What are some functions that respect the following criteria? : $f(1/x) = f(x)$ and $\int_{0}^{+\infty} f(x) dx = 1$ I'm looking for some functions that respect these six criteria:


*

*$f$ is defined on $[0 ; +\infty[$

*$f$ is differentiable everywhere in $[0 ; +\infty[$

*$f(0) = 0$

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

*$f(1/x) = f(x)$

*$\int_{0}^{+\infty} f(x) dx = 1$
 A: $$
\begin{align}
1
&=\int_0^\infty f(x)\,\mathrm{d}x\\
&=\int_0^1f(x)\,\mathrm{d}x+\int_1^\infty f(x)\,\mathrm{d}x\\
&=\int_0^1f(x)\,\mathrm{d}x+\int_0^1\frac{f\left(\frac1x\right)}{x^2}\,\mathrm{d}x\\
&=\int_0^1f(x)\left(1+\frac1{x^2}\right)\mathrm{d}x\tag{1}
\end{align}
$$
Any function that satisfies $(1)$ can be extended to $(1,\infty)$ via $f(x)=f\!\left(\frac1x\right)$.
A: By the reciprocal condition, $f$ is determined by its restriction to $[0,1]$.
We have
$$\int_1^\infty f(x)\,\mathrm dx=\int_0^1f(\tfrac 1x)x^{-2}\,\mathrm dx $$
so you need
$$\int_0^1(1+x^{-2})f(x)\,\mathrm dx=1. $$
There are a lot of such functions.
A: Your function must satisfy
$$1 = \int_0^1 f(x)\, dx + \int_1^\infty f(x)\,dx
= \int_0^1 f(x)\, dx + \int_0^1 f(1/x){dx\over x^2}
= \int_0^1 f(x) \left(1 + {1\over x^2}\right) dx.
$$
The mapping $x\mapsto x^p$, $x\in[0,1]$ can for any $p > 0$
be extended to satisify $f(x) = f(1/x)$, $x > 0$. If $f(x) = x^p$,
$x\in [0,1]$, the only $p$ that works is $p = 1 + \sqrt{2}$.
I think you can cook up some other functions as well tht do the job.
A: Let's search for some function like this. The usual way is to use the properties you want the function to satisfy and try to find some stronger conditions which lead you to an appropriate candidate. By the first and fourth properties, it suffices to know how $f$ is defined on $[0,1]$. Indeed, if $x\geq 1$, we have $f(x)=f(1/x)$, and $1/x\leq 1$.
Moreover, let's try to look at continuous functions, so the third property should be easily obtainable, and the second one follows from this as well (and the fourth property).
Let's look at the fifth property: If $f$ satisfies the required properties, then
$$1=\int_0^\infty f(x)dx=\int_0^1 f(x)dx+\int_1^\infty f(x)dx$$
Making the substitution $x=1/y$ in the second term, we get
$$1=\int_0^1 f(x)dx+\int_0^1\frac{1}{y^2}f(1/y)dy=\int_0^1 f(x)dx+\int_0^1\frac{1}{y^2}f(y)dy=\int_0^1(1+\frac{1}{x^2})f(x)dx$$
So if we have $(1+\frac{1}{x^2})f(x)=1$ in $[0,1]$, we are done.
Therefore, this leads us to define $f(x)=\frac{x^2}{1+x^2}$ if $0\leq x\leq 1$, and $f(x)=f(1/x)=\frac{1}{1+x^2}$ if $x\geq 1$. We have all the desired properties (the fact that $\int_0^\infty f(x)dx=1$ follows from the previous calculations, since we basically chose $f$ to satisfy this).
A: If $f$ is such a function then the function
$$g(t):=f(e^t)\qquad(-\infty<t<\infty)$$
is even, satisfies $$\lim_{t\to\pm\infty} g(t)=0\ ,\tag{1}$$ and
$$\int_{-\infty}^\infty g(t)e^t\>dt=\int_0^\infty f(x)\>dx=1\ .\tag{2}$$
As $g$ is even the condition $(2)$ can be rewritten as
$$\int_0^\infty g(t)\,\cosh(t)\>dt={1\over2}\ .\tag{3}$$
Conversely, if $g$ is even and satisfies $(1)\wedge(3)$ then
$$f(x):=g(\log x)\qquad(x>0)$$
has the required properties. A simple example would be the function
$$g(t):={1\over\cosh t}\>1_{|t|\leq1/2}\ .$$
