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The problem $2176$ in Math. Mag. 96 (3) 2023 require to show that $\int_{0}^{1} \frac{\log\bigl(x^2 + x + 1\bigr)}{x^2 + 1}\,\mathrm dx \;=\; \frac{\pi}{6}\,\log\!\bigl(\sqrt{3} + 2\bigr) \;-\;\frac{C}{3}$ where $C$ is the Catalan's constant, and you can find the nice solution on page $144$ here.

Now if we consider the sequence $ I(n) = \int_{0}^{1} \frac{\log\bigl( \sum_{i=0}^n x^i \bigr)}{x^2 + 1}\,\mathrm dx \ $

it's simple to show that $ I(1) = \frac{\pi}{8}\,\log\!\bigl(2\bigr)$ and $I(3) = \frac{5\pi}{8}\,\log\!\bigl(2\bigr) \;-\;C$.

Is it possible to find a closed form for the limit $\lim_{n\to \infty} I(n)$?

user967210
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2 Answers2

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we have:$$\sum_{i=0}^{n}x^i=\frac{1-x^{n+1}}{1-x}$$

\begin{align*} I&=\lim_{n\to \infty} \int_{0}^{1} \frac{\log\bigl( \sum_{i=0}^n x^i \bigr)}{x^2 + 1}dx \\ &=-\int_{0}^{1} \frac{\log\bigl( 1-x \bigr)}{x^2 + 1}dx+\lim_{n\to \infty}\int_{0}^{1} \frac{\log\bigl( 1-x^{n+1}\bigr)}{x^2 + 1}dx \end{align*}

Let:$f_n(x)=\frac{\log\bigl( 1-x^{n+1}\bigr)}{x^2 + 1}$ We have :$\forall x\in [0,1[ $ $\lim_{n\to \infty}f_n(x)=0$ and $|f_n(x)|<\frac{1}{1+x^2}$ integrable Using Dominated convergence theorem

$$\lim_{n\to \infty}\int_{0}^{1} \frac{\log\bigl( 1-x^{n+1}\bigr)}{x^2 + 1}dx =\int_{0}^{1} 0 dx=0$$

Therfore: $$I=-\int_{0}^{1} \frac{\log\bigl( 1-x \bigr)}{x^2 + 1}dx= \operatorname{Im}\left( \operatorname{Li}_2\left( \frac{1+i}{2} \right) \right) $$

Delta
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Substituting $x\mapsto \frac{1-x}{1+x}$ gives $$\int_0^{1}\frac{\log(1-x)}{1+x^2}\, dx=\int_0^1 \frac{\log 2+\log x-\log(1+x)}{1+x^2}\, dx$$ $$=\frac{\pi}{4}\log(2)-G-\frac{\pi}{8}\log(2)=-G+\frac{\pi}{8}\log(2)$$