Proof that function is not Lebesgue integrable.

Question

Dears,

my aim is to prove that function given as $$ f(x)=\sum\limits_{n=2}^{\infty} \frac{\sin(nx)}{\ln(n)} $$ is not Lebesgue integrable. That exercise is on sheet related with Fourier Transform so probably that is good way, but I don't know how to do that.

If you have suggestions, you're welcome.

Have a nice day,

Answer 1

$f^2$ is not integrable over $[0,2\pi]$ due to Parseval's Identity and divergence of $\frac 1 {(\ln x)^2}$ (Bertrand's series). Then there is some non-negligible set $A \subset [0,2 \pi]$ where $|f| >M$ for some real positive number $M$. Since $f$ is $2 \pi$ periodic : $|f(x)| > M$ for every $x \in \cup_{k \in \mathbb{Z}} \left( A+k2 \pi \right)$ and $f$ cannot be integrable over $\mathbb{R}$.

Note : the integrability over $[0,2 \pi]$ is still a mistery to me, however it does not seem to be part of your question. But if someone has an idea ...

Answer 2

By the Dirichlet Test, we know that $f$ converges uniformly over $[\delta, \pi-\delta]$ for any $\delta \in (0, \pi/2)$. So

$$ \int_{\delta}^{\pi-\delta} f(x) \, \mathrm{d}x = \sum_{n=2}^{\infty} \int_{\delta}^{\pi-\delta} \frac{\sin(nx)}{\log n} \, \mathrm{d}x = \sum_{n=2}^{\infty} \frac{1+(-1)^{n-1}}{n\log n}\cos(n\delta). $$

Now, fix a large even integer $N \geq 1$ and consider $\delta = \pi/N$. Then we claim that

$$ \sum_{n=N}^{\infty} \frac{1+(-1)^{n-1}}{n\log n}\cos(n\pi/N) = \mathcal{O}\left(\frac{1}{\log N}\right). \tag{*}$$

Assuming $\text{(*)}$ for a moment, we get

\begin{align*} \int_{\pi/N}^{\pi(1-1/N)} f(x) \, \mathrm{d}x &= \sum_{\substack{1 < n < N \\ n\text{ odd}}} \frac{2}{n \log n}\cos(n\pi/N) + \mathcal{O}\left(\frac{1}{\log N}\right) \\ &\geq \sum_{\substack{1 < n < N/3 \\ n\text{ odd}}} \frac{2\cos(\pi/3)}{n \log n} + \mathcal{O}\left(\frac{1}{\log N}\right). \end{align*}

Using this, it is easy to check that

$$ \lim_{N\to\infty} \int_{\pi/N}^{\pi(1-1/N)} f(x) \, \mathrm{d}x = \infty. $$

In view of Dominated Convergence Theorem, this cannot happen if $f$ is Lebesgue-integrable on $[0, \pi]$, and so, $f$ is not Lebesgue-integrable on $[0, \pi]$.

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Proof of $\text{(*)}$. By writing $N = qN + r$ and grouping terms according to the value of $r$,

\begin{align*} \sum_{n=N}^{\infty} \frac{1+(-1)^{n-1}}{n\log n}\cos(n\pi/N) &= \sum_{\substack{0 \leq r < N \\ r \text{ odd}}} \left( \sum_{q=1}^{\infty} (-1)^q \frac{2}{(qN+r)\log (qN+r)} \right)\cos(r\pi/N) \end{align*}

Now, for $n \geq N$, Mean-Value Theorem tells that

\begin{align*} \left| \frac{1}{n \log n} - \frac{1}{(n+N)\log(n+N)} \right| \leq \frac{2N}{n^2\log n}, \end{align*}

and so,

$$ \left| \sum_{q=1}^{\infty} (-1)^q \frac{2}{(qN+r)\log (qN+r)} \right| \leq \frac{C}{N\log N} $$

uniformly in $q$ and $r$. Therefore

$$ \left| \sum_{n=N}^{\infty} \frac{1+(-1)^{n-1}}{n\log n}\cos(n\pi/N) \right| \leq \frac{C}{\log N} \sum_{r=0}^{N-1} \frac{|\cos(r\pi/N)|}{N} \leq \frac{C'}{\log N} $$

for some absolute constant $C'$.