Integral equations

Question

Are there functions $f(z)$ that satisfy the following:

$$\int_1^\infty \frac{\mathrm{d}z}{f(z)} = \frac{1}{\int_{1}^{\infty} f(z) \, \mathrm{d}z} $$

Answer 1

I will assume that $f > 0$. Then by the Cauchy-Schwarz inequality, for $a > 1$,

$$ \left( \int_{1}^{a} 1 \, \mathrm{d}x \right)^2 = \left( \int_{1}^{a} f(x)^{1/2} f(x)^{-1/2} \, \mathrm{d}x \right)^2 \leq \left( \int_{1}^{a} f(x) \, \mathrm{d}x \right)\left( \int_{1}^{a} \frac{1}{f(x)} \, \mathrm{d}x \right). $$

So by letting $a \to \infty$,

$$ \left( \int_{1}^{\infty} f(x) \, \mathrm{d}x \right)\left( \int_{1}^{\infty} \frac{1}{f(x)} \, \mathrm{d}x \right) = \infty. $$

In other words, it is impossible to make both integrals simultaneously finite.

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If we allow sign changes and interpret both integrals in improper integral sense, then we actually have an example.

In order to describe the example, for each bounded open interval $I = (a, b)$ we define

$$ \phi_{I}(x) = \frac{\sqrt{(x-a)(b-x)}}{b-a}. $$

Then it is not hard to check that

$$ \int_{I} \phi_I(x) \, \mathrm{d}x = \frac{\pi}{8}(b-a) \qquad\text{and}\qquad \int_{I} \frac{1}{\phi_I(x)} \, \mathrm{d}x = \pi(b-a). $$

Now let $H_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n} = \sum_{k=1}^{n} \frac{1}{k} $ denote the $n$th harmonic number, and define

$$ f(x) = \begin{cases} (-1)^{n-1} A \phi_{(H_n, H_{n+1})}(x) & \text{if $x \in (H_n, H_{n+1})$ for some $n$}, \\ 0, & \text{otherwise}, \end{cases} $$

where $A$ is a non-zero constant to be determined later. Then it follows that

$$ \int_{1}^{\infty} f(x) \, \mathrm{d}x = \frac{\pi A}{8} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n+1} = \frac{\pi}{8}(1-\log 2) A $$

and

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

So by solving the equation

$$ \left( \frac{\pi}{8}(1-\log 2) A \right) \left( \pi (1-\log 2) A \right) = 1, $$

we find that $f$ with $A = \frac{\sqrt{8}}{\pi(1-\log 2)}$ satisfies the given condition.

Remark. A similar construction should work if we begin with some function $\phi$ over an interval $I$ such that both $\int_{I} \phi$ and $\int_{I} \frac{1}{\phi}$ converge.

Answer 2

Partial Answer: There are no solutions for $f$ continuous.

First we can say WLOG that $f>0$ in $[1,\infty]$ (since $f$ has to be of constant sign) and so therefore for both integrals to converge we have, on the LHS and RHS respectively: $$\lim_{x\to\infty}\frac{1}{f(x)}=0\quad \text{and}\quad \lim_{x\to\infty}f(x)=0$$

which is a contradiction.

Seems much harder for discontinuous functions...