Show that $\int_{0}^{\infty} u^n u^{-\log u}f(\log(u))\,\mathrm{d}u=0$ where $f$ is odd satisfying $f(u+1/2)=\pm f(u)$ How to show that, if $f$ is odd satisfying $f(u+1/2)=\pm f(u)$, then
$$\int_{0}^{\infty} u^n u^{-\log u}f(\log(u))\,\mathrm{d}u=0$$
where $n\in \mathbb{Z}$?
I found this in ScienceDirect page, just page no. 1, where it again took from a work by Stieltjes. I checked it myself, but it did not provide any proof. I tried to verify it myself, but I do not know how or where to start, because $f$ is arbitrary with a certain condition.
 A: Here, we consider the substitution $u = e^t$. By simple calculation, the left hand size is:
$\int_{-\infty}^\infty e^{(n+1)t-t^2}f(t)dt$.
Since $n$ is arbitrary in $\mathbb{Z}$, we only need to consider $L(n) := \int_{-\infty}^\infty e^{nt-t^2}f(t)dt = e^{\frac{n^2}{4}}\int_{-\infty}^\infty e^{-(t-\frac{n}{2})^2}f(t)dt = e^{\frac{n^2}{4}}\int_{-\infty}^\infty e^{-t^2}f(t+\frac{n}{2})dt$.
By your condition, $f(t+\frac{n}{2}) = (-1)^n f(t) = (-1)^{n+1}f(-t) = (-1)^{2n+1}f(-t+\frac{n}{2})$, hence $f(t+\frac{n}{2})$ is an odd function with respect to $t$. Therefore, the integrated is odd, hence $L(n) = 0$ for $\forall z \in \mathbb{Z}$.
A: So first of all we make the straightforward substitution $x=\log u$ hence $e^xdx=du$ which leads to
$$
I=\int_0^\infty u^nu^{-\log u}f(\log u)=\int_{-\infty}^\infty e^{nx}e^{-x^2}f(x)e^xdx\\
=\int_{-\infty}^\infty e^{-(x^2-nx-x)}f(x)dx\\
=e^{\left(\tfrac{n+1}{2}\right)^2}\int_{-\infty}^\infty e^{-\left(x-\tfrac{n+1}{2}\right)^2}f(x)dx\\
$$
No we shift the variable $x$ so the gaussian is centered around $0$ via $y=x-\tfrac{n+1}{2}$ and make use of periodic nature of $f$.
$$
I=e^{\left(\tfrac{n+1}{2}\right)^2}\int_{-\infty}^\infty e^{-y^2}f(y+\tfrac{n+1}{2})dy\\
=e^{\left(\tfrac{n+1}{2}\right)^2}\int_{-\infty}^\infty e^{-y^2}(\pm1)^{n+1}f(y)dy\\
=(\pm1)^{n+1}e^{\left(\tfrac{n+1}{2}\right)^2}\underbrace{\int_{-\infty}^\infty e^{-y^2}f(y)dy}_{=0}
$$
Since the remaining integrand is a product of an even and an odd function the last integral yields $0$.
