A integral involving binomial coefficient

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.