Solving the functional equation $x[f(x+1)-f(x-1)]=1$ 
Possible Duplicate:
Solving the functional equation $f(x+1) - f(x-1) = g(x)$ 

How do I approach this problem $x[f(x+1)-f(x-1)]=1$.
 A: Pick any function $f(x)$ on $(0,2]$.  Then $f(x)$ on $(2,\infty )$ is determined-just step downward by 2's until you get into $(0,2]$.  A similar technique works for $x<0$ but you step up.
A: Let
$$
f(x)=\sum_{k=1}^\infty\left(\frac{1}{2k-1}-\frac{1}{2k+x-1}\right)\tag{1}
$$
then
$$
\begin{align}
f(x+1)-f(x-1)
&=\sum_{k=1}^\infty\left(\frac{1}{2k+x-2}-\frac{1}{2k+x}\right)\\
&=\lim_{N\to\infty}\left(\frac1x-\frac{1}{2N+x}\right)\\
&=\frac1x\tag{2}
\end{align}
$$
and
$$
\begin{align}
f(x)
&=\frac12\sum_{k=1}^\infty\left(\frac{1}{k-\frac12}-\frac{1}{k+\frac{x-1}{2}}\right)\\[6pt]
&=\frac12\psi\left(\frac{x+1}{2}\right)-\frac12\psi\left(\frac{1}{2}\right)\\
&=\frac12\psi\left(\frac{x+1}{2}\right)+\log(2)+\frac{\gamma}{2}\tag{3}
\end{align}
$$
where $\psi$ is the Digamma function.
Two functions defined by $f(x+1)-f(x-1)$ differby a function which is $2$-periodic. Thus, a function that is defined by $(2)$, would differ from $(3)$ by a $2$-periodic function.
Therefore,
$$
f(x)=\frac12\psi\left(\frac{x+1}{2}\right)+\varphi(x)\tag{4}
$$
where $\varphi(x)$ is any $2$-periodic function.
A: $x(f(x+1)-f(x-1))=1$
$f(x+1)-f(x-1)=\dfrac{1}{x}$
$x\to x+1$:
$f(x+2)-f(x)=\dfrac{1}{x+1}$
$x\to2x$:
$f(2x+2)-f(2x)=\dfrac{1}{2x+1}$
$f(2(x+1))-f(2x)=\dfrac{1}{2x+1}$
$f(2x)=\sum_x\dfrac{1}{2x+1}+\Theta_1(x)$, where $\Theta_1(x)$ is an arbitrary periodic function with unit period
Since $\lim_{x\to+\infty}\dfrac{1}{2x+1}=0$, so according to http://en.wikipedia.org/wiki/Indefinite_sum#Mueller.27s_formula, the result can be further simplified to
$f(2x)=\sum_{n=0}^\infty\left(\dfrac{1}{2n+1}-\dfrac{1}{2x+2n+1}\right)+\Theta_1(x)$, where $\Theta_1(x)$ is an arbitrary periodic function with unit period
$f(x)=\sum_{n=0}^\infty\left(\dfrac{1}{2n+1}-\dfrac{1}{x+2n+1}\right)+\Theta(x)$, where $\Theta(x)$ is an arbitrary periodic function with period $2$
