Estimate ${\int_0^1 \sinh(ax) f(x) dx}/{\int_0^1 \cosh(ax) f(x) dx}$

Question

Let $f$ be a continuous function defined on $[0,1]$ which is not identically zero and such that $$ \int_0^1 f(x) dx = 0. $$ Consider, for "small" $a$, $$F(a) := \frac{\int_0^1 \sinh(ax) f(x) dx}{\int_0^1 \cosh(ax) f(x) dx}.$$

1. Is this definition valid, i.e. does there exist $a_0 > 0$ such that the denominator is $\neq 0$ for all $a \in (0,a_0)$?

2. If such $a_0$ exists, is $F$ bounded in $(0,a_0)$?

Answer 1

Write

$$ c_n = \int_{0}^{1} x^n f(x) \, \mathrm{d}x. $$

Then the following claim holds:

Proposition. If $f$ is continuous and not identically zero, then there exist an even integer $n_{e}$ and an odd integer $n_{o}$ such that $c_{n_e} \neq 0$ and $c_{n_o} \neq 0$.

This claim is an easy corollary of the Stone-Weierstrass Theorem. Now let $n_e$ and $n_o$ be as in the proposition, and assume in addition they are chosen as the smallest such ones. Now writing

$$ \begin{aligned} G(a) &:= \int_{0}^{1} \sinh(ax) f(x) \, \mathrm{d}x = \sum_{k=0}^{\infty} \frac{c_{2k+1}}{(2k+1)!} a^{2k+1}, \\ H(a) &:= \int_{0}^{1} \cosh(ax) f(x) \, \mathrm{d}x = \sum_{k=0}^{\infty} \frac{c_{2k}}{(2k)!} a^{2k}, \end{aligned} $$

we find that

$$ G(a) \sim \frac{c_{n_o}}{n_o!} a^{n_0} \qquad \text{and} \qquad H(a) \sim \frac{c_{n_e}}{n_e!} a^{n_e} $$

as $a \to 0$. In particular, this shows that there exists $a_0 > 0$ such that $H(a) \neq 0$ for $0 < |a| < a_0$, proving the well-definedness of $F(a)=G(a)/H(a)$ near $a = 0$. Moreover, since

$$F(a) \sim \text{constant} \cdot a^{n_o - n_e} \qquad \text{as} \quad a \to 0,$$

it follows that $F$ is bounded near $a = 0$ if and only if $n_e < n_o$.

Answer 2

As a suggestion, you may try using L'Hospital rule. For instance, if $\int^1_0 x^2 f(x)\,dx\neq0$ you get \begin{aligned} \lim_{a\rightarrow0}\frac{\int^1_0\sinh(ax)f(x)\,dx}{\int^1_0 \cosh(ax)f(x)\,dx}&= \lim_{a\rightarrow0}\frac{\int^1_0x\cosh(ax)f(x)\,dx}{\int^1_0 x\sinh(ax)f(x)\,dx}\\ &=\lim_{a\rightarrow0}\frac{\int^1_0x^2\sinh(ax)f(x)\,dx}{\int^1_0 x^2\cosh(ax)f(x)\,dx}=0 \end{aligned} The change of order of integration and differentiation can be explained by dominated convergence. That there is at least one integer $n\leq 1$ such that $\int^1_0x^{2n}f(x)\,dx>0$ follows from the Stone--Weierstrass theorem for $\{1,x^2\}$ separates points of $[0,1]$ and so the set of polynomials with only even powers is dense in $\mathcal{C}[0,1]$