Asymptotic behaviour of $\int_{0}^{1}f(x)x^ndx $ as $n\to \infty $

Solution 1:

Integration by parts works here: For $n \in \mathbb{N}_0$ we have $$ I_n \equiv \int \limits_0^1 f(x) x^n \mathrm{d} x = \frac{f(1)}{n+1} - \frac{1}{n+1} \int \limits_0^1 f'(x) x^{n+1} \mathrm{d} x \, . $$ Since $f \in \mathrm{C}^1([0,1])$, we can estimate the remaining integral according to $$ \left\lvert \int \limits_0^1 f'(x) x^{n+1} \mathrm{d} x \right \rvert \leq \lVert f' \rVert_\infty \int \limits_0^1 x^{n+1} \mathrm{d} x = \frac{\lVert f' \rVert_\infty}{n+2} \, , \, n \in \mathbb{N}_0 \, . $$ This shows that $$ I_n = \frac{f(1)}{n} + \mathcal{O}\left(\frac{1}{n^2}\right) \, , \, n \to \infty \, , $$ as conjectured by Daniel Schepler in the comments.

We can't integrate by parts again, but we can still find the next term in the asymptotic expansion by computing \begin{align} \lim_{n \to \infty} n^2 \left[I_n - \frac{f(1)}{n}\right] &= \lim_{n \to \infty} \left[-\frac{n^2 f(1)}{n (n+1)} - \frac{n^2}{n+1} \int \limits_0^1 f'(x) x^{n+1} \mathrm{d} x\right] \\ &\!\!\!\!\!\overset{u = x^{n+1}}{=} - f(1) - \lim_{n \to \infty} \frac{n^2}{(n+1)^2} \int \limits_0^1 f'\left(u^{1/(n+1)}\right) u^{1/(n+1)} \mathrm{d} u \\ &= - f(1) - \int \limits_0^1 f'(1) \, \mathrm{d} u \\ &= - f(1) - f'(1) \, . \end{align} Here we used the continuity of $f'$ and the dominated convergence theorem. This implies $$ I_n = \frac{f(1)}{n} - \frac{f(1)+f'(1)}{n^2} + \mathcal{o} \left(\frac{1}{n^2}\right) $$ for $n \to \infty$.

Note that the assumption $f(x) > 0$ for $x \in [0,1]$ is not needed (though for example the first term vanishes for $f(1) = 0$, of course).

Solution 2:

We know that $f(x)\leq\max f$ and so $\int_{0}^{1}f(x)x^{n}dx\leq(\max f)\int_{0}^{1} x^{n}dx=(\max f)\frac{1}{n+1}$.

Solution 3:

Since $f$ is continuous on a closed bounded interval, it has a maximum absolute value $M\ge0$ on that interval, so you have $$ \left\vert \int_0^1 f(x)x^n\,dx \right\vert \le \int_0^1 \left\vert f(x) x^n\right\vert \, dx \le M\int_0^1x^n\, dx = \frac M{n+1}. $$

Solution 4:

To spice it up a bit:

We can use Lebesgue's dominated convergence theorem to get $$ \lim_{n \to \infty} \int_0^1 f(x)x^ndx = \int_0^1 \lim_{n \to \infty} f(x)x^n dx = \int_0^1 0 dx = 0. $$ Here $f(x)x^n$ is dominated by the continuous and hence integrable function $f(x)$.

Technically for $x=1$ we get $f(1)$ as limit, but the equality above is still true since we are allowed to change the integrand on a null set.