Demonstrate that a limit is zero (Fourier series)

Question

Can anyone help me to proof the equality $$\lim_{n\to\infty}\int_0^π\ log(x)sin(nx)\, dx=0$$ I know that is equal to the coefficient of Fourier sine series $$\int_0^π\ log(x)sin(nx)\, dx=\frac{π}{2}A_n$$ But I don't know how to proof which the limit is zero, I tried to proof it solving the integral by parts, but I have an indeterminate $0\times\ (-\infty)$

Answer 1

A direct proof, without referring to the Riemann-Lebesgue lemma (but following its proof) would be this. Given $\epsilon>0$, split the integral into $[0,a]$ and $[a,1]$, where $a$ is chosen so that $a(1-\log a)<\epsilon/2$. This choice is made so that $$ \left|\int_0^a \log x \sin nx \, dx\right| \le \int_0^a |\log x|\, dx = x(1-\log x )\bigg|_0^a = a(1-\log a)<\frac{\epsilon}{2} $$
for all $n$.

Over $[a,1]$, integrate by parts using $u=\log x$ and $v = -\cos nx/n$;
$$ \int_a^\pi \log x \sin nx \, dx = \int_a^\pi u\,dv = uv\,\bigg|_a^\pi - \int_a^\pi v\,du $$ where both terms tend to zero as $n\to\infty$ because $v\to 0$ uniformly. Hence, for all sufficienly large $n$ we have $$ \left|\int_a^1 \log x \sin nx \, dx\right| <\epsilon/2$$ which yields $$ \left|\int_0^1 \log x \sin nx \, dx\right| <\epsilon $$