I don't fully understand what you are asking. In any case, here is a different way to prove the theorem, it depends heavily on the Gamma function, and particular integrals. I doubt it is what you want, but it may be useful to someone:
We want to prove the generalized binomial theorem, namely that $$(1+x)^{\alpha}=\sum_{k=0}^{\infty}\binom{\alpha}{k}x^{k}$$where $\binom{\alpha}{k}=\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}.$ Since increasing $\alpha$ by an integer amount is simply multiplying by $(1+x)^{n}$, and is easily dealt with, we need only consider some interval of length $1$. Suppose $\alpha\in[-1,0).$ We can write $$\binom{\alpha}{k}=\frac{(-1)^{k}(-\alpha)(1-\alpha)\cdots(k-1-\alpha)}{k!}.$$Multiplying the top and bottom by $\Gamma\left(-\alpha\right),$and using the properties of Gamma gives $$\binom{\alpha}{k}=\frac{(-1)^{k}\Gamma(-\alpha+k)}{k!\Gamma(-\alpha)}.$$ Recall that for $s>0$, $\Gamma(s)=\int_{0}^{\infty}t^{s-1}e^{-t}dt.$ Then $$\sum_{k=0}^{\infty}\binom{\alpha}{k}x^{k}=\frac{1}{\Gamma(-\alpha)}\sum_{k=0}^{\infty}x^{k}\frac{(-1)^{k}\Gamma(-\alpha+k)}{k!}$$ and by using this definition of $\Gamma(s)$, and switching the order we have
$$\frac{1}{\Gamma(-\alpha)}\int_{0}^{\infty}e^{-t}t^{-\alpha-1}\sum_{k=0}^{\infty}x^{k}\frac{(-1)^{k}t^{k}}{k!}dt.$$Recognizing the series, we have$$\frac{1}{\Gamma(-\alpha)}\int_{0}^{\infty}e^{-t(1+x)}t^{-\alpha-1}dt.$$Substituting $u=t(1+x)$, this becomes $$(1+x)^{\alpha}\frac{\int_{0}^{\infty}e^{-u}u^{-\alpha-1}du}{\Gamma(-\alpha)}=(1+x)^{\alpha},$$which was the desired result.
Maybe that helps,
