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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.

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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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