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Is anything known about these integrals? Textbook suggestions are welcome

$\begin{equation*} f(n,p)=\int_{x=-0.5}^p \frac{n!}{x!(n-x)!} dx, \end{equation*}$

$n>0, p\le n+0.5$.

For instance, as $n$ grows $2^n-f(n,n+1/2)$ seems to be positive and monotonically decreasing...how would I find the limit?

 ListPlot[Table[2^n - NIntegrate[n!/(x! (n - x)!), {x, -1/2, n + 1/2}], {n, Range[30]}], PlotRange -> All] 

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    ok changed it....btw restriction p<=n+1/2 is pretty arbitrary, I picked it simply because that range seemed interesting2010-08-31

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Not a full solution, but maybe enough for you to get a handle on the problem:

Using $\Gamma (x+1) = x \Gamma (x)$ and $\Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin \pi x}$, one obtains $ \Gamma(x + 1) \Gamma(n - x + 1) = \frac{(-1)^n \pi}{\sin \pi x} \prod_{k=0}^{n} (x - k) . $ Thus, $ f(n, p) = \frac{(-1)^n n!}{\pi} \int_{-0.5}^p \frac{\sin \pi x}{\prod_{k=0}^n (x-k)} d x . $ Using partial fractions, you can now transform $f$ into a sum of values of the sine integral.

Addendum: Computing the sums yields $ f(n, n+\frac{1}{2}) = \frac{2}{\pi} \sum_{k = 0}^n \binom{n}{k} Si (\pi k + \frac{\pi}{2}) $ Note, that $Si (x) = \frac{\pi}{2} + O(x^{- 2})$. However, I have no idea how to leverage this to find the limit.

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    That looks even nastier than the original.2010-08-31