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I could use somebody's help to understand a calculation:

$$\begin{array}{lcl} \sum\limits_{s=0}^{\infty} z^s\int_0^1 \frac{t(t+1)..(t+s-1)}{s!}dt &=& \sum\limits_{s=0}^{\infty} z^s \int_0^1 {{t+s-1}\choose{s}}dt \\ &=& \sum\limits_{s=0}^{\infty} z^s(-1)^s \int_0^1 {{-t}\choose{s}}dt\\ &=& \int_0^1\sum\limits_{s=0}^{\infty} {{-t}\choose{s}}(-z)^sdt\\ &=& \int_0^1(1-z)^{-t}dt \end{array} $$

I can take the first equality to be a definition, even if I knew this kind of expression for the binomial coefficients only for integers ${n}\choose{k}$.The second equality is a bit mysterious to me. The inversion of the summation and the integral I would have done it from the beginning provided $z \geq 0$ by monotone convergence theorem. The last step is also not clear to me.

Any help will be appreciated, thank you!

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Let's go step by step and see what the calculation is.

1.First is simply using the definition of $\binom{n}{r}$

  1. The second one used $\int_a^b f(x)dx= \int_a^b f(a+b-x) dx$, which gives us $\binom{s-t}{s}$, which clearly is $(-1)^s \binom{-t}{s}$

  2. The third step simply involves taking the entire thing together, which brings the summation inside the integral.

  3. The last uses the binomial theorem for negative powers$ (1+x)^{-r}=1+rx+ \frac{r(r-1)x}{2}...$ or simply $\sum_{r=1}^{\infty}\binom{-n}{r} (x)^r$

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  1. The interchange of infinite summation and integration is possible if $|z|<1$, this is called uniform continuity.
  2. If you have an expression of the form $\binom{-k}{s} $, just set $-k=\alpha$ and use the Binomial expansion (As it's done in your first step). You get $(-1)^s \binom{k+s}{s}$.
  3. This is called Generalized Binomial expansion: $(1+a)^{-r} = \sum_{k=0}^{\infty} \binom{-r}{k} a^k$. This is valid for $|a|<1$.
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