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My question is having following one as a root.

On one side, for $f : [0,1] \to \mathbb R$ continuous, one can prove that $$\lim\limits_{n \to +\infty} n \int_0^1 x^n f(x) dx =f(1)$$

On the other side, in the post mentionned, it was proven that $$\lim_{n\to +\infty}\left(2n\int_{0}^{1}\dfrac{x^n}{1+x^2}dx\right)^n$$ exists using integration by parts.

Therefore my question is the following.

What can we say of $$\lim\limits_{n \to +\infty} \left(\frac{n}{f(1)} \int_0^1 x^n f(x) dx \right)^n$$ if $f$ is only supposed continuous with $f(1) \neq 0$?

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1 Answers1

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It need not converge. For example, consider $f(x) = 1 + \sqrt{1-x}$.

$$ n \int_0^1 x^n f(x)\; dx = n \left( \dfrac{1}{n+1} + \dfrac{\sqrt{\pi} \;\Gamma(n+1)}{2\; \Gamma(n + 5/2)}\right) \sim 1 + \dfrac{\sqrt{\pi}}{2\sqrt{n}}$$

so $$ \lim_{n \to \infty} \left( n \int_0^1 x^n f(x)\; dx\right)^n = \infty$$

On the other hand, if $f \in C^1$, integration by parts gives

$$ \int_0^1 x^n f(x)\; dx = \dfrac{f(1)}{n+1} - \dfrac{1}{n+1} \int_0^1 x^{n+1} f'(x)\; dx \sim \dfrac{f(1)}{n+1} - \dfrac{f'(1)}{(n+1)^2} $$

and then (if $f(1) \ne 0$)

$$\lim_{n \to \infty} \left( \dfrac{n}{f(1)} \int_0^1 x^n f(x)\; dx \right)^n = \lim_{n \to \infty} \left(1 + \dfrac{f'(1) - f(1)}{f(1) n}\right)^n = \exp\left( \dfrac{f'(1) - f(1)}{f(1)}\right)$$

Robert Israel
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