Moments of the log-normal distribution

Counterexample

Let $f$ be the density of the standard log-normal distribution: $$f(x) = \frac{1}{x \sqrt{2\pi}} \exp\left(-\frac{(\ln x)^2}{2}\right) \quad (x > 0), \qquad f(x) = 0 \quad (x \le 0).$$ Its moments are $\int_0^\infty x^n f(x)\, dx = e^{n^2/2}$ for every integer $n \ge 0$. For a real number $a$ with $-1 \le a \le 1$, put $$f_a(x) = f(x)\bigl(1 + a \sin(2\pi \ln x)\bigr) \quad (x > 0), \qquad f_a(x) = 0 \quad (x \le 0).$$ Each $f_a$ is a probability density, because $f_a \ge 0$ and it has the same total mass as $f$ by the computation below. Check that $f_a$ has the same moments as $f$. It suffices to show $\int_0^\infty x^n f(x) \sin(2\pi \ln x)\, dx = 0$ for every integer $n \ge 0$. The substitution $x = e^y$ turns this integral into $\int_\R e^{ny} \varphi(y) \sin(2\pi y)\, dy$, where $\varphi(y) = e^{-y^2/2}/\sqrt{2\pi}$ is the standard normal density. Since $e^{ny} \varphi(y) = e^{n^2/2} \varphi(y - n)$, the substitution $y = u + n$ gives $$e^{n^2/2} \int_\R \varphi(u) \sin(2\pi u + 2\pi n)\, du = e^{n^2/2} \int_\R \varphi(u) \sin(2\pi u)\, du = 0,$$ because $n$ is an integer and the last integrand is odd. So the densities $f_a$, $-1 \le a \le 1$, are pairwise different and all have the moments $e^{n^2/2}$. Hence the moment problem for this moment sequence is not determinate: it is indeterminate.