Evaluate $\lim\limits_{n\to+\infty}\int_{0}^{1}\ln(1-x^2+2x^{n}+x^{2n}){\rm d}x$.

Question

Problem

Evaluate $$\lim_{n\to+\infty}\int_{0}^{1}\ln(1-x^2+2x^{n}+x^{2n}){\rm d}x.$$

I guess it needs to apply Integral Mean Value Theorem or Integral Inequality . But I fail to find the upper-bound function and the lower-bound function. Who can offer a hint?Thanks!

Answer 1

The integrals at hand can be rewritten as $$\int_0^1 \log(1 - x^2 + 2x^n + x^{2n}) dx = 2\int_0^1 \log(1+x^n) dx - \int_0^1 f_n(x) dx$$ where $f_n(x) = -\log\left(1 - \left(\frac{x}{1+x^n}\right)^2 \right)$.

For the first piece, we can bound it as

$$0 \le \int_0^1 \log(1+x^n) dx \le \int_0^1 x^n dx = \frac{1}{n+1}$$

By squeezing, this leads to $\lim_{n\to\infty} \int_0^1 \log(1+x^n) dx = 0$.

For the second piece, notice for any fixed $x \in (0,1)$, we have

$$0 \le f_n(x) \le f_{n+1}(x) \le \infty \quad\text{ and }\quad \lim_{n\to\infty} f_n(x) = -\log(1-x^2)$$

Using Monotone converge theorem, we find

$$\lim_{n\to\infty} \int_0^1 f_n(x) dx = \int_0^1 \lim_{n\to\infty} f_n(x) dx = -\int_0^1\log(1-x^2) dx = 2-\log 4$$

Combine these, we can conclude the integrals at hand converge and

$$\lim_{n\to\infty}\int_0^1\log(1-x^2 + 2x^n + x^{2n}) dx = 2\cdot 0 - (2-\log 4) = \log 4 - 2$$

Answer 2

An addendum to Achille's answer on how to evaluate the integral once the convergence difficulties have been addressed. In the region $|x| < 1$, \begin{align*} \int \log(1-x^2)\, dx &= x \log (1-x^2) + \int \frac{2x^2}{1-x^2}\, dx \tag{integration by parts} \\ &= x \log (1-x^2) + \int \left(\frac{1}{1+x} + \frac{1}{1-x} - 2\, \right) \tag{partial fractions}dx \\ &= (x+1) \log (1+x) + (x+1) \log (1-x) - 2x + C. \end{align*} Taking $0$ and $1$ as our limits of integration, we note that the last line equals $C$ at $x = 0$. At $x = 1$, the term $(x-1) \log (1-x)$ vanishes as $\lim_{x \to 0} x \log x$, and the remaining terms give $2 \log 2 - 2 + C$, so we have $$\int_0^1 \log(1-x^2)\, dx = \log 4 - 2.$$