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\[\int_0^1 x^n\ln |x|\ln |x+a|{\rm d}x\quad (n\in \mathbb{N},a\in\mathbb{R}\setminus \{0\}) \]

Solution: 置

\[I_n(a):=\int_0^1 x^n\ln |x|\ln |x+a|{\rm d}x \]

於是

\[\begin{aligned} I_{n}(a) &=\left[(\ln |x| \ln |x+a|) \frac{x^{n+1}}{n+1}\right]_{0}^{1}-\int_{0}^{1} \frac{x^{n+1}}{n+1}\left(\frac{\ln |x+a|}{x}+\frac{\ln |x|}{x+a}\right) {\rm d} x \\ &=-\frac{1}{n+1}\left[\underbrace{\int_{0}^{1} x^{n} \ln |x+a| {\rm d} x}_{A_{n}(a)}+\underbrace{\int_{0}^{1} \frac{x^{n+1}}{x+a} \ln |x| {\rm d}x}_{B_{n}(a)}\right] \end{aligned} \]

計算 \(A_n(a)\)

\[A_{n}(a)=\left[\ln |x+a| \cdot \frac{x^{n+1}}{n+1}\right]_{0}^{1}-\int_{0}^{1} \frac{x^{n+1}}{n+1} \cdot \frac{1}{x+a} {\rm d}x=\frac{\ln |a+1|}{n+1}-\frac{1}{n+1} \int_{0}^{1} \frac{x^{n+1}}{x+a} {\rm d} x \]

注意到

\[\int \frac{x^{m}}{x+a} {\rm d}x=\frac{x^{m}}{m}+\sum_{j=0}^{m-2}\left(\frac{x^{j+1}}{j+1}(-1)^{m-2-j} a^{m-2-j}\right)+(-1)^{m} a^{m} \ln |x+a| \]

令 \(m=n+1\)

, 於是

\[\begin{aligned} A_{n}(a) &=\frac{\ln |a+1|}{n+1}-\frac{1}{n+1} \int_{0}^{1} \frac{x^{n+1}}{x+a} {\rm d} x \\ &=\frac{\ln |a+1|}{n+1}-\frac{1}{n+1}\left[\frac{x^{n+1}}{n+1}+\sum_{j=0}^{n+1-2}\left(\frac{x^{j+1}}{j+1}(-1)^{n+1-2-j} a^{n+1-2-j}\right)+(-1)^{n+1} a^{n+1} \ln |x+a|\right]_{0}^{1} \\ &=\frac{\ln |a+1|}{n+1}-\frac{1}{n+1}\left[\frac{x^{n+1}}{n+1}+\sum_{j=0}^{n-1} \frac{(-1)^{n-1-j} a^{n-1-j}}{j+1}+(-1)^{n+1} a^{n+1} \ln \left|\frac{a+1}{a}\right|\right] \end{aligned} \]

計算 \(B_n(a)\)

\[\begin{aligned} B_{n}(a) &=\int_{0}^{1} \frac{x^{n+1}}{x+a} \ln |x| {\rm d} x=\int_{0}^{1} x^{n}\left(\frac{x+a-a}{x+a}\right) \ln |x| {\rm d}x=\int_{0}^{1} x^{n}\left(1-\frac{a}{x+a}\right) \ln |x| {\rm d}x \\ &=\underbrace{\int_{0}^{1} x^{n} \ln |x| {\rm d}x}_{C_{n}}-a \int_{0}^{1} \frac{x^{n}}{x+a} \ln |x| {\rm d}x=C_{n}-a B_{n-1}(a) \end{aligned} \]

於是得到遞推關係

\[B_{n+1}(a)=C_{n+1}-a B_{n}(a) \]

易知

\[C_{n+1}=\int_{0}^{1} x^{n+1} \ln |x| {\rm d}x=-\frac{1}{(n+2)^2} \]

計算初值 \(B_{0}(a)\)

\[\begin{aligned} B_{0}(a) &=\int_{0}^{1} \frac{x}{x+a} \ln |x| {\rm d}x=\int_{0}^{1} \frac{x+a-a}{x+a} \ln |x| {\rm d}x\\ &=\int_{0}^{1} \ln |x| {\rm d} x-a \underbrace{\int_{0}^{1} \frac{\ln |x|}{x+a} {\rm d}x}_{D(a)} \\ &=[x(\ln |x|-1)]_{0}^{1}-a D(a)=-a D(a) \end{aligned} \]

計算 \(D(a)\)

\[\begin{aligned} D(a) &=\frac1a\int_0^1\frac{\ln |x|}{x/a+1}{\rm d} x=\frac{1}{a} \int_{0}^{\frac{1}{a}} \frac{\ln |a u|}{u+1} \cdot a {\rm d} u=\int_{0}^{\frac{1}{a}} \frac{\ln |a|+\ln |u|}{u+1}{\rm d} u\\ &=\ln |a| \int_{0}^{\frac{1}{a}} \frac{1}{u+1} {\rm d}u+\underbrace{\int_{0}^{\frac{1}{a}} \frac{\ln |u|}{u+1} {\rm d}u}_{E(a)}=\ln |a|\cdot \ln \left|\frac{a+1}{a}\right|+E(a) \end{aligned} \]

計算 \(E(a)\)

\[\begin{aligned} E(a) &=\{\ln |u| \cdot \ln |u+1|\}_{0}^{1/a}-\int_{0}^{\frac{1}{a}} \ln |u+1| \cdot \frac{1}{u} {\rm d}u=-\ln |a| \ln \left|\frac{a+1}{a}\right|-\int_{0}^{\frac{1}{a}} \frac{\ln |u+1|}{u} {\rm d}u \\ &=-\ln |a| \cdot\ln \left|\frac{a+1}{a}\right|-\left\{-\operatorname{Li}_{2}(-u)\right\}_{0}^{1/a}=-\ln |a|\cdot \ln \left|\frac{a+1}{a}\right|+\operatorname{Li}_{2}\left(-\frac{1}{a}\right) \end{aligned} \]

於是

\[D(a)=\ln |a|\cdot \ln \left|\frac{a+1}{a}\right|+E(a)=\operatorname{Li}_{2}\left(-\frac{1}{a}\right) \]\[B_{0}(a)=-a D(a)=-a\operatorname{Li}_{2}\left(-\frac{1}{a}\right) \]

現有如下遞推關係

\[B_{n+1}(a)=C_{n+1}-a B_{n}(a)=-\frac{1}{(n+1)^2}-aB_n(a) \]

這裡我們利用廣義遞推式

\[I_{n+1}(a)=\theta_{n}(a)+\phi_{n}(a) I_{n}(a) \]

的解

\[I_{n}(a)=\theta_{n-1}(a)+\sum_{j=0}^{n-2} \theta_{j}(a) \prod_{i=j+1}^{n-1} \phi_{i}+I_{0}(a) \prod_{k=0}^{n-1} \phi_{k}(a) \]

這裡

\[\begin{align*} I_n(a)&=B_n(a)\\ \theta_{n}(a)&=-\frac{1}{(n+2)^2}\\ \phi_{n}(a)&=-a\\ I_0(a)&=B_{0}(a)=-a\operatorname{Li}_{2}\left(-\frac{1}{a}\right) \end{align*} \]

因此

\[\begin{aligned} B_{n}(a) &=-\frac{1}{(n-1+2)^{2}}+\sum_{j=0}^{n-2}\left(-\frac{1}{(j+2)^{2}} \prod_{i=j+1}^{n-1}(-a)\right)+-a \operatorname{Li}_{2}\left(-\frac{1}{a}\right) \prod_{k=0}^{n-1}(-a) \\ &=-\frac{1}{(n+1)^{2}}-\sum_{j=0}^{n-2} \frac{1}{(j+2)^{2}} \cdot(-1)^{n-1-(j+1)+1} \cdot a^{n-1-(j+1)+1}-a \operatorname{Li}_{2}\left(-\frac{1}{a}\right)(-1)^{n-1-0+1} a^{n-1-0+1} \\ &=-\frac{1}{(n+1)^{2}}-\sum_{j=0}^{n-2} \frac{(-1)^{n-(j+1)} \cdot a^{n-(j+1)}}{(j+2)^{2}}+(-1)^{n+1} a^{n+1} \operatorname{Li}_{2}\left(-\frac{1}{a}\right) \end{aligned} \]

綜上所述

\[\begin{align*} I_{n}(a)&=-\frac{1}{n+1}\left[A_{n}(a)+B_{n}(a)\right]\\ &=-\frac{1}{n+1}\left[\frac{\ln |a+1|}{n+1}-\frac{1}{n+1}\left[\frac{a^{n+1}}{n+1}+\sum_{j=0}^{n-1} \frac{(-1)^{n-1-j} a^{n-1-j}}{j+1}+(-1)^{n+1} a^{n+1} \ln \left|\frac{a+1}{a}\right|\right]\right.\\ &-\frac{1}{(n+1)^{2}}-\left.\sum_{j=0}^{n-2} \frac{(-1)^{n-(j+1)} a^{n-(j+1)}}{(j+2)^{2}}+(-1)^{n+1} a^{n+1} \mathrm{Li}_{2}\left(-\frac{1}{a}\right)\right] \\ &=-\frac{\ln |a+1|}{(n+1)^{2}}+\frac{a^{n+1}}{(n+1)^{3}}-\frac{1}{(n+1)^{2}} \sum_{j=0}^{n-1} \frac{(-1)^{n-1-j} a^{n-1-j}}{j+1}-\frac{(-1)^{n+1} a^{n+1}}{(n+1)^{2}} \ln \left|\frac{a+1}{a}\right|+\frac{1}{(n+1)^{3}} \\ &+\frac{1}{n+1} \sum_{j=0}^{n-2} \frac{(-1)^{n-(j+1)} a^{n-(j+1)}}{(j+2)^{2}}-\frac{(-1)^{n+1} a^{n+1}}{n+1} \mathrm{Li}_{2}\left(-\frac{1}{a}\right) \end{align*} \]

\(\square\)