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asin x acos x atan x 的积分处理思路

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积分∫arcsinx arccosx arctanx dx的处理思路

计算这个积分还是挺烧脑的...第一次算这么复杂的积分

过程有些多有些杂，也许不能保证完全正确，但是结构和处理方法应该是没有问题...

如果有误请指出

1. 代换处理，干掉几个反三角​

知道 x∈(−1,1)x\in(-1,1) ，不妨用代换 x=cos⁡tx=\cos t , 可以直接得到 arccos⁡cos⁡t=t\arccos \cos t=t ， arcsin⁡cos⁡t=π2−t\arcsin \cos t=\frac{\pi}{2}-t

你说为什么不用别的三角函数代换，比如 tan⁡\tan ？嗯...你想， arcsin⁡cos⁡x\arcsin\cos x 这种起码消得掉，一消消俩，如果用了正切就会留下 arcsin⁡tan⁡x\arcsin\tan x 和 arccos⁡tan⁡x\arccos\tan x 这俩没法化简的，不顺手

∫arcsin⁡xarccos⁡xarctan⁡xdx=∫arcsin⁡(cos⁡t)arccos⁡(cos⁡t)arctan⁡(cos⁡t)d(cos⁡t)=∫t(t−π2)sin⁡t⋅arctan⁡(cos⁡t)dt\begin{aligned} \int \arcsin x\arccos x\arctan x\mathrm{d}x&=\int\arcsin (\cos t)\arccos( \cos t )\arctan (\cos t)\mathrm{d}(\cos t)\\ &=\int t\left( t-\frac{\pi}{2}\right)\sin t\cdot \arctan(\cos t)\mathrm{d}t \end{aligned}

这个东西可以分解为俩积分的权和

It=∫tsin⁡t⋅arctan⁡(cos⁡t)dtIt2=∫t2sin⁡t⋅arctan⁡(cos⁡t)dtI=It2−π2It\begin{aligned} I_{t}&=\int t\sin t\cdot \arctan(\cos t)\mathrm{d}t\\ I_{t^2}&=\int t^2\sin t\cdot \arctan(\cos t)\mathrm{d}t\\ I&=I_{t^2}-\frac{\pi}{2}I_{t} \end{aligned}

从低次开始研究

2. ItI_t 的计算​

先计算次数比较低的 ItI_{t} . 注意到 arctan⁡cos⁡t\arctan \cos t 无法化成初等函数, 而代换法刚才已经用过, 那就对它分部积分, 试图通过求导转化 arctan⁡\arctan .

∫tsin⁡tdt=sin⁡t−tcos⁡t \int t\sin t\mathrm{d}t=\sin t-t\cos t

分部后三角函数是二次的，所以可以化成二倍角

∫tsin⁡t⋅arctan⁡(cos⁡t)dt=(sin⁡t−tcos⁡t)arctan⁡(cos⁡t)−∫(sin⁡t−tcos⁡t)(−sin⁡tcos⁡2t+1)dt=(sin⁡t−tcos⁡t)arctan⁡(cos⁡t)−∫(2tsin⁡(2t)−82cos⁡(2t)+6+1)dt=(sin⁡t−tcos⁡t)arctan⁡(cos⁡t)−t−14∫2tsin⁡(2t)−8cos⁡(2t)+3d(2t)\begin{aligned} \int t\sin t\cdot \arctan (\cos t)\mathrm{dt}&=(\sin t-t\cos t)\arctan(\cos t)-\int(\sin t-t\cos t)\left( -\frac{\sin t}{\cos^2t+1} \right)\mathrm{d}t\\ &=(\sin t-t\cos t)\arctan(\cos t)-\int\left( \frac{2t \sin (2 t)-8}{2\cos (2 t)+6}+1 \right)\mathrm{d}t\\ &=(\sin t-t\cos t)\arctan(\cos t)-t-\frac{1}{4}\int\frac{2t \sin (2 t)-8}{\cos (2 t)+3}\mathrm{d}(2t)\\ \end{aligned}

令 T=2tT=2t , T∈(0,π2)T\in\left( 0,\frac{\pi}{2}\right) , 注意到常数能分离成以 cos⁡T\cos T 为分母的积分, 通过万能代换即可求出, 这里省略了过程

∫Tsin⁡T−8cos⁡T+3dT=∫Tsin⁡Tcos⁡T+3dT−∫8cos⁡T+3dT=∫Tsin⁡Tcos⁡T+3dT−42arctan⁡(tan⁡T22)\begin{aligned} \int \frac{T\sin T-8}{\cos T+3}\mathrm{d}T&=\int \frac{T\sin T}{\cos T+3}\mathrm{d}T-\int \frac{8}{\cos T+3}\mathrm{d}T\\ &=\int \frac{T\sin T}{\cos T+3}\mathrm{d}T-4\sqrt{2}\arctan\left( \frac{\tan \frac{T}{2}}{\sqrt{2}} \right) \end{aligned}

左侧积分看起来与一个有趣的特殊积分 ∫0π2xsin⁡x1+cos⁡2x\int_{0}^{\frac{\pi}{2}} \frac{x\sin x}{1+\cos^2x} 是近亲，所以我搜索了一下资料，找到了一些资料

xsinx/(1+cos²x)的不定积分怎么求？

求完积分之后我才发现里面有一些现成结论，当时应该看得仔细一点（雾）

左侧积分有一个明显的因子 TT , 所以我们尝试分部积分一下, 看看是否有好的结构出现

∫Tsin⁡Tcos⁡T+3dT=∫T⋅sin⁡Tcos⁡T+3dT=T∫sin⁡Tcos⁡T+3dT−∫1(∫sin⁡Tcos⁡T+3dT)dT=−Tlog⁡(cos⁡T+3)+∫log⁡(cos⁡T+3)dT\begin{aligned} \int \frac{T\sin T}{\cos T+3}\mathrm{d}T&=\int T\cdot \frac{\sin T}{\cos T+3}\mathrm{d}T\\ &=T\int \frac{\sin T}{\cos T+3}\mathrm{d}T-\int 1\left( \int \frac{\sin T}{\cos T+3}\mathrm{d}T \right)\mathrm{d}T\\ &=-T\log(\cos T+3)+\int \log(\cos T+3)\mathrm{d}T \end{aligned}

熟悉的味道? 右侧积分看着就是特殊积分 ∫log⁡cos⁡xdx\int \log \cos x\mathrm{d}x 的变体, 似乎叫 LL 函数 (From 《积分的方法与技巧》)

由于两个不同类型的函数复合不方便处理, 回退到上一步. 现在, 分部积分和代换法都用过了, 如何处理这个三角和非三角混合的积分?

想了许久，想起来机器积分结果里出现了一堆复数，想是把 sin⁡\sin 和 cos⁡\cos 用复数展开，在复数域 C\mathbb{C} 上处理这个积分.

cos⁡x=eix+e−ix2sin⁡x=eix−e−ix2i\begin{aligned} \cos x&=\frac{\mathrm{e}^{\mathrm{i}x}+\mathrm{e}^{-\mathrm{i}x}}{2}\\ \sin x&=\frac{\mathrm{e}^{\mathrm{i}x}-\mathrm{e}^{-\mathrm{i}x}}{2\mathrm{i}}\\ \end{aligned}

令 u=eiTu=\mathrm{e}^{\mathrm{i}T} , T=log⁡ui=−ilog⁡uT=\frac{\log u}{\mathrm{i}}=-\mathrm{i}\log u , 接下来发现作为因子的分数可以有理分解

∫Tsin⁡Tcos⁡T+3dT=∫T⋅eiT−e−iT2ieiT+e−iT2+3dT=−∫iT(e2iT−1)6eiT+e2iT+1dT=−∫log⁡u⋅(u2−1)u2+6u+1d(−ilog⁡u)=i∫u2−1u((u+3)2−8)⋅log⁡udu=i∫((u+3)2−8)−6u+2u((u+3)2−8)⋅log⁡udu=i∫(−log⁡uu+2(3+u)log⁡uu2+6u+1)du=−i2log⁡2u+2i∫3+uu2+6u+1⋅log⁡udu\begin{aligned} \int \frac{T\sin T}{\cos T+3}\mathrm{d}T&=\int \frac{T\cdot\frac{\mathrm{e}^{\mathrm{i}T}-\mathrm{e}^{-\mathrm{i}T}}{2\mathrm{i}}}{\frac{\mathrm{e}^{\mathrm{i}T}+\mathrm{e}^{-\mathrm{i}T}}{2}+3}\mathrm{d}T\\ &=-\int\frac{\mathrm{i}T \left(\mathrm{e}^{2 \mathrm{i} T}-1\right)}{6 \mathrm{e}^{\mathrm{i} T}+\mathrm{e}^{2 i T}+1}\mathrm{d}T\\ &=-\int \frac{\log u\cdot(u^2-1)}{u^2+6u+1}\mathrm{d}(-\mathrm{i}\log u)\\ &=\mathrm{i}\int \frac{u^2-1}{u((u+3)^2-8)}\cdot \log u\mathrm{d}u\\ &=\mathrm{i}\int \frac{((u+3)^2-8)-6u+2}{u((u+3)^2-8)}\cdot \log u\mathrm{d}u\\ &=\mathrm{i}\int\left( -\frac{\log u}{u}+\frac{2(3+u)\log u}{u^2+6u+1} \right)\mathrm{d}u\\ &=-\frac{\mathrm{i}}{2}\log^2 u+2\mathrm{i}\int \frac{3+u}{u^2+6u+1}\cdot \log u\mathrm{d}u\\ \end{aligned}

非常惊喜，可以发现 (u2+6u+1)′=2(u+3)(u^2+6u+1)'=2(u+3) 正好是分子，那么右侧积分可以凑微分进行分部积分：

∫u+3u2+6u+1du=12∫1u2+6u+1d(u2+6u+1)=12log⁡(u2+6u+1)\begin{aligned} \int \frac{u+3}{u^2+6u+1}\mathrm{d}u&=\frac{1}{2}\int \frac{1}{u^2+6u+1}\mathrm{d}(u^2+6u+1)\\ &=\frac{1}{2}\log(u^2+6u+1) \end{aligned}

∫3+uu2+6u+1log⁡udu=12log⁡(u2+6u+1)log⁡u−12∫log⁡(u2+6u+1)udu\begin{aligned} \int \frac{3+u}{u^2+6u+1}\log u\mathrm{d}u&=\frac{1}{2}\log(u^2+6u+1)\log u-\frac{1}{2}\int \frac{\log(u^2+6u+1)}{u}\mathrm{d}u \end{aligned}

上面可以因式分解. 因为形式上已经很接近特殊积分了, 这里我们正式引入多对数函数 Lin(x)\mathrm{Li}_{n}(x) . Lin(x)\mathrm{Li}_{n}(x) 表示n重对数, 特别地, 其中

Li2(x)=∫−log⁡(1−x)xdx\begin{aligned} \mathrm{Li}_{2}(x)&=\int-\frac{\log(1-x)}{x}\mathrm{d}x \end{aligned}

这样，对于原式右侧积分我们就可以因式分解后凑一凑

∫log⁡(u2+6u+1)udu=∫log⁡(u+3+22)+log⁡(u+3−22)udu=∫log⁡(u+3+22)udu+∫log⁡(u+3−22)udu=∫log⁡(u3+22+1)+log⁡(3+22)udu+∫log⁡(u3−22+1)+log⁡(3−22)udu=∫log⁡(1+u3+22)u3+22d(u3+22)+log⁡(3+22)∫duu+∫log⁡(1+u3−22)u3−22d(u3−22)+log⁡(3−22)∫duu=log⁡ulog⁡((3+22)(3−22))−Li2(−(3+22)u)−Li2(−(3−22)u)=−Li2(−(3+22)u)−Li2(−(3−22)u)\begin{aligned} &\int \frac{\log(u^2+6u+1)}{u}\mathrm{d}u\\ &=\int \frac{\log(u+3+2\sqrt{2})+\log(u+3-2\sqrt{2})}{u}\mathrm{d}u\\ &=\int \frac{\log(u+3+2\sqrt{2})}{u}\mathrm{d}u+\int \frac{\log(u+3-2\sqrt{2})}{u}\mathrm{d}u\\ &=\int \frac{\log\left( \frac{u}{3+2\sqrt{2}}+1 \right)+\log(3+2\sqrt{2})}{u}\mathrm{d}u+\int \frac{\log\left( \frac{u}{3-2\sqrt{2}}+1 \right)+\log(3-2\sqrt{2})}{u}\mathrm{d}u\\ &=\int \frac{\log\left( 1+\frac{u}{3+2\sqrt{2}} \right)}{\frac{u}{3+2\sqrt{2}}}\mathrm{d}\left( \frac{u}{3+2\sqrt{2}} \right)+\log(3+2\sqrt{2})\int \frac{\mathrm{d}u}{u}+\int \frac{\log\left( 1+\frac{u}{3-2\sqrt{2}} \right)}{\frac{u}{3-2\sqrt{2}}}\mathrm{d}\left( \frac{u}{3-2\sqrt{2}} \right)+\log(3-2\sqrt{2})\int \frac{\mathrm{d}u}{u}\\ &=\log u\log((3+2\sqrt{2})(3-2\sqrt{2}))-\mathrm{Li}_{2}(-(3+2\sqrt{2})u)-\mathrm{Li}_{2}(-(3-2\sqrt{2})u)\\ &=-\mathrm{Li}_{2}(-(3+2\sqrt{2})u)-\mathrm{Li}_{2}(-(3-2\sqrt{2})u) \end{aligned}

其实很超标, 竟然化出来了. 所以我们有:

∫3+uu2+6u+1log⁡udu=12log⁡ulog⁡(u2+6u+1)−12(−Li2(−(3+22)u)−Li2(−(3−22)u))=12log⁡ulog⁡(u2+6u+1)+12Li2(−(3+22)u)+12Li2(−(3−22)u)\begin{aligned} \int \frac{3+u}{u^2+6u+1}\log u\mathrm{d}u&=\frac{1}{2}\log u\log(u^2+6u+1)-\frac{1}{2}\left( -\mathrm{Li}_{2}(-(3+2\sqrt{2})u)-\mathrm{Li}_{2}(-(3-2\sqrt{2})u) \right) \\ &=\frac{1}{2}\log u\log(u^2+6u+1)+\frac{1}{2}\mathrm{Li}_{2}(-(3+2\sqrt{2})u)+\frac{1}{2}\mathrm{Li}_{2}(-(3-2\sqrt{2})u) \end{aligned}

到这里积分 ItI_{t} 就被解决了.

3. It2I_{t^2} 的计算​

It2=∫t2sin⁡t⋅arctan⁡(cos⁡t)dt\begin{aligned} I_{t^2}=\int t^2\sin t\cdot \arctan (\cos t)\mathrm{dt} \end{aligned}

可能有人疑惑为啥一开始不就把 t2−π2tt^2-\frac{\pi}{2}t 作为一个整体进行分部积分, 而要分成 ItI_{t} 和 It2I_{t^2} , 其实先用低次的较简单积分试试水，有助于理清思路.

t2t^2 看着有点难受，照猫画虎一通处理掉 arctan⁡\arctan .

∫t2sin⁡tdt=2tsin⁡t+(2−t2)cos⁡t\begin{aligned} \int t^2\sin t\mathrm{d}t&=2t\sin t+(2-t^2)\cos t \end{aligned}

∫t2sin⁡t⋅arctan⁡(cos⁡t)dt=(2tsin⁡t+(2−t2)cos⁡t)arctan⁡(cos⁡t)−∫(2tsin⁡t+(2−t2)cos⁡t)(−sin⁡tcos⁡2t+1)dt\begin{aligned} \int t^2\sin t\cdot \arctan (\cos t)\mathrm{dt}&=(2t\sin t+(2-t^2)\cos t)\arctan(\cos t)-\int(2t\sin t+(2-t^2)\cos t)\left( -\frac{\sin t}{\cos^2t+1} \right)\mathrm{d}t\\ \end{aligned}

看到了二次式, 凑一凑二倍角吧（这里偷懒，用软件化的）

∫(2tsin⁡t+(2−t2)cos⁡t)(−sin⁡t1+cos⁡2t)dt=∫t2sin⁡2t−2t−2sin⁡2t+2tcos⁡2t2cos⁡2t+6d(2t)\begin{aligned} &\int(2t\sin t+(2-t^2)\cos t)\left( -\frac{\sin t}{1+\cos^2t} \right)\mathrm{d}t\\ =&\int\frac{t^2 \sin 2 t-2 t-2 \sin 2 t+2 t \cos 2 t}{2\cos 2 t+6}\mathrm{d}(2t)\\ \end{aligned}

有点无计可施，只能复数展开观察情况

∫(T2−8)sin⁡T+4Tcos⁡T−4T8cos⁡T+24dT=∫(T2−8)⋅eiT−e−iT2i+4T⋅eiT+e−iT2−4T8eiT+e−iT2+24dT=∫−i(T2−8)(eiT−e−iT)+4T(eiT+e−iT)−8T8(eiT+e−iT)+48dT=∫−8i+4T−8TeiT+iT2+(8i+4T−iT2)e2iT8(1+6eit+e2iT)dT\begin{aligned} &\int \frac{(T^2-8)\sin T+4T\cos T-4T}{8\cos T+24}\mathrm{d}T\\ =&\int \frac{(T^2-8)\cdot\frac{\mathrm{e}^{\mathrm{i}T}-\mathrm{e}^{-\mathrm{i}T}}{2\mathrm{i}}+4T\cdot\frac{\mathrm{e}^{\mathrm{i}T}+\mathrm{e}^{-\mathrm{i}T}}{2}-4T}{8\frac{\mathrm{e}^{\mathrm{i}T}+\mathrm{e}^{-\mathrm{i}T}}{2}+24}\mathrm{d}T\\ =&\int \frac{-\mathrm{i}(T^2-8)(\mathrm{e}^{\mathrm{i}T}-\mathrm{e}^{-\mathrm{i}T})+4T(\mathrm{e}^{\mathrm{i}T}+\mathrm{e}^{-\mathrm{i}T})-8T}{8(\mathrm{e}^{\mathrm{i}T}+\mathrm{e}^{-\mathrm{i}T})+48}\mathrm{d}T\\ =&\int \frac{-8\mathrm{i}+4T-8T\mathrm{e}^{\mathrm{i}T}+\mathrm{i}T^2+(8\mathrm{i}+4T-\mathrm{i}T^2)\mathrm{e}^{2\mathrm{i}T}}{8(1+6\mathrm{e}^{\mathrm{i}t}+\mathrm{e}^{2\mathrm{i}T})}\mathrm{d}T\\ \end{aligned}

顶不住了，实在不想化简，用软件辅助一下，换元 u=eiTu=\mathrm{e}^{\mathrm{i}T} , T=−ilog⁡uT=-\mathrm{i}\log u

=∫u2(ilog⁡2u−4ilog⁡u+8i)−ilog⁡2u+8iulog⁡u−4ilog⁡u−8i8(u2+6u+1)du=i8∫(u−1)(8(u+1)+(u+1)log⁡2u−4(u−1)log⁡u)u2+6u+1du\begin{aligned} &=\int\frac{u^2 \left(\mathrm{i} \log ^2u-4 \mathrm{i} \log u+8 \mathrm{i}\right)-\mathrm{i} \log ^2u+8 \mathrm{i} u \log u-4 \mathrm{i} \log u-8 \mathrm{i}}{8 \left(u^2+6 u+1\right)}\mathrm{d}u\\ &=\frac{\mathrm{i}}{8}\int\frac{(u-1) \left(8 (u+1)+(u+1) \log ^2u-4 (u-1) \log u\right)}{u^2+6 u+1}\mathrm{d}u \end{aligned}

终于有一点可以分解的意思了，观察这个式子, 最显著的特征是 log⁡0\log^0 , log⁡1\log^1 , log⁡2\log^2 , 既然是多项式, 便可以尝试按照这种方法对右侧积分分组，逐个击破.

∫(u−1)(8(u+1)+(u+1)log⁡2u−4(u−1)log⁡u)u2+6u+1du=∫8(u+1)(u−1)u2+6u+1du+∫(u+1)(u−1)log⁡2uu2+6u+1du−∫4(u−1)2log⁡uu2+6u+1du\begin{aligned} &\int\frac{(u-1) \left(8 (u+1)+(u+1) \log ^2u-4 (u-1) \log u\right)}{u^2+6 u+1}\mathrm{d}u\\ =&\int \frac{8(u+1)(u-1)}{u^2+6u+1}\mathrm{d}u+\int \frac{(u+1)(u-1)\log^2u}{u^2+6u+1}\mathrm{d}u-\int \frac{4(u-1)^2\log u}{u^2+6u+1}\mathrm{d}u \end{aligned}

4. 击破 log⁡0\log^0​

对于 log⁡0\log^0 ，可以用三角换元简单积出

∫8(u2−1)u2+6u+1du=∫(u+3)2−8−6(u+3)+16(u+3)2−8du=∫du−3∫d(u+3)2(u+3)2−8+16∫1(u+3)2−8d(u+3)=u−3log⁡((u+3)2−8)+22log⁡22−(u+3)22+(u+3)\begin{aligned} \int \frac{8(u^2-1)}{u^2+6u+1}\mathrm{d}u&=\int \frac{(u+3)^2-8-6(u+3)+16}{(u+3)^2-8}\mathrm{d}u\\ &=\int \mathrm{d}u-3\int \frac{\mathrm{d}(u+3)^2}{(u+3)^2-8}+16\int \frac{1}{(u+3)^2-8}\mathrm{d}(u+3)\\ &=u-3\log((u+3)^2-8)+2\sqrt{2}\log \frac{2\sqrt{2}-(u+3)}{2\sqrt{2}+(u+3)} \end{aligned}

5. 击破 log⁡2\log^2​

中间的 log⁡2\log^2 算是魔王积分，式子比较复杂，况且我们的工具本来也不多，只有有理分解能简化一些计算量.

∫(u2−1)log⁡2uu2+6u+1du=∫[(u+3)2−8−6(u+3)+16]log⁡2u(u+3)2−8du=∫log⁡2udu+∫(16−6u−18)log⁡2u(u+3)2−8du=2x−2xlog⁡x+xlog⁡2x−2∫(3u+1)log⁡2uu2+6u+1du\begin{aligned} \int \frac{(u^2-1)\log^2u}{u^2+6u+1}\mathrm{d}u&=\int \frac{[(u+3)^2-8-6(u+3)+16]\log^2u}{(u+3)^2-8}\mathrm{d}u\\ &=\int \log^2u\mathrm{d}u+\int \frac{(16-6u-18)\log^2u}{(u+3)^2-8}\mathrm{d}u\\ &=2 x-2 x \log x+x \log ^2x-2\int \frac{(3u+1)\log^2u}{u^2+6u+1}\mathrm{d}u \end{aligned}

化掉了一个二次, 还算有些胜利的曙光.

死胡同一例
注意右边的积分式，上面是什么？woc, 这我们做过类似的啊！看原算式我们知道
∫3+uu2+6u+1log⁡udu=12log⁡(u2+6u+1)log⁡u−12∫log⁡(u2+6u+1)udu=12log⁡(u2+6u+1)log⁡u−12(log⁡ulog⁡(u2+6u+1)−Li2(−(3+22)u)−Li2(−(3−22)u)=Li2(−(3+22)u)+Li2(−(3−22)u)\begin{aligned} \int \frac{3+u}{u^2+6u+1}\log u\mathrm{d}u&=\frac{1}{2}\log(u^2+6u+1)\log u-\frac{1}{2}\int \frac{\log(u^2+6u+1)}{u}\mathrm{d}u\\ &=\frac{1}{2}\log(u^2+6u+1)\log u-\frac{1}{2}\left(\log u\log(u^2+6u+1)-\mathrm{Li}_{2}(-(3+2\sqrt{2})u)-\mathrm{Li}_{2}(-(3-2\sqrt{2})u\right)\\ &=\mathrm{Li}_{2}(-(3+2\sqrt{2})u)+\mathrm{Li}_{2}(-(3-2\sqrt{2})u) \end{aligned}
嘿, 不看不知道, 一看就想笑, 您瞧瞧这不就剩了一个 Li2\mathrm{Li}_{2} 函数？而且我们知道多重对数积分毕竟有对数俩字, 能处理对数积分的方法自然也能处理多重对数:
∫Li2(x)dx=∫1⋅Li2(x)dx=xLi2(x)−∫x⋅(−log⁡(1−x)x)dx=xLi2(x)+∫log⁡(1−x)dx=xLi2(x)+(x−1)log⁡(1−x)−x\begin{aligned} \int \mathrm{Li}_{2}(x)\mathrm{d}x&=\int 1\cdot\mathrm{Li}_{2}(x)\mathrm{d}x\\ &=x\mathrm{Li}_{2}(x)-\int x\cdot\left( -\frac{\log (1-x)}{x} \right)\mathrm{d}x\\ &=x\mathrm{Li}_{2}(x)+\int \log(1-x)\mathrm{d}x\\ &=x\mathrm{Li}_{2}(x)+(x-1) \log (1-x)-x \end{aligned}
∫(Li2(−(3+22)u)+Li2(−(3−22)u))du=∫Li2(−(3+22)u)−(3+22)d(−(3+22)u)+∫Li2(−(3−22)u)−(3−22)d(−(3−22)u)=uLi2(−(3−2)u)+uLi2(−(3+22)u)−2u+(u−22+3)log⁡(22u+3u+1)+(u+22+3)log⁡(−22u+3u+1)\begin{aligned} \int \left( \mathrm{Li}_{2}(-(3+2\sqrt{2})u)+\mathrm{Li}_{2}(-(3-2\sqrt{2})u) \right)\mathrm{d}u&=\int\frac{\mathrm{Li}_{2}(-(3+2\sqrt{2})u)}{-(3+2\sqrt{2})}\mathrm{d}(-(3+2\sqrt{2})u)+\int\frac{\mathrm{Li}_{2}(-(3-2\sqrt{2})u)}{-(3-2\sqrt{2})}\mathrm{d}(-(3-2\sqrt{2})u)\\ &=u \text{Li}_2\left(-(3-\sqrt{2}) u\right)+u\text{Li}_2\left(-\left(3+2 \sqrt{2}\right) u\right)-2 u\\ &\quad+\left(u-2 \sqrt{2}+3\right) \log \left(2 \sqrt{2} u+3 u+1\right)+\left(u+2 \sqrt{2}+3\right) \log \left(-2 \sqrt{2} u+3 u+1\right) \end{aligned}
(看得我都不想做了)

这里又卡住了很久，想了分部积分的办法，然而积上去虽说可以却无比复杂，不像人干的. 有些灰心，但是用软件处理了一下积分式，发现了一个非常重要的变换：

(3u+1)log⁡2uu2+6u+1=(22+3)log⁡2u2u+42+6+(3−22)log⁡2u2u−42+6\begin{aligned} \frac{(3 u+1) \log ^2u}{u^2+6 u+1}&=\frac{\left(2 \sqrt{2}+3\right) \log ^2u}{2 u+4 \sqrt{2}+6}+\frac{\left(3-2 \sqrt{2}\right) \log ^2u}{2 u-4 \sqrt{2}+6} \end{aligned}

哥们被干懵了，这还能分...果然还是要养成分解式子的思维直觉（悲）. 所以说一早这样做多好，还弄些七奇八怪的

∫(3u+1)log⁡2uu2+6u+1du=(22+32)∫log⁡2uu+22+3du+(3−222)∫log⁡2uu−22+3du\begin{aligned} \int\frac{(3 u+1) \log ^2u}{u^2+6 u+1}\mathrm{d}u&=\left(\frac{2 \sqrt{2}+3}{2}\right) \int\frac{\log ^2u}{ u+2 \sqrt{2}+3}\mathrm{d}u+\left(\frac{3-2 \sqrt{2}}{2}\right)\int\frac{ \log ^2u}{u-2 \sqrt{2}+3}\mathrm{d}u\\ \end{aligned}

这个会积了, 是吧? 研究两侧积分的一般形式

∫log⁡2uu+cdu=∫log⁡uu+c⋅log⁡udu=∫log⁡uu+cdu⋅log⁡u−∫(∫log⁡uu+cdu)1udu\begin{aligned} \int \frac{\log^2u}{u+c}\mathrm{d}u&=\int \frac{\log u}{u+c}\cdot \log u\mathrm{d}u\\ &=\int \frac{\log u}{u+c}\mathrm{d}u\cdot \log u-\int \left( \int \frac{\log u}{u+c}\mathrm{d}u \right) \frac{1}{u}\mathrm{d}u\\ \end{aligned}

u+cu+c 那里可以分部积分，对吧

∫log⁡uu+cdu=log⁡(u+c)log⁡u−∫log⁡(u+c)udu=log⁡(u+c)log⁡u−∫log⁡(u+cc)+log⁡cudu=log⁡(u+c)log⁡u−log⁡c⋅∫duu+∫−log⁡(1−(−uc))−ucd(−uc)=log⁡(u+c)log⁡u−log⁡clog⁡u+Li2(−uc)=log⁡ulog⁡(uc+1)+Li2(−uc)\begin{aligned} \int \frac{\log u}{u+c}\mathrm{d}u&=\log(u+c)\log u-\int \frac{\log(u+c)}{u}\mathrm{d}u\\ &=\log(u+c)\log u-\int \frac{\log\left( \frac{u+c}{c} \right)+\log c}{u}\mathrm{d}u\\ &=\log(u+c)\log u-\log c\cdot \int \frac{\mathrm{d}u}{u}+\int -\frac{\log\left( 1-\left(- \frac{u}{c} \right) \right)}{-\frac{u}{c}}\mathrm{d}\left( -\frac{u}{c} \right)\\ &=\log(u+c)\log u-\log c\log u+\mathrm{Li}_{2}\left( -\frac{u}{c} \right)\\ &=\log u\log\left( \frac{u}{c}+1 \right)+\mathrm{Li}_{2}\left( -\frac{u}{c} \right) \end{aligned}

回到原式

∫log⁡2uu+cdu=log⁡2ulog⁡(uc+1)+Li2(−uc)log⁡u−∫log⁡ulog⁡(uc+1)+Li2(−uc)udu=log⁡2ulog⁡(uc+1)+Li2(−uc)log⁡u−∫log⁡ulog⁡(uc+1)udu−∫Li2(−uc)udu\begin{aligned} \int \frac{\log^2u}{u+c}\mathrm{d}u&=\log^2 u\log\left( \frac{u}{c}+1 \right)+\mathrm{Li}_{2}\left(- \frac{u}{c} \right)\log u-\int \frac{\log u\log\left( \frac{u}{c}+1 \right)+\mathrm{Li}_{2}\left(- \frac{u}{c} \right)}{u}\mathrm{d}u\\ &=\log^2 u\log\left( \frac{u}{c}+1 \right)+\mathrm{Li}_{2}\left( -\frac{u}{c} \right)\log u-\int \frac{\log u\log\left( \frac{u}{c}+1 \right)}{u}\mathrm{d}u-\int\frac{\mathrm{Li}_{2}\left( -\frac{u}{c} \right)}{u}\mathrm{d}u\\ \end{aligned}

为什么要把一个特殊函数丢到右边的积分里? 因为这就是 Li3\mathrm{Li}_{3} 的定义呀~

( Li3\mathrm{Li}_{3} : 你好.)

Li3(x)=∫Li2(x)xdx\begin{aligned} \mathrm{Li}_{3}(x)&=\int \frac{\mathrm{Li}_{2}(x)}{x}\mathrm{d}x \end{aligned}

∫Li2(−uc)udu=∫Li2(−uc)−ucd(−uc)=Li3(−uc)\begin{aligned} \int \frac{\mathrm{Li}_{2}\left( -\frac{u}{c} \right)}{u}\mathrm{d}u&=\int \frac{\mathrm{Li}_{2}\left( -\frac{u}{c} \right)}{-\frac{u}{c}}\mathrm{d}\left(- \frac{u}{c} \right)\\ &=\mathrm{Li}_{3}\left( -\frac{u}{c} \right) \end{aligned}

左边的积分还可以把 log⁡u\log u 拿出来分部积分

∫log⁡(uc+1)u⋅log⁡udu=−Li2(−uc)log⁡u+∫Li2(−uc)udu=−Li2(−uc)log⁡u+∫Li2(−uc)−ucd(−uc)=−Li2(−uc)log⁡u+Li3(−uc)\begin{aligned} \int \frac{\log\left( \frac{u}{c}+1 \right)}{u}\cdot\log u\mathrm{d}u&=-\mathrm{Li}_{2}\left( -\frac{u}{c} \right)\log u+\int \frac{\mathrm{Li}_{2}\left( -\frac{u}{c} \right)}{u}\mathrm{d}u\\ &=-\mathrm{Li}_{2}\left( -\frac{u}{c} \right)\log u+\int \frac{\mathrm{Li}_{2}\left( -\frac{u}{c} \right)}{-\frac{u}{c}}\mathrm{d}\left( -\frac{u}{c} \right)\\ &=-\mathrm{Li}_{2}\left( -\frac{u}{c} \right)\log u+\mathrm{Li}_{3}\left( -\frac{u}{c} \right) \end{aligned}

那么最终结果即

∫log⁡2uu+cdu=log⁡(uc+1)log⁡2u+Li2(−uc)log⁡u−(−Li2(−uc)log⁡u+Li3(−uc))−Li3(−uc)=log⁡(uc+1)log⁡2u+2Li2(−uc)log⁡u−2Li3(−uc)\begin{aligned} \int \frac{\log^2u}{u+c}\mathrm{d}u&=\log\left( \frac{u}{c}+1 \right)\log^2u+\mathrm{Li}_{2}\left( -\frac{u}{c} \right)\log u-\left( -\mathrm{Li}_{2}\left( -\frac{u}{c} \right)\log u+\mathrm{Li}_{3}\left( -\frac{u}{c} \right) \right) -\mathrm{Li}_{3}\left( -\frac{u}{c} \right)\\ &=\log\left( \frac{u}{c}+1 \right)\log^2u+2\mathrm{Li}_{2}\left( -\frac{u}{c} \right) \log u-2\mathrm{Li}_{3}\left( -\frac{u}{c} \right) \end{aligned}

∫log⁡2uu+3+22du=log⁡(u3+22+1)log⁡2u+2Li2(−u3+22)log⁡u−2Li3(−u3+22)\begin{aligned} \int \frac{\log^2u}{u+3+2\sqrt{2}}\mathrm{d}u&=\log\left( \frac{u}{3+2\sqrt{2}}+1 \right)\log^2u+2\mathrm{Li}_{2}\left( -\frac{u}{3+2\sqrt{2}} \right) \log u-2\mathrm{Li}_{3}\left( -\frac{u}{3+2\sqrt{2}} \right) \end{aligned}

∫log⁡2uu+3−22du=log⁡(u3−22+1)log⁡2u+2Li2(−u3−22)log⁡u−2Li3(−u3−22)\begin{aligned} \int \frac{\log^2u}{u+3-2\sqrt{2}}\mathrm{d}u&=\log\left( \frac{u}{3-2\sqrt{2}}+1 \right)\log^2u+2\mathrm{Li}_{2}\left( -\frac{u}{3-2\sqrt{2}} \right) \log u-2\mathrm{Li}_{3}\left( -\frac{u}{3-2\sqrt{2}} \right) \end{aligned}

好感动...终于终结了最难的那个对数积分. 是该计算最后剩下的 log⁡1\log^1 了.

6. 击破 log⁡1\log^1​

∫4(u−1)2log⁡uu2+6u+1du\begin{aligned} \int \frac{4(u-1)^2\log u}{u^2+6u+1}\mathrm{d}u \end{aligned}

吸取处理 log⁡2\log^2 的经验，先尝试拆碎这个积分

4(u−1)2u2+6u+1=4(u+3)2−8−32(u+3)+72(u+3)2−8=1+72−32(u+3)(u+3)2−8=1−32u+24u2+6u+1\begin{aligned} \frac{4(u-1)^2}{u^2+6u+1}&=\frac{4(u+3)^2-8-32(u+3)+72}{(u+3)^2-8}\\ &=1+\frac{72-32(u+3)}{(u+3)^2-8}\\ &=1-\frac{32u+24}{u^2+6u+1} \end{aligned}

这里的有理分解得列个方程求, 结果有点丑

1−32u+24u2+6u+1=1−(12−52)u−482+72u−22+3−(52−12)uu+22+3\begin{aligned} 1-\frac{32u+24}{u^2+6u+1}&=1-\frac{\left(12-5 \sqrt{2}\right) u-48 \sqrt{2}+72}{u-2 \sqrt{2}+3}-\frac{\left(5 \sqrt{2}-12\right) u}{u+2 \sqrt{2}+3} \end{aligned}

（说实话，总觉得这一步神神鬼鬼的）

原积分转化为

∫4(u−1)2log⁡uu2+6u+1du=∫(1−(12−52)u−482+72u−22+3−(52−12)uu+22+3)log⁡udu=∫log⁡udu−(12−52)∫u−1247(92−16)u−22+3log⁡udu−(52−12)∫uu+22+3log⁡udu\begin{aligned} \int \frac{4(u-1)^2\log u}{u^2+6u+1}\mathrm{d}u&=\int \left( 1-\frac{\left(12-5 \sqrt{2}\right) u-48 \sqrt{2}+72}{u-2 \sqrt{2}+3}-\frac{\left(5 \sqrt{2}-12\right) u}{u+2 \sqrt{2}+3} \right) \log u\mathrm{d}u\\ &=\int \log u\mathrm{d}u-(12-5\sqrt{2})\int \frac{u-\frac{12}{47}\left(9 \sqrt{2}-16\right)}{u-2\sqrt{2}+3}\log u\mathrm{d}u-(5\sqrt{2}-12)\int \frac{u}{u+2\sqrt{2}+3}\log u\mathrm{d}u \end{aligned}

可以看到后面两项实际上形式是统一的，所以实质上我们就是在处理以下积分

∫u+au+blog⁡udu=∫u+b+(a−b)u+blog⁡udu=∫(1+a−bu+b)log⁡udu=∫log⁡udu+(a−b)∫log⁡uu+bdu=ulog⁡u−u+(a−b)∫log⁡uu+bdu\begin{aligned} \int \frac{u+a}{u+b}\log u\mathrm{d}u&=\int \frac{u+b+(a-b)}{u+b}\log u\mathrm{d}u\\ &=\int\left( 1+\frac{a-b}{u+b} \right)\log u\mathrm{d}u\\ &=\int \log u\mathrm{d}u+(a-b)\int \frac{\log u}{u+b}\mathrm{d}u\\ &=u\log u-u+(a-b)\int \frac{\log u}{u+b}\mathrm{d}u\\ \end{aligned}

右边那一项是老朋友了，这里直接给出结果

∫log⁡uu+bdu=log⁡ulog⁡(ub+1)+Li2(−ub)\begin{aligned} \int \frac{\log u}{u+b}\mathrm{d}u&=\log u\log\left( \frac{u}{b}+1 \right)+\mathrm{Li}_{2}\left( -\frac{u}{b} \right) \end{aligned}

于是有

∫u+au+blog⁡udu=ulog⁡u−u+(a−b)log⁡ulog⁡(ub+1)+(a−b)Li2(−ub)\begin{aligned} \int \frac{u+a}{u+b}\log u\mathrm{d}u&=u\log u-u+(a-b)\log u\log\left( \frac{u}{b}+1 \right)+(a-b)\mathrm{Li}_{2}\left( -\frac{u}{b} \right) \end{aligned}

∫u−1247(92−16)u−22+3log⁡udu=ulog⁡u−u+147(51−142)log⁡ulog⁡(u3−22+1)+147(51−142)Li2(−u3−22)\begin{aligned} \int \frac{u-\frac{12}{47}\left(9 \sqrt{2}-16\right)}{u-2\sqrt{2}+3}\log u\mathrm{d}u&=u\log u-u+\frac{1}{47} \left(51-14 \sqrt{2}\right)\log u\log\left( \frac{u}{3-2 \sqrt{2}}+1 \right)\\&\quad+\frac{1}{47} \left(51-14 \sqrt{2}\right)\mathrm{Li}_{2}\left( -\frac{u}{3-2 \sqrt{2}} \right) \end{aligned}

∫uu+22+3log⁡udu=ulog⁡u−u−(22+3)log⁡ulog⁡(u22+3+1)−(22+3)Li2(−u22+3)\begin{aligned} \int \frac{u}{u+2\sqrt{2}+3}\log u\mathrm{d}u&=u\log u-u-(2\sqrt{2}+3)\log u\log\left( \frac{u}{2\sqrt{2}+3}+1 \right)\\ &\quad-(2\sqrt{2}+3)\mathrm{Li}_{2}\left( -\frac{u}{2\sqrt{2}+3} \right) \end{aligned}

至此，所有的积分都已经求完！

到这个时候我已经在桌子前求了大半天的积分。是时候把它们代回到原式了...

7. 代回原式​

算力有限，化简是不可能的...

先是 log⁡1\log^1 积分的整合

∫4(u−1)2log⁡uu2+6u+1du=  ulog⁡u−u−(12−52)(ulog⁡u−u+147(51−142)log⁡ulog⁡(u3−22+1)+147(51−142)Li2(−u3−22))−(52−12)(ulog⁡u−u−(22+3)log⁡ulog⁡(u22+3+1)−(22+3)Li2(−u22+3))\begin{aligned} \int \frac{4(u-1)^2\log u}{u^2+6u+1}\mathrm{d}u=\,\,&u\log u-u\\ &-(12-5\sqrt{2})\left( u\log u-u+\frac{1}{47} \left(51-14 \sqrt{2}\right)\log u\log\left( \frac{u}{3-2 \sqrt{2}}+1 \right)+\frac{1}{47} \left(51-14 \sqrt{2}\right)\mathrm{Li}_{2}\left( -\frac{u}{3-2 \sqrt{2}} \right) \right)\\ &-(5\sqrt{2}-12)\left( u\log u-u-(2\sqrt{2}+3)\log u\log\left( \frac{u}{2\sqrt{2}+3}+1 \right)-(2\sqrt{2}+3)\mathrm{Li}_{2}\left(- \frac{u}{2\sqrt{2}+3} \right) \right) \end{aligned}

然后是 log⁡0\log^0 积分

∫8(u+1)(u−1)u2+6u+1du=u−3log⁡((u+3)2−8)+22log⁡22−(u+3)22+(u+3)\begin{aligned} &\int \frac{8(u+1)(u-1)}{u^2+6u+1}\mathrm{d}u =u-3\log((u+3)^2-8)+2\sqrt{2}\log \frac{2\sqrt{2}-(u+3)}{2\sqrt{2}+(u+3)} \end{aligned}

以及多得可怕的 log⁡2\log^2 积分

∫(u2−1)log⁡2uu2+6u+1du=  2u−2ulog⁡u+ulog⁡2u−2∫(3u+1)log⁡2uu2+6u+1du=  2u−2ulog⁡u+ulog⁡2u−2((22+32)∫log⁡2uu+22+3du+(3−222)∫log⁡2uu−22+3du)=  2u−2ulog⁡u+ulog⁡2u−(22+3)∫log⁡2uu+22+3du−(3−22)∫log⁡2uu−22+3du=  2u−2ulog⁡u+ulog⁡2u−(22+3)(log⁡(u3+22+1)log⁡2u+2Li2(−u3+22)log⁡u−2Li3(−u3+22))−(3−22)(log⁡(u3−22+1)log⁡2u+2Li2(−u3−22)log⁡u−2Li3(−u3−22))\begin{aligned} &\int \frac{(u^2-1)\log^2u}{u^2+6u+1}\mathrm{d}u\\ =\,\,&2 u-2 u \log u+u \log ^2u-2\int \frac{(3u+1)\log^2u}{u^2+6u+1}\mathrm{d}u\\ =\,\,&2 u-2 u \log u+u \log ^2u-2\left( \left(\frac{2 \sqrt{2}+3}{2}\right) \int\frac{\log ^2u}{ u+2 \sqrt{2}+3}\mathrm{d}u+\left(\frac{3-2 \sqrt{2}}{2}\right)\int\frac{ \log ^2u}{u-2 \sqrt{2}+3}\mathrm{d}u \right) \\ =\,\,&2 u-2 u \log u+u \log ^2u- \left(2 \sqrt{2}+3\right) \int\frac{\log ^2u}{ u+2 \sqrt{2}+3}\mathrm{d}u-\left(3-2 \sqrt{2}\right)\int\frac{ \log ^2u}{u-2 \sqrt{2}+3}\mathrm{d}u \\ =\,\,&2 u-2 u \log u+u \log ^2u\\ &- \left(2 \sqrt{2}+3\right) \left( \log\left( \frac{u}{3+2\sqrt{2}}+1 \right)\log^2u+2\mathrm{Li}_{2}\left( -\frac{u}{3+2\sqrt{2}} \right) \log u-2\mathrm{Li}_{3}\left( -\frac{u}{3+2\sqrt{2}} \right)\right) \\ &-\left(3-2 \sqrt{2}\right)\left( \log\left( \frac{u}{3-2\sqrt{2}}+1 \right)\log^2u+2\mathrm{Li}_{2}\left( -\frac{u}{3-2\sqrt{2}} \right) \log u-2\mathrm{Li}_{3}\left( -\frac{u}{3-2\sqrt{2}} \right)\right) \\ \end{aligned}

三个对数积分加起来，乘上系数 i8\frac{\mathrm{i}}{8} ,

i8[u−3log⁡((u+3)2−8)+22log⁡22−(u+3)22+(u+3)+ulog⁡u−u−(12−52)(ulog⁡u−u+147(51−142)log⁡ulog⁡(u3−22+1)+147(51−142)Li2(−u3−22))−(52−12)(ulog⁡u−u−(22+3)log⁡ulog⁡(u22+3+1)−(22+3)Li2(−u22+3))+2u−2ulog⁡u+ulog⁡2u−(22+3)(log⁡(u3+22+1)log⁡2u+2Li2(−u3+22)log⁡u−2Li3(−u3+22))−(3−22)(log⁡(u3−22+1)log⁡2u+2Li2(−u3−22)log⁡u−2Li3(−u3−22))]\begin{aligned} &\frac{\mathrm{i}}{8}\bigg[u-3\log((u+3)^2-8)+2\sqrt{2}\log \frac{2\sqrt{2}-(u+3)}{2\sqrt{2}+(u+3)}+u\log u-u\\ &-(12-5\sqrt{2})\left( u\log u-u+\frac{1}{47} \left(51-14 \sqrt{2}\right)\log u\log\left( \frac{u}{3-2 \sqrt{2}}+1 \right)+\frac{1}{47} \left(51-14 \sqrt{2}\right)\mathrm{Li}_{2}\left( -\frac{u}{3-2 \sqrt{2}} \right) \right)\\ &-(5\sqrt{2}-12)\left( u\log u-u-(2\sqrt{2}+3)\log u\log\left( \frac{u}{2\sqrt{2}+3}+1 \right)-(2\sqrt{2}+3)\mathrm{Li}_{2}\left(- \frac{u}{2\sqrt{2}+3} \right) \right) \\ &+2 u-2 u \log u+u \log ^2u\\ &- \left(2 \sqrt{2}+3\right) \left( \log\left( \frac{u}{3+2\sqrt{2}}+1 \right)\log^2u+2\mathrm{Li}_{2}\left( -\frac{u}{3+2\sqrt{2}} \right) \log u-2\mathrm{Li}_{3}\left( -\frac{u}{3+2\sqrt{2}} \right)\right) \\ &-\left(3-2 \sqrt{2}\right)\left( \log\left( \frac{u}{3-2\sqrt{2}}+1 \right)\log^2u+2\mathrm{Li}_{2}\left( -\frac{u}{3-2\sqrt{2}} \right) \log u-2\mathrm{Li}_{3}\left( -\frac{u}{3-2\sqrt{2}} \right)\right) \bigg] \end{aligned}

要换回三角元咯，还记得 u=eiTu=\mathrm{e}^{\mathrm{i}T} , T=2tT=2t 麽，原式变成

i8[e2it−3log⁡((e2it+3)2−8)+22log⁡22−(e2it+3)22+(e2it+3)+e2itlog⁡e2it−e2it−(12−52)(e2itlog⁡e2it−e2it+147(51−142)log⁡e2itlog⁡(e2it3−22+1)+147(51−142)Li2(−e2it3−22))−(52−12)(e2itlog⁡e2it−e2it−(22+3)log⁡e2itlog⁡(e2it22+3+1)−(22+3)Li2(−e2it22+3))+2e2it−2e2itlog⁡e2it+e2itlog⁡2e2it−(22+3)(log⁡(e2it3+22+1)log⁡2e2it+2Li2(−e2it3+22)log⁡e2it−2Li3(−e2it3+22))−(3−22)(log⁡(e2it3−22+1)log⁡2e2it+2Li2(−e2it3−22)log⁡e2it−2Li3(−e2it3−22))]\begin{aligned} &\frac{\mathrm{i}}{8}\bigg[{\mathrm{e}^{2\mathrm{i}{t}}}-3\log(({\mathrm{e}^{2\mathrm{i}{t}}}+3)^2-8)+2\sqrt{2}\log \frac{2\sqrt{2}-({\mathrm{e}^{2\mathrm{i}{t}}}+3)}{2\sqrt{2}+({\mathrm{e}^{2\mathrm{i}{t}}}+3)}+{\mathrm{e}^{2\mathrm{i}{t}}}\log {\mathrm{e}^{2\mathrm{i}{t}}}-{\mathrm{e}^{2\mathrm{i}{t}}}\\ &-(12-5\sqrt{2})\left( {\mathrm{e}^{2\mathrm{i}{t}}}\log {\mathrm{e}^{2\mathrm{i}{t}}}-{\mathrm{e}^{2\mathrm{i}{t}}}+\frac{1}{47} \left(51-14 \sqrt{2}\right)\log {\mathrm{e}^{2\mathrm{i}{t}}}\log\left( \frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3-2 \sqrt{2}}+1 \right)+\frac{1}{47} \left(51-14 \sqrt{2}\right)\mathrm{Li}_{2}\left( -\frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3-2 \sqrt{2}} \right) \right)\\ &-(5\sqrt{2}-12)\left( {\mathrm{e}^{2\mathrm{i}{t}}}\log {\mathrm{e}^{2\mathrm{i}{t}}}-{\mathrm{e}^{2\mathrm{i}{t}}}-(2\sqrt{2}+3)\log {\mathrm{e}^{2\mathrm{i}{t}}}\log\left( \frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{2\sqrt{2}+3}+1 \right)-(2\sqrt{2}+3)\mathrm{Li}_{2}\left(- \frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{2\sqrt{2}+3} \right) \right) \\ &+2 {\mathrm{e}^{2\mathrm{i}{t}}}-2 {\mathrm{e}^{2\mathrm{i}{t}}} \log {\mathrm{e}^{2\mathrm{i}{t}}}+{\mathrm{e}^{2\mathrm{i}{t}}} \log ^2{\mathrm{e}^{2\mathrm{i}{t}}}\\ &- \left(2 \sqrt{2}+3\right) \left( \log\left( \frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3+2\sqrt{2}}+1 \right)\log^2{\mathrm{e}^{2\mathrm{i}{t}}}+2\mathrm{Li}_{2}\left( -\frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3+2\sqrt{2}} \right) \log {\mathrm{e}^{2\mathrm{i}{t}}}-2\mathrm{Li}_{3}\left( -\frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3+2\sqrt{2}} \right)\right) \\ &-\left(3-2 \sqrt{2}\right)\left( \log\left( \frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3-2\sqrt{2}}+1 \right)\log^2{\mathrm{e}^{2\mathrm{i}{t}}}+2\mathrm{Li}_{2}\left( -\frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3-2\sqrt{2}} \right) \log {\mathrm{e}^{2\mathrm{i}{t}}}-2\mathrm{Li}_{3}\left( -\frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3-2\sqrt{2}} \right)\right) \bigg] \end{aligned}

别忘了分部积分的分部哦，这就是我们最开始次数为 t2t^2 的积分的结果

∫t2sin⁡t⋅arctan⁡(cos⁡t)dt=  (2tsin⁡t+(2−t2)cos⁡t)arctan⁡(cos⁡t)−i8[e2it−3log⁡((e2it+3)2−8)+22log⁡22−(e2it+3)22+(e2it+3)+e2itlog⁡e2it−e2it−(12−52)(e2itlog⁡e2it−e2it+147(51−142)log⁡e2itlog⁡(e2it3−22+1)+147(51−142)Li2(−e2it3−22))−(52−12)(e2itlog⁡e2it−e2it−(22+3)log⁡e2itlog⁡(e2it22+3+1)−(22+3)Li2(−e2it22+3))+2e2it−2e2itlog⁡e2it+e2itlog⁡2e2it−(22+3)(log⁡(e2it3+22+1)log⁡2e2it+2Li2(−e2it3+22)log⁡e2it−2Li3(−e2it3+22))−(3−22)(log⁡(e2it3−22+1)log⁡2e2it+2Li2(−e2it3−22)log⁡e2it−2Li3(−e2it3−22))]\begin{aligned} &\int t^2\sin t\cdot \arctan (\cos t)\mathrm{dt}\\ =\,\,&(2{t}\sin {t}+(2-{t}^2)\cos {t})\arctan(\cos {t})-\\ &\frac{\mathrm{i}}{8}\bigg[{\mathrm{e}^{2\mathrm{i}{t}}}-3\log(({\mathrm{e}^{2\mathrm{i}{t}}}+3)^2-8)+2\sqrt{2}\log \frac{2\sqrt{2}-({\mathrm{e}^{2\mathrm{i}{t}}}+3)}{2\sqrt{2}+({\mathrm{e}^{2\mathrm{i}{t}}}+3)}+{\mathrm{e}^{2\mathrm{i}{t}}}\log {\mathrm{e}^{2\mathrm{i}{t}}}-{\mathrm{e}^{2\mathrm{i}{t}}}\\ &-(12-5\sqrt{2})\left( {\mathrm{e}^{2\mathrm{i}{t}}}\log {\mathrm{e}^{2\mathrm{i}{t}}}-{\mathrm{e}^{2\mathrm{i}{t}}}+\frac{1}{47} \left(51-14 \sqrt{2}\right)\log {\mathrm{e}^{2\mathrm{i}{t}}}\log\left( \frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3-2 \sqrt{2}}+1 \right)+\frac{1}{47} \left(51-14 \sqrt{2}\right)\mathrm{Li}_{2}\left( -\frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3-2 \sqrt{2}} \right) \right)\\ &-(5\sqrt{2}-12)\left( {\mathrm{e}^{2\mathrm{i}{t}}}\log {\mathrm{e}^{2\mathrm{i}{t}}}-{\mathrm{e}^{2\mathrm{i}{t}}}-(2\sqrt{2}+3)\log {\mathrm{e}^{2\mathrm{i}{t}}}\log\left( \frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{2\sqrt{2}+3}+1 \right)-(2\sqrt{2}+3)\mathrm{Li}_{2}\left(- \frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{2\sqrt{2}+3} \right) \right) \\ &+2 {\mathrm{e}^{2\mathrm{i}{t}}}-2 {\mathrm{e}^{2\mathrm{i}{t}}} \log {\mathrm{e}^{2\mathrm{i}{t}}}+{\mathrm{e}^{2\mathrm{i}{t}}} \log ^2{\mathrm{e}^{2\mathrm{i}{t}}}\\ &- \left(2 \sqrt{2}+3\right) \left( \log\left( \frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3+2\sqrt{2}}+1 \right)\log^2{\mathrm{e}^{2\mathrm{i}{t}}}+2\mathrm{Li}_{2}\left( -\frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3+2\sqrt{2}} \right) \log {\mathrm{e}^{2\mathrm{i}{t}}}-2\mathrm{Li}_{3}\left( -\frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3+2\sqrt{2}} \right)\right) \\ &-\left(3-2 \sqrt{2}\right)\left( \log\left( \frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3-2\sqrt{2}}+1 \right)\log^2{\mathrm{e}^{2\mathrm{i}{t}}}+2\mathrm{Li}_{2}\left( -\frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3-2\sqrt{2}} \right) \log {\mathrm{e}^{2\mathrm{i}{t}}}-2\mathrm{Li}_{3}\left( -\frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3-2\sqrt{2}} \right)\right) \bigg] \end{aligned}

回过神来，整理一下次数为 tt 积分的结果. 对于最底下那一条

∫3+uu2+6u+1log⁡udu=12log⁡ulog⁡(u2+6u+1)+12Li2(−(3+22)u)+12Li2(−(3−22)u)\begin{aligned} &\int \frac{3+u}{u^2+6u+1}\log u\mathrm{d}u\\ &=\frac{1}{2}\log u\log(u^2+6u+1)+\frac{1}{2}\mathrm{Li}_{2}(-(3+2\sqrt{2})u)+\frac{1}{2}\mathrm{Li}_{2}(-(3-2\sqrt{2})u) \end{aligned}

回到三角积分, 即

−i2log⁡2u+2i∫3+uu2+6u+1⋅log⁡udu=i(log⁡ulog⁡(u2+6u+1)−12log⁡2u+Li2(−(3+22)u)+Li2(−(3−22)u))\begin{aligned} &-\frac{\mathrm{i}}{2}\log^2 u+2\mathrm{i}\int \frac{3+u}{u^2+6u+1}\cdot \log u\mathrm{d}u\\ &=\mathrm{i}\left( \log u\log(u^2+6u+1)-\frac{1}{2}\log^2 u+\mathrm{Li}_{2}(-(3+2\sqrt{2})u)+\mathrm{Li}_{2}(-(3-2\sqrt{2})u)\right) \end{aligned}

代回一开始的分部积分（咳，第一版还漏了）

(sin⁡t−tcos⁡t)arctan⁡(cos⁡t)−14(i(log⁡ulog⁡(u2+6u+1)−12log⁡2u+Li2(−(3+22)u)+Li2(−(3−22)u))−42arctan⁡(tan⁡T22))\begin{aligned} &(\sin t-t\cos t)\arctan(\cos t)-\frac{1}{4}\left(\mathrm{i}\left( \log u\log(u^2+6u+1)-\frac{1}{2}\log^2 u+\mathrm{Li}_{2}(-(3+2\sqrt{2})u)+\mathrm{Li}_{2}(-(3-2\sqrt{2})u)\right) -4\sqrt{2}\arctan\left(\frac{\tan\frac{T}{2}}{\sqrt{2}}\right) \right)\end{aligned}

换元， u=eiTu=\mathrm{e}^{\mathrm{i}T} , T=2tT=2t , 易得原式等于

(sin⁡t−tcos⁡t)arctan⁡(cos⁡t)−14(i(log⁡e2itlog⁡(e4it+6e2it+1)−12log⁡2e2it+Li2(−(3+22)e2it)+Li2(−(3−22)e2it))−42arctan⁡(tan⁡T22))\begin{aligned}(\sin t-t\cos t)\arctan(\cos t)-\frac{1}{4} \left(\mathrm{i}\left( \log {\mathrm{e}^{2\mathrm{i}{t}}}\log({\mathrm{e}^{4\mathrm{i}{t}}}+6{\mathrm{e}^{2\mathrm{i}{t}}}+1)-\frac{1}{2}\log^2 {\mathrm{e}^{2\mathrm{i}{t}}}+\mathrm{Li}_{2}(-(3+2\sqrt{2}){\mathrm{e}^{2\mathrm{i}{t}}})+\mathrm{Li}_{2}(-(3-2\sqrt{2}){\mathrm{e}^{2\mathrm{i}{t}}})\right)-4\sqrt{2}\arctan\left(\frac{\tan\frac{T}{2}}{\sqrt{2}}\right)\right) \end{aligned}

最开始的积分，是这两个不同次数的积分的权和

∫t(t−π2)sin⁡t⋅arctan⁡(cos⁡t)dt=It2−π2It=(2tsin⁡t+(2−t2)cos⁡t)arctan⁡(cos⁡t)−i8[e2it−3log⁡((e2it+3)2−8)+22log⁡22−(e2it+3)22+(e2it+3)+e2itlog⁡e2it−e2it−(12−52)(e2itlog⁡e2it−e2it+147(51−142)log⁡e2itlog⁡(e2it3−22+1)+147(51−142)Li2(−e2it3−22))−(52−12)(e2itlog⁡e2it−e2it−(22+3)log⁡e2itlog⁡(e2it22+3+1)−(22+3)Li2(−e2it22+3))+2e2it−2e2itlog⁡e2it+e2itlog⁡2e2it−(22+3)(log⁡(e2it3+22+1)log⁡2e2it+2Li2(−e2it3+22)log⁡e2it−2Li3(−e2it3+22))−(3−22)(log⁡(e2it3−22+1)log⁡2e2it+2Li2(−e2it3−22)log⁡e2it−2Li3(−e2it3−22))]−π2[(sin⁡t−tcos⁡t)arctan⁡(cos⁡t)−14(i(log⁡e2itlog⁡(e4it+6e2it+1)−12log⁡2e2it+Li2(−(3+22)e2it)+Li2(−(3−22)e2it))−42arctan⁡(tan⁡T22))]\begin{aligned} &\int t\left( t-\frac{\pi}{2}\right)\sin t\cdot \arctan(\cos t)\mathrm{d}t\\ &=I_{t^2}-\frac{\pi}{2}I_{t}\\ &=(2{t}\sin {t}+(2-{t}^2)\cos {t})\arctan(\cos {t})-\\ &\frac{\mathrm{i}}{8}\bigg[{\mathrm{e}^{2\mathrm{i}{t}}}-3\log(({\mathrm{e}^{2\mathrm{i}{t}}}+3)^2-8)+2\sqrt{2}\log \frac{2\sqrt{2}-({\mathrm{e}^{2\mathrm{i}{t}}}+3)}{2\sqrt{2}+({\mathrm{e}^{2\mathrm{i}{t}}}+3)}+{\mathrm{e}^{2\mathrm{i}{t}}}\log {\mathrm{e}^{2\mathrm{i}{t}}}-{\mathrm{e}^{2\mathrm{i}{t}}}\\ &-(12-5\sqrt{2})\left( {\mathrm{e}^{2\mathrm{i}{t}}}\log {\mathrm{e}^{2\mathrm{i}{t}}}-{\mathrm{e}^{2\mathrm{i}{t}}}+\frac{1}{47} \left(51-14 \sqrt{2}\right)\log {\mathrm{e}^{2\mathrm{i}{t}}}\log\left( \frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3-2 \sqrt{2}}+1 \right)+\frac{1}{47} \left(51-14 \sqrt{2}\right)\mathrm{Li}_{2}\left( -\frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3-2 \sqrt{2}} \right) \right)\\ &-(5\sqrt{2}-12)\left( {\mathrm{e}^{2\mathrm{i}{t}}}\log {\mathrm{e}^{2\mathrm{i}{t}}}-{\mathrm{e}^{2\mathrm{i}{t}}}-(2\sqrt{2}+3)\log {\mathrm{e}^{2\mathrm{i}{t}}}\log\left( \frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{2\sqrt{2}+3}+1 \right)-(2\sqrt{2}+3)\mathrm{Li}_{2}\left(- \frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{2\sqrt{2}+3} \right) \right) \\ &+2 {\mathrm{e}^{2\mathrm{i}{t}}}-2 {\mathrm{e}^{2\mathrm{i}{t}}} \log {\mathrm{e}^{2\mathrm{i}{t}}}+{\mathrm{e}^{2\mathrm{i}{t}}} \log ^2{\mathrm{e}^{2\mathrm{i}{t}}}\\ &- \left(2 \sqrt{2}+3\right) \left( \log\left( \frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3+2\sqrt{2}}+1 \right)\log^2{\mathrm{e}^{2\mathrm{i}{t}}}+2\mathrm{Li}_{2}\left( -\frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3+2\sqrt{2}} \right) \log {\mathrm{e}^{2\mathrm{i}{t}}}-2\mathrm{Li}_{3}\left( -\frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3+2\sqrt{2}} \right)\right) \\ &-\left(3-2 \sqrt{2}\right)\left( \log\left( \frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3-2\sqrt{2}}+1 \right)\log^2{\mathrm{e}^{2\mathrm{i}{t}}}+2\mathrm{Li}_{2}\left( -\frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3-2\sqrt{2}} \right) \log {\mathrm{e}^{2\mathrm{i}{t}}}-2\mathrm{Li}_{3}\left( -\frac{{\mathrm{e}^{2\mathrm{i}{t}}}}{3-2\sqrt{2}} \right)\right) \bigg]\\ &-\frac{\pi}{2}\left[(\sin t-t\cos t)\arctan(\cos t)\\-\frac{1}{4} \left(\mathrm{i}\left( \log {\mathrm{e}^{2\mathrm{i}{t}}}\log({\mathrm{e}^{4\mathrm{i}{t}}}+6{\mathrm{e}^{2\mathrm{i}{t}}}+1)-\frac{1}{2}\log^2 {\mathrm{e}^{2\mathrm{i}{t}}}+\mathrm{Li}_{2}(-(3+2\sqrt{2}){\mathrm{e}^{2\mathrm{i}{t}}})+\mathrm{Li}_{2}(-(3-2\sqrt{2}){\mathrm{e}^{2\mathrm{i}{t}}})\right)-4\sqrt{2}\arctan\left(\frac{\tan\frac{T}{2}}{\sqrt{2}}\right)\right) \right] \end{aligned}

换回最后一个元，我们热烈而又忠诚地看到最终表达式的诞生

8. Final Answer​

(2arccos⁡xsin⁡arccos⁡x+(2−arccos⁡x2)cos⁡arccos⁡x)arctan⁡(cos⁡arccos⁡x)−i8[e2i⋅arccos⁡x−3log⁡((e2i⋅arccos⁡x+3)2−8)+22log⁡22−(e2i⋅arccos⁡x+3)22+(e2i⋅arccos⁡x+3)+e2i⋅arccos⁡xlog⁡e2i⋅arccos⁡x−e2i⋅arccos⁡x−(12−52)(e2i⋅arccos⁡xlog⁡e2i⋅arccos⁡x−e2i⋅arccos⁡x+147(51−142)log⁡e2i⋅arccos⁡xlog⁡(e2i⋅arccos⁡x3−22+1)+147(51−142)Li2(−e2i⋅arccos⁡x3−22))−(52−12)(e2i⋅arccos⁡xlog⁡e2i⋅arccos⁡x−e2i⋅arccos⁡x−(22+3)log⁡e2i⋅arccos⁡xlog⁡(e2i⋅arccos⁡x22+3+1)−(22+3)Li2(−e2i⋅arccos⁡x22+3))+2e2i⋅arccos⁡x−2e2i⋅arccos⁡xlog⁡e2i⋅arccos⁡x+e2i⋅arccos⁡xlog⁡2e2i⋅arccos⁡x−(22+3)(log⁡(e2i⋅arccos⁡x3+22+1)log⁡2e2i⋅arccos⁡x+2Li2(−e2i⋅arccos⁡x3+22)log⁡e2i⋅arccos⁡x−2Li3(−e2i⋅arccos⁡x3+22))−(3−22)(log⁡(e2i⋅arccos⁡x3−22+1)log⁡2e2i⋅arccos⁡x+2Li2(−e2i⋅arccos⁡x3−22)log⁡e2i⋅arccos⁡x−2Li3(−e2i⋅arccos⁡x3−22))]−π2[(sin⁡arccos⁡x−arccos⁡x⋅x)arctan⁡(cos⁡arccos⁡x)−14(i(log⁡e2i⋅arccos⁡xlog⁡(e4iarccos⁡x+6e2i⋅arccos⁡x+1)−12log⁡2e2i⋅arccos⁡x+Li2(−(3+22)e2i⋅arccos⁡x)+Li2(−(3−22)e2i⋅arccos⁡x))−42arctan⁡(tan⁡arccos⁡x22))]\begin{aligned} &(2{\arccos x}\sin {\arccos x}+(2-{\arccos x}^2)\cos {\arccos x})\arctan(\cos {\arccos x})-\\ &\frac{\mathrm{i}}{8}\bigg[{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}-3\log(({\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}+3)^2-8)+2\sqrt{2}\log \frac{2\sqrt{2}-({\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}+3)}{2\sqrt{2}+({\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}+3)}+{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}\log {\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}-{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}\\ &-(12-5\sqrt{2})\left( {\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}\log {\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}-{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}+\frac{1}{47} \left(51-14 \sqrt{2}\right)\log {\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}\log\left( \frac{{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}}{3-2 \sqrt{2}}+1 \right)+\frac{1}{47} \left(51-14 \sqrt{2}\right)\mathrm{Li}_{2}\left( -\frac{{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}}{3-2 \sqrt{2}} \right) \right)\\ &-(5\sqrt{2}-12)\left( {\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}\log {\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}-{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}-(2\sqrt{2}+3)\log {\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}\log\left( \frac{{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}}{2\sqrt{2}+3}+1 \right)-(2\sqrt{2}+3)\mathrm{Li}_{2}\left(- \frac{{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}}{2\sqrt{2}+3} \right) \right) \\ &+2 {\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}-2 {\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}} \log {\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}+{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}} \log ^2{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}\\ &- \left(2 \sqrt{2}+3\right) \left( \log\left( \frac{{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}}{3+2\sqrt{2}}+1 \right)\log^2{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}+2\mathrm{Li}_{2}\left( -\frac{{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}}{3+2\sqrt{2}} \right) \log {\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}-2\mathrm{Li}_{3}\left( -\frac{{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}}{3+2\sqrt{2}} \right)\right) \\ &-\left(3-2 \sqrt{2}\right)\left( \log\left( \frac{{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}}{3-2\sqrt{2}}+1 \right)\log^2{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}+2\mathrm{Li}_{2}\left( -\frac{{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}}{3-2\sqrt{2}} \right) \log {\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}-2\mathrm{Li}_{3}\left( -\frac{{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}}{3-2\sqrt{2}} \right)\right) \bigg]\\ &-\frac{\pi}{2}\left[(\sin {\arccos x}-{\arccos x}\cdot x)\arctan(\cos {\arccos x})\\-\frac{1}{4} \left(\mathrm{i}\left( \log {\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}\log({\mathrm{e}^{4\mathrm{i}{\arccos x}}}+6{\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}+1)-\frac{1}{2}\log^2 {\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}}+\mathrm{Li}_{2}(-(3+2\sqrt{2}){\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}})+\mathrm{Li}_{2}(-(3-2\sqrt{2}){\mathrm{e}^{2\mathrm{i}\cdot{\arccos x}}})\right)-4\sqrt{2}\arctan\left(\frac{\tan\frac{\arccos x}{2}}{\sqrt{2}}\right)\right) \right] \end{aligned}

乐~