Is $\sum_{k=2}^\infty \frac{\cos((2k^2-k+7)x)}{k(\ln k)^2}$ a Fourier series?

Question

Is the following series a Fourier series?

$$\sum_{k=2}^\infty \frac{\cos((2k^2-k+7)x)}{k(\ln k)^2}$$

My attempt:

Suppose it is a Fourier series.

Suppose $f(x)=\displaystyle\sum_{k=2}^\infty \frac{\cos((2k^2-k+7)x)}{k(\ln k)^2}$.

By definition ,the coefficient of the Fourier series are $$a_k=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\cos(kx)\,dx$$ $$b_k=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\sin(kx)\,dx$$.

Also, we know that $$\int_{-\pi}^\pi \sin(mx)\cos(nx)\,dx=0~(m\neq n)$$ and

$$\int_{-\pi}^\pi \cos(mx)\cos(nx)\,dx=0~(m\neq n)$$.

So, $$a_k= \frac{1}{\pi}\int_{-\pi}^\pi \left(\sum_{k=2}^\infty \frac{\cos((2k^2-k+7)x)}{k(\ln k)^2}\right)\cos(kx)\,dx$$.

Let $2k^2-k+7=k$. There are no integer solutions. So, $a_k=0$.

Similarly, $b_k=0.$

We get a contradiction. So it is not a Fourier series.

Is this correct? Are there any other methods?

Answer 1

You are overusing the letter $k$. Here's what we can say:

$$ a_\ell = \frac{1}{\pi} \int_{-\pi}^\pi \left(\sum_{k=2}^\infty \frac{\cos((2k^2-k+7)x)}{k(\ln k)^2} \right)\cos(\ell x)\,\mathrm{d}x \\ =\sum_{k=2}^\infty \frac{1}{\pi k (\ln k)^2}\int_{-\pi}^{\pi} \cos((2k^2-k+7)x)\cos(\ell x)\, \mathrm{d}x. $$

Indeed, since the sum is already an infinite linear combination of $\cos(rx)$s for integers $r$, it is quite clearly a Fourier series, with $a_r$ being the sum of all $\frac{1}{k(\ln k)^2}$s for which $2k^2-k+7$ is $r$.

Answer 2

Note that $2k^2 - k + 7$ is an integer for each $k$, so this is a trigonometric series. Moreover, it converges uniformly and absolutely by the integral test. So this is indeed the Fourier series of a continuous periodic function with period $2\pi$.