Integrand of an integral equation. In the integral equation assuming $y\gg 0$ $$\large\int_{y^r}^{y^{2r}}f(x)dx=\frac {y^{2r}}2$$ what is a good candidate for $f(x)$ if $r\in\big(\frac12,1\big)$ holds?
Assume $f(x)>0$ and is monotone with $f(y^r)\rightarrow c= O(1)$ and $f(y^{2r})\rightarrow g(y^{2r})= O(y^{2r})$.
$f(x)=\frac x{y^{2r}}$ comes pretty close.
 A: I'll put the problem into a more general context (the present problem
being a special case):
Let $g:{\Bbb R}_+\rightarrow  {\Bbb R}$ be  Lipschitz continuous
and consider the problem of finding $f$ so that:
$$ \int_{y^\alpha}^y f(x)\; dx = g(y) $$
Suppose first that $g(1)=0$ and let
$F$ be a primitive of $f$. Then we want to solve
$$ F(y) - F(y^\alpha)= g(y) $$
which has the formal solution:
$$ F(y) = g(y)+F(y^\alpha) = g(y) + g(y^\alpha) + g(y^{\alpha^2}) = 
... =
  \sum_{k\geq 0} g(y^{\alpha^k}) .$$
As $k\rightarrow \infty$ we have $y^{\alpha^k}=1+\alpha^k \log(y)+ ...$ so a sufficient condition for the
series is that e.g. $g$ is Lipschitz continuous at $1$ and $g(1)=0$.
The solution in this case exists for all $y>0$.
Consider now for $y\neq 1$, $y>0$ the problem of finding $f_0$ so that:
$$ \int_{y^\alpha}^y f_0(x)\; dx = 1 $$
The solution is given by:
$$ F_0(y) = \frac{\log |\log y|}{-\log \alpha} $$
as is seen by inspection since:
$${\log |\log y|} - \log|\log (y^\alpha)| = 
{\log |\log y|} - \log \alpha (|\log y|) = -\log \alpha  $$
The general solution is thus (and we need $y\neq 1$ when $g(1)\neq 0$):
$$ F(y) = 
  \left( \sum_{k\geq 0} \left(g(y^{\alpha^k})-g(1) \right) \right) 
   + g(1)   \left( \frac{\log |\log y|}{\log 1/\alpha} \right) .
$$
To find $f$ it suffices that $g$ is continuously differentiable in which case:
$$ f(y) = F'(y)=
  \frac{1}{y} \left( \sum_{k\geq 0} g'(y^{\alpha^k}) 
    \alpha^k  y^{\alpha^k} \right) 
   + g(1)   \left( \frac{{\rm sign}(y)}{y |\log y|}
    \frac{1}{\log \frac{1}{\alpha}} \right) .
$$
The first sum converges when $g'$ is assumed continuous.
In the present context we are looking at 
$\alpha=1/2$ and $g(y)=y/2$ (details omitted).
A: With $t=y^r$ and $F$ the antiderivative of $f$, this is a functional equation
$$F(t^2)-F(t)=\frac{t^2}2.$$
Then with $p:=\log\log t$ (in base $2$) and $G(p):=F(2^{2^p})$,
$$F(2^{2^{\log\log t+1}})-F(2^{2^{\log\log t}})=G(p+1)-G(p)=\frac{2^{2^{p+1}}}2$$ which is a simple recurrence.
Then $G$ can be any function $g(p)$ defined in the range $p\in[0,1)$ and 
$$G(p)=g(\{p\})+\frac12\sum_{k=0}^{\lfloor p\rfloor}2^{2^{\{p\}+k+1}}$$ elsewhere.
By backsubstitution,
$$F(t)=g(\{\log\log t\})+\frac12\sum_{k=0}^{\lfloor \log\log t\rfloor}\left(2^{2^{\{\log\log t\}}}\right)^{k+1}$$ and $f(t)$ comes by derivation (painfully).
A: $y>1$ ?
$$
f(x) =  g(x) +1/2 
\int_{y^r}^y g(x) dx = 1/2 * y^r
$$
And so on... 
