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Is anything known about these integrals? Textbook suggestions are welcome

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

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

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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The equation in your Mathematica code does not match the one in TeX. :) –  J. M. Aug 31 '10 at 5:02
    
To simplify things a bit, your $f(n,p)$ is $\int_{-1/2}^p \binom{n}{x}\mathrm{d}x$ –  J. M. Aug 31 '10 at 5:10
    
good catch, fixed –  Yaroslav Bulatov Aug 31 '10 at 5:27
    
You missed my hint: check your definition for $f(n,p)$ –  J. M. Aug 31 '10 at 6:07
    
@J. M. : Is there a typo in the $p\le n$ condition, in the definition of $f(n,p)$ ? In your example $f(n,n+1/2)$, $p=n+1/2>n$. –  Américo Tavares Aug 31 '10 at 10:19

1 Answer 1

up vote 1 down vote accepted

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. –  J. M. Aug 31 '10 at 7:27

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