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For what function (or functions) is the following true:

1) $f(x)$ is positive for $x>0$

2) $\lim\limits_{x\to 0}{f(x)} = \infty$

3) $\lim\limits_{x\to\infty}f(x) = 0$

4) $\int_{0}^{\infty} {f(x)} dx = C$

5) $f(x)$ is symmetric over $y=x$

6) $f(x)$ isn't written in case structure

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I am not sure I wrote the condition with an integral in the right way. If it is what you meant? – Ilya Aug 15 '11 at 16:51
If condition $4$ were $\int_0^{\infty}f(x)\,dx = C$, I think that taking any function defined on $[1,\infty)$ with $f(1)=1$, $f$ strictly decreasing, and $\int_1^{\infty}f(x)\,dx$ converges, and then extending to $(0,1)$ to ensure symmetry, will do. – Arturo Magidin Aug 15 '11 at 16:59
6) is a nonsense condition, why do you want that? – Jonas Teuwen Aug 15 '11 at 17:33
@bob: "$f$ is defined by exactly one equation" is not a well-defined notion; condition (6) is either empty or incoherent. – Arturo Magidin Aug 15 '11 at 18:09
@bob: You are still not giving a well-defined notion. What does it mean for a function to "contain conditional equations"? These are not well-defined concepts! – Arturo Magidin Aug 15 '11 at 18:38
up vote 9 down vote accepted

A simple general recipe for these conditions is the implicit equation $$g(x)g(y) = 1,$$ with $g$ being an increasing function such that $g(0) = 0$ and $g(x) \to \infty$ as $x \to \infty$. Then $f$ is given by $$f(x) = g^{-1}\left(\frac 1 {g(x)}\right).$$

To get a finite integral for $\int _0^\infty f(x)$ under the given conditions, all you need is for $f(x)$ to decay faster than $1/x$ as $x \to \infty$. This is guaranteed as long as, informally, $g(x)$ grows "faster" towards $\infty$ as $x \to \infty$ than it decays to $0$ as $x \to 0$. (Formally, if $g(x)$ goes as $x^n$ as $x\to\infty$ and as $x^m$ as $x\to 0$, then you need $n > m$.)

You can recover Ilmari's example by taking $$g(x) = \begin{cases} x & \text{if }0 < x \le 1, \\ x^2 & \text{if }1 < x. \end{cases}$$

For a smooth example, let $g(x) = e^x - 1$. Then you get $$f(x) = -\ln\left(1 - e^{-x}\right),$$ for which WolframAlpha reports that $\int_0^\infty f(x) = \pi^2/6.$

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To adjust the value of the integral of $f$ at a given value $C$, one can simply use the solutions $f_a(x)=a^{-1}f(ax)$ where $a$ is positive. The integral of $f_a$ being $\pi^2/(6a^2)$, it is $C$ for $a=\pi/\sqrt{6C}$. – Did Aug 15 '11 at 21:33

As a particular example of a function satisfying Arturo Magidin's conditions,

$$f(x) = \begin{cases} 1/\sqrt x & \text{if }0 < x \le 1 \\ 1/x^2 & \text{if }1 < x \end{cases}$$

ought to work. In particular, $\displaystyle \int_0^\infty f(x)\;dx = 3$ and $f(f(x)) = x$.

Addendum: If you don't want a piecewise defined function, you could write $f$ above as $f(x) = \exp\;g(\log x)$, where

$$g(u) = -\frac34 |u| - \frac54 u.$$

Of course, this is just a notational trick; $f$ still has a "kink" at $x=1$ due to the non-differentiability of $|u| = |\log x|$ there. However, it's a trick that points in a useful direction: if you also want the function to be everywhere differentiable, you can replace $|u| = \sqrt{u^2}$ in the definition of $g$ above with the hyperbola $\sqrt{1+u^2}$ to get

$$\tilde g(u) = -\frac 34 \sqrt{1 + u^2} - \frac 54 u,$$

and thus

$$\tilde f(x) = \exp \left( -\frac 34 \sqrt{1 + (\log x)^2} - \frac 54 \log x \right).$$

This function $\tilde f$ still satisfies $\tilde f(\tilde f(x)) = x$, and since $\tilde f(x) < f(x)$ for all $x$, we know that its integral from 0 to infinity must be less than 3. (Actually, it is about 2.19574343.)

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This is correct, but I need f(x) to be represented by only one equation. Edited for reference. – user474632 Aug 15 '11 at 17:27
Is a piecewise definition consistent with condition (6) in the question. Disambiguating condition (6) is a tough one; all the rest are simple. – Michael Hardy Aug 15 '11 at 17:48
"One equation" is a poorly-defined condition, but you might ask for $f$ to be analytic. The first example that comes to mind (of an involution on $(0,\infty)$ that is analytic) is $f(x) = 1/x$, but of course that does not satisfy the integrability condition. – Robert Israel Aug 15 '11 at 18:14
@bob: I wonder why you have the restriction to "no piecewise functions"? I showed here how to "join up" a piecewise equation, and if you're allowing the construction $\sqrt{x^2}$, well... – J. M. Aug 16 '11 at 4:14
That's definitely the term my engrish was unable to translate (I accepted this answer because it is correct for both of my misspoken questions). If anyone knows how to meet these criteria with a non-piecewise function, I'd like to know (I believe its construction is not possible within the limitations of exponential form). – user474632 Aug 16 '11 at 10:38

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