frac12\int_0^\infty\frac{t\,\mathrm{d}t}{\sinh(t)} \end{align*}\]
其中
\[\int_0^\infty\frac{t\,\mathrm{dt}}{\sinh(t)}=\int_0^\infty\sum_{k=0}^\infty2t\,e^{-(2k+1)t}\,\mathrm{d}t=\sum_{k=0}^\infty\frac2{(2k+1)^2}=\frac{\pi^2}4 \]
所以
\[\color{red}{\sum_{n=0}^\infty\frac{(-4)^n}{(2n+1)^2\displaystyle \binom{2n}{n}}=\frac{\pi^2}8-\frac12\mathrm{arcsinh}^2(1)} \]
即
\[\Large\boxed{\displaystyle \int_{0}^{1}\frac{\arctan x}{\sqrt{1-x^{2}}}\, \mathrm{d}x=\color{blue}{\frac{\pi^2}8-\frac12\mathrm{arcsinh}^2(1)}} \]