Integral equations Are there functions $f(z)$ that satisfy the following:
$$\int_1^\infty \frac{\mathrm{d}z}{f(z)}  = \frac{1}{\int_{1}^{\infty} f(z) \, \mathrm{d}z} $$
 A: 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.

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.
A: 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...
