A integral involving binomial coefficient Is there a function $f(x)$ ($f(x)$ and $n$ are irrelevant) that makes the following relationship true? 
$$\int\limits_{0}^1 x^{n-1}f(x)dx=\binom{2n}{n}=\frac{\Gamma(2n+1)}{\Gamma(n+1)\Gamma(n+1)}\quad ?$$
I tried to get the function by using the Merlin transform, but 
I failed.
 A: We know the Wallis formula where $$\int_0^{\frac{\pi}{2}} \cos^{2n}(x)dx=\int_0^{\frac{\pi}{2}} \sin^{2n}(x)dx=\frac{\pi}{2^{2n+1}}{2n\choose n}$$ thus one of the functions can be $f(x)=2^{2n}
\frac{cos^{2n}(\frac{\pi}{2}x)}{x^{n-1}}$ there can be many other functions. Look up the Wallis formula for sin and cosine raised to integer powers.
A: Looking for a $f(x)$ independent from $n$, 
let's put the whole into discrete form, as a Riemann Sum and thus
in matrix form
$$
\eqalign{
  & \int_0^1 {x^{\,n} f(x)\,dx} \quad  \to \quad \sum\limits_{0\, \le \,k\, \le \,h - 1} {\left( {{k \over h}} \right)^{\,n} f\left( {{k \over h}} \right){1 \over h}}  =   \cr 
  &  = {1 \over {h^{\,n + 1} }}\sum\limits_{0\, \le \,k\, \le \,h - 1} {k^{\,n} \varphi _h \left( k \right)}  =   \cr 
  &  = \left( {{1 \over {h^{\,n + 1} }} \circ {\bf I}_{\,h} } \right)\left( {\matrix{
   1 & 1 & 1 & 1 &  \cdots   \cr 
   0 & 1 & {2^1 } & {3^1 } &  \cdots   \cr 
   0 & 1 & {2^2 } & {3^2 } &  \cdots   \cr 
    \vdots  &  \vdots  &  \vdots  &  \vdots  &  \ddots   \cr 
   {\left( {h - 1} \right)^0 } & {\left( {h - 1} \right)^1 } & {\left( {h - 1} \right)^2 } & {\left( {h - 1} \right)^3 } &  \cdots   \cr 
 } } \right)\left( {\matrix{
   {\varphi _h \left( 0 \right)}  \cr 
   {\varphi _h \left( 1 \right)}  \cr 
    \vdots   \cr 
    \vdots   \cr 
   {\varphi _h \left( {h - 1} \right)}  \cr 
 } } \right) = \left( {\matrix{
   1  \cr 
   {\left( \matrix{
  2 \cr 
  1 \cr}  \right)}  \cr 
    \vdots   \cr 
    \vdots   \cr 
   {\left( \matrix{
  2h - 2 \cr 
  h - 1 \cr}  \right)}  \cr 
 } } \right) \cr} 
$$
where $\left( {{1 \over {h^{\,n + 1} }} \circ {\bf I}_{\,h} } \right)$ is a diagonal matrix with elements  $1/h^{n+1}$
The Vandermonde matrix is invertible, so the system is solvable, leading to a $\varphi _h(k)$.
It remains to see if for $h \to \infty$ $\varphi _h(k/h)$ tends to a definite function.
But applying the formula for the Vandermonde Determinant it is easy to see that the determinant of the two
LHS matrices diminishes rapidly with $h$, so that the determinant of the inverse is rapidly increasing, and so
does the absolute value of $\varphi (k)$, which moreover alternates in sign.
So we can conclude that    

a $f(x)$, independent of $n$, does not exist.

A: Suppose you have a Taylor series: $$f(x) = \sum_{k\ge 0}a_kx^k$$
Then the integral:
$$\int_0^1 x^{n-1}f(x)dx = \int_0^1 \sum_{k\ge 0} a_kx^{k+n-1}dx = \left(\sum_{k\ge 0} \dfrac{a_k}{k+n}\right)$$
Note: This integral is only defined for $n\ge 1$. So, just choose any infinite sum that gives you $\dbinom{2n}{n}$, and you have your answer.
