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Evaluate $$\lim_{n \to \infty} \int_{1-\epsilon}^1 \ln(1+x+...+x^{n-1})dx$$ for $0<\epsilon<1$.

The problem is at the point $x=1$, because there the formula $$1+x+...+x^{n-1}=\frac{1-x^n}{1-x}$$ can't be applied. I think of using the mean value theorem for integrals, so there would be $c_n \in [1-\epsilon,1]$ so that $$\int_{1-\epsilon}^1 \ln(1+x+...+x^{n-1})dx=\epsilon\ln(1+c_n+...+c_n^{n-1})$$ but taking limits gives an indeterminate form.

I do not know Lebesgue's dominated convergence theorem. I would appreciate an elementary proof.

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  • $\begingroup$ Note that $$ \left|\int_{1-\epsilon}^1 \log(1-x^n) \, {\rm d}x\right| \leq \left|\int_{0}^1 \frac{\log(1-x^n)}{x} \, {\rm d}x\right| = \left|\int_0^1 \sum_{k=1}^\infty \frac{x^{nk-1}}{k} \, {\rm d}x\right| = \frac{1}{n} \sum_{k=1}^\infty \frac{1}{k^2} $$ which obviously vanishes for $n\rightarrow \infty$ since the sum converges. So the evaluation of the original integral is actually trivial. $\endgroup$ – Diger Mar 3 '19 at 16:34
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Here is a solution which only uses basic properties of Riemann integral. Although this proof implicitly uses the idea of improper Riemann integral or Lebesgue integral, all such instances will be properly dealt with limit of Riemann integrals instead. You may shorten the proof if improper Riemann integral is available to you.


  • Bounding limsup. Let $r\in(1-\epsilon, 1)$. Since $1+\cdots+x^{n-1} \leq \frac{1}{1-x}$ on $x \in [1-\epsilon, r]$, we have

    \begin{align*} \int_{1-\epsilon}^{r} \log\left( \frac{1-x^n}{1-x} \right) \, \mathrm{d}x &\leq \int_{1-\epsilon}^{r} \log\left( \frac{1}{1-x}\right) \, \mathrm{d}x \\ &= \left[x + (x-1)\log\left(\frac{1}{1-x}\right) \right]_{1-\epsilon}^{r} \\ &\leq \epsilon - \epsilon \log \epsilon. \end{align*}

    So, letting $r \uparrow 1$ gives

    $$ \int_{1-\epsilon}^{1} \log\left( \frac{1-x^n}{1-x} \right) \, \mathrm{d}x \leq \epsilon - \epsilon \log \epsilon. $$

    Then letting $n \to \infty$, we obtain

    $$ \limsup_{n\to\infty} \int_{1-\epsilon}^{1} \log\left( \frac{1-x^n}{1-x} \right) \, \mathrm{d}x \leq \epsilon - \epsilon \log \epsilon. $$

  • Bounding liminf. Again, fix $r \in (1-\epsilon, 1)$. Then

    \begin{align*} \int_{1-\epsilon}^{1} \log\left( \frac{1-x^n}{1-x} \right) \, \mathrm{d}x &\geq \int_{1-\epsilon}^{r} \log\left( \frac{1-x^n}{1-x} \right) \, \mathrm{d}x \\ &\geq \int_{1-\epsilon}^{r} \log\left( \frac{1 - r^n}{1-x} \right) \, \mathrm{d}x \\ &= (r-1+\epsilon) \log(1 - r^n) + \left[x + (x-1)\log\left(\frac{1}{1-x}\right) \right]_{1-\epsilon}^{r}. \end{align*}

    Letting $n \to \infty$ gives

    $$ \liminf_{n\to\infty} \int_{1-\epsilon}^{1} \log\left( \frac{1-x^n}{1-x} \right) \, \mathrm{d}x \geq \left[x + (x-1)\log\left(\frac{1}{1-x}\right) \right]_{1-\epsilon}^{r}, $$

    and since this lower bound is independent of $r$, letting $r \uparrow 1$ gives

    $$ \liminf_{n\to\infty} \int_{1-\epsilon}^{1} \log\left( \frac{1-x^n}{1-x} \right) \, \mathrm{d}x \geq \epsilon - \epsilon \log \epsilon. $$

  • Conclusion. Combining altogether, we find that

    $$ \lim_{n\to\infty} \int_{1-\epsilon}^{1} \log\left( \frac{1-x^n}{1-x} \right) \, \mathrm{d}x = \epsilon - \epsilon \log \epsilon. $$

    Of course, once the dominated convergence theorem is available, the proof will be one-liner.

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    $\begingroup$ Thank you so much! This is perfect! $\endgroup$ – sgc Mar 3 '19 at 10:55
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You can replace the upper limit with $k$ and apply $\lim_{k\rightarrow 1}$ to the whole integral. Then that formula can be applied.

So the integral becomes

$$\int_{c}^{k} (\log(1-x^n) - \log (1-x))dx$$

Since $x<1$ in the given interval $\log(1-x^n)$ is $0$ as n tends to infinity.

So \begin{align*} & =-\int_{c}^{k} \log (1-x))dx \\ & =(1-x)\log(1-x)-(1-x) \\ \end{align*} from $c$ to $k$

$$=(1-k)\log(1-k)-(1-k)-((1-c)\log(1-c)-(1-c))$$

The limit of first term is $0$ as $k\rightarrow 1$. The second term is

$$=(1-c)(1-\log(1-c))$$

Now $c=1-\varepsilon$ like in the question

$$=\varepsilon(1-\log\varepsilon)$$

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