Prove that the series $\sum_{n=2}^{\infty} \frac{\sin(nx)}{\log(n)}$ represented a function $f(x)$ that is not Lebesgue integrable.
My anwser: Suppose that $f(x)=\sum_{n=2}^{\infty} \frac{\sin(nx)}{\log(n)}$ is Lebesgue integrable. According to Parseval's theorem, there is integrable function $f$ such that $$\int_{-L}^{L} |f(x)|^{2} dx=2L \sum_{n=2}^{\infty} \frac{1}{\log^{2} n}.$$ Though, the series is not convergent.
Is my anwser correct?
$$f(x)=\sum_{n=2}^{\infty} \frac{\sin(nx)}{\log n}$$
Let $$g(x)=\frac{x}{2\pi}- \lfloor \frac{x}{2\pi}\rfloor = \frac12-\sum_{n=1}^\infty \frac{\sin(n x)}{\pi n}$$ If $f \in L^1(0,2\pi)$ then $\int_0^{2\pi} f(x)g(x)dx$ would converge. But $$\int_0^{2\pi} f(x)g(x)dx =\lim_{k \to \infty} \int_0^{2\pi} f(x)g_k(x)dx= \sum_{n=2}^\infty \frac{1}{\pi n} \frac{1}{\log n} \int_0^{2\pi} \sin^2(n x) dx= \infty $$
Where $g_k(x) = g \ast k e^{-\pi k^2 x^2} =\frac12-\sum_{n=1}^\infty \frac{\sin(n x)}{\pi n} e^{-\pi (2\pi n)^2/k^2}$ is a smooth version of $g$ needed to justify the inversion $\int \sum \sum = \sum\sum \int$.
Note a partial summation shows that $f(x) = \sum_{n=2}^{\infty} \Im(\sum_{m=2}^n e^{imx}) (\frac{1}{\log n}-\frac{1}{\log n+1})=\sum_{n=2}^{\infty}\Im(e^{2ix}\frac{1-e^{i(n-1) x}}{1-e^{ix}}) \frac{1/n+O(1/n^2)}{(\log n+1) (\log n)}$ converges and is continuous on $(0,2\pi)$.