I have been trying to determine the series expansion of the beta function, but so far I haven't been successful. The two results I wish to obtain are the following:

$$ B(x,y) = \sum_{n=0}^{\infty} \frac{\binom{n-y}{n} }{x+n} $$

and \begin{equation} B(x,y) = \sum_{n=0}^{\infty} \frac{1}{n! (n+x)} \frac{\Gamma(n-y+1)}{\Gamma(1- y)} \end{equation}

For the second result, it follows directly from (Eq. 16.79 of Basic Concepts of String Theory)

$$B(x,y) = \frac{1}{\Gamma(1-y)} \int_{0}^{\infty} ds \int_0^1 dz \; s^{-y} e^{-(1-z)s} z^{x-1} $$

However, I don't understand what were the steps to obtain this equality.

Therefore, any hints on how to proceed in obtaining the first result and this last equation would be highly appreciated.


For the first identity, you can show that your expression for the beta function is equivalent to the integral formula

$$ B(x,y) = \int_0^1 t^{x-1} (1-t)^{y-1} dt $$ Then expand the second factor in the integrand,

$$ B(x,y) = \int_0^1 t^{x-1} \sum_{k=0}^{\infty} (-1)^k \left( \begin{array}{c} y-1 \\ k \end{array} \right) t^k dt $$

Given that each term of the summand is measurable we can use apply the monotone convergence theorem and switch the orders of integration and summation to give

$$ B(x,y) = \sum_{k=0}^{\infty} (-1)^k \left( \begin{array}{c} y-1 \\ k \end{array} \right) \int_0^1 t^{x+k-1} dt $$ $$ = \sum_{k=0}^\infty \left( \begin{array}{c} k-y \\ k \end{array} \right) \frac{1}{x+n} $$

where in the last line we used an identity for the binomial coefficient and computed the integral.

The second result is merely a restatement of the first, writing the binomial coefficient in terms of Gamma functions.


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