计算积分

河中巨怪

\[\int_{0}^{1}{\frac{{{\ln }^{n }}x}{{{x}^{2}}+x+1}}\,\mathrm{d} x, \,\text{for}\, \,n \in\mathbb{N^*}.\]

\[\int_{0}^{1}{\frac{{{\ln }^{n }}x}{{{x}^{2}}-x+1}}\,\mathrm{d} x, \,\text{for}\, \,n \in\mathbb{N^*}.\]

Renascence_5

for the first one,give you a hint. Using $$\frac{1}{x^2+x+1}=\frac{1-x}{1-x^3}$$ and then the geometric series.maybe you will get a series related to the digamma function or polygarithm.

青青子衿

某群一道积分:计算
\begin{align*}
\int_{0}^{1}{\frac{\ln\,\!x}{{{x}^{2}}+x+1}}\mathrm{d}x&=\dfrac{4\pi^3}{81\sqrt3}\\
\int_{0}^{1}{\frac{\ln\,\!x}{{{x}^{2}}-x-1}}\mathrm{d}x&=\dfrac{\pi^2}{5\sqrt5}\\
\end{align*}

hbghlyj

某群一道积分: 计算
$$
\int_0^1 \frac{\ln ^2 x}{x^2+x+1} d x
$$
解
$$
\begin{aligned}
\int_0^1 \frac{\ln ^2 x}{x^2+x+1} \mathrm{~d} x & =\int_0^1\left(\frac{\ln ^2 x}{1-x^3}-\frac{x \ln ^2 x}{1-x^3}\right) \mathrm{d} x \\
& =\int_0^{\infty}\left(\frac{t^2}{1-\mathrm{e}^{-3 t}}-\frac{\mathrm{e}^{-t} t^2}{1-\mathrm{e}^{-3 t}}\right) \mathrm{e}^{-t} \mathrm{~d} t \\
& =\int_0^{\infty} t^2 \sum_{n=0}^{\infty}\left(\mathrm{e}^{-(3 n+1) t}-\mathrm{e}^{-(3 n+2) t}\right) \mathrm{d} t \\
& =2 \sum_{n=0}^{\infty}\left(\frac{1}{(3 n+1)^3}-\frac{1}{(3 n+2)^3}\right) \\
& =2 \sum_{n=-\infty}^{\infty} \frac{1}{(3 n+1)^3} \\
& =\frac{8 \pi^3}{81 \sqrt{3}}
\end{aligned}
$$
其中, 由
$$
\sum_{n=-\infty}^{\infty} f(n)=-\pi \sum_{k=1}^m \operatorname{Res}\left\{f(z) \cot \pi z, a_k\right\}
$$
可得
$$
\begin{aligned}
\sum_{n=-\infty}^{\infty} \frac{1}{(3 n+1)^3} & =-\pi \operatorname{Res}\left\{\frac{\cot \pi z}{(3 z+1)^3}, z=-\frac{1}{3}\right\} \\
& =-\frac{1}{2 !} \frac{\mathrm{d}^2}{\mathrm{~d} z^2}\left[\left(z+\frac{1}{3}\right)^3 \cdot \frac{\pi \cot \pi z}{(3 z+1)^3}\right]_{z=-1 / 3} \\
& =\frac{4 \pi^3}{81 \sqrt{3}}
\end{aligned}
$$