Gamma integrals












4












$begingroup$


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*}



$n>0, ple n+0.5$.



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]







gamma-function special-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    The equation in your Mathematica code does not match the one in TeX. :)
    $endgroup$
    – J. M. is a poor mathematician
    Aug 31 '10 at 5:02










  • $begingroup$
    To simplify things a bit, your $f(n,p)$ is $int_{-1/2}^p binom{n}{x}mathrm{d}x$
    $endgroup$
    – J. M. is a poor mathematician
    Aug 31 '10 at 5:10










  • $begingroup$
    good catch, fixed
    $endgroup$
    – Yaroslav Bulatov
    Aug 31 '10 at 5:27










  • $begingroup$
    You missed my hint: check your definition for $f(n,p)$
    $endgroup$
    – J. M. is a poor mathematician
    Aug 31 '10 at 6:07










  • $begingroup$
    @J. M. : Is there a typo in the $ple n$ condition, in the definition of $f(n,p)$ ? In your example $f(n,n+1/2)$, $p=n+1/2>n$.
    $endgroup$
    – Américo Tavares
    Aug 31 '10 at 10:19


















4












$begingroup$


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*}



$n>0, ple n+0.5$.



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]







gamma-function special-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    The equation in your Mathematica code does not match the one in TeX. :)
    $endgroup$
    – J. M. is a poor mathematician
    Aug 31 '10 at 5:02










  • $begingroup$
    To simplify things a bit, your $f(n,p)$ is $int_{-1/2}^p binom{n}{x}mathrm{d}x$
    $endgroup$
    – J. M. is a poor mathematician
    Aug 31 '10 at 5:10










  • $begingroup$
    good catch, fixed
    $endgroup$
    – Yaroslav Bulatov
    Aug 31 '10 at 5:27










  • $begingroup$
    You missed my hint: check your definition for $f(n,p)$
    $endgroup$
    – J. M. is a poor mathematician
    Aug 31 '10 at 6:07










  • $begingroup$
    @J. M. : Is there a typo in the $ple n$ condition, in the definition of $f(n,p)$ ? In your example $f(n,n+1/2)$, $p=n+1/2>n$.
    $endgroup$
    – Américo Tavares
    Aug 31 '10 at 10:19
















4












4








4


2



$begingroup$


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*}



$n>0, ple n+0.5$.



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]







gamma-function special-functions







share|cite|improve this question











$endgroup$




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*}



$n>0, ple n+0.5$.



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]







gamma-function special-functions




gamma-function special-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 10:08









Glorfindel

3,41381930




3,41381930










asked Aug 31 '10 at 4:25









Yaroslav BulatovYaroslav Bulatov

1,88411526




1,88411526












  • $begingroup$
    The equation in your Mathematica code does not match the one in TeX. :)
    $endgroup$
    – J. M. is a poor mathematician
    Aug 31 '10 at 5:02










  • $begingroup$
    To simplify things a bit, your $f(n,p)$ is $int_{-1/2}^p binom{n}{x}mathrm{d}x$
    $endgroup$
    – J. M. is a poor mathematician
    Aug 31 '10 at 5:10










  • $begingroup$
    good catch, fixed
    $endgroup$
    – Yaroslav Bulatov
    Aug 31 '10 at 5:27










  • $begingroup$
    You missed my hint: check your definition for $f(n,p)$
    $endgroup$
    – J. M. is a poor mathematician
    Aug 31 '10 at 6:07










  • $begingroup$
    @J. M. : Is there a typo in the $ple n$ condition, in the definition of $f(n,p)$ ? In your example $f(n,n+1/2)$, $p=n+1/2>n$.
    $endgroup$
    – Américo Tavares
    Aug 31 '10 at 10:19




















  • $begingroup$
    The equation in your Mathematica code does not match the one in TeX. :)
    $endgroup$
    – J. M. is a poor mathematician
    Aug 31 '10 at 5:02










  • $begingroup$
    To simplify things a bit, your $f(n,p)$ is $int_{-1/2}^p binom{n}{x}mathrm{d}x$
    $endgroup$
    – J. M. is a poor mathematician
    Aug 31 '10 at 5:10










  • $begingroup$
    good catch, fixed
    $endgroup$
    – Yaroslav Bulatov
    Aug 31 '10 at 5:27










  • $begingroup$
    You missed my hint: check your definition for $f(n,p)$
    $endgroup$
    – J. M. is a poor mathematician
    Aug 31 '10 at 6:07










  • $begingroup$
    @J. M. : Is there a typo in the $ple n$ condition, in the definition of $f(n,p)$ ? In your example $f(n,n+1/2)$, $p=n+1/2>n$.
    $endgroup$
    – Américo Tavares
    Aug 31 '10 at 10:19


















$begingroup$
The equation in your Mathematica code does not match the one in TeX. :)
$endgroup$
– J. M. is a poor mathematician
Aug 31 '10 at 5:02




$begingroup$
The equation in your Mathematica code does not match the one in TeX. :)
$endgroup$
– J. M. is a poor mathematician
Aug 31 '10 at 5:02












$begingroup$
To simplify things a bit, your $f(n,p)$ is $int_{-1/2}^p binom{n}{x}mathrm{d}x$
$endgroup$
– J. M. is a poor mathematician
Aug 31 '10 at 5:10




$begingroup$
To simplify things a bit, your $f(n,p)$ is $int_{-1/2}^p binom{n}{x}mathrm{d}x$
$endgroup$
– J. M. is a poor mathematician
Aug 31 '10 at 5:10












$begingroup$
good catch, fixed
$endgroup$
– Yaroslav Bulatov
Aug 31 '10 at 5:27




$begingroup$
good catch, fixed
$endgroup$
– Yaroslav Bulatov
Aug 31 '10 at 5:27












$begingroup$
You missed my hint: check your definition for $f(n,p)$
$endgroup$
– J. M. is a poor mathematician
Aug 31 '10 at 6:07




$begingroup$
You missed my hint: check your definition for $f(n,p)$
$endgroup$
– J. M. is a poor mathematician
Aug 31 '10 at 6:07












$begingroup$
@J. M. : Is there a typo in the $ple n$ condition, in the definition of $f(n,p)$ ? In your example $f(n,n+1/2)$, $p=n+1/2>n$.
$endgroup$
– Américo Tavares
Aug 31 '10 at 10:19






$begingroup$
@J. M. : Is there a typo in the $ple n$ condition, in the definition of $f(n,p)$ ? In your example $f(n,n+1/2)$, $p=n+1/2>n$.
$endgroup$
– Américo Tavares
Aug 31 '10 at 10:19












1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    That looks even nastier than the original.
    $endgroup$
    – J. M. is a poor mathematician
    Aug 31 '10 at 7:27












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    That looks even nastier than the original.
    $endgroup$
    – J. M. is a poor mathematician
    Aug 31 '10 at 7:27
















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    That looks even nastier than the original.
    $endgroup$
    – J. M. is a poor mathematician
    Aug 31 '10 at 7:27














1












1








1





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 31 '10 at 7:10

























answered Aug 31 '10 at 5:27









Michael UlmMichael Ulm

1,2101810




1,2101810








  • 2




    $begingroup$
    That looks even nastier than the original.
    $endgroup$
    – J. M. is a poor mathematician
    Aug 31 '10 at 7:27














  • 2




    $begingroup$
    That looks even nastier than the original.
    $endgroup$
    – J. M. is a poor mathematician
    Aug 31 '10 at 7:27








2




2




$begingroup$
That looks even nastier than the original.
$endgroup$
– J. M. is a poor mathematician
Aug 31 '10 at 7:27




$begingroup$
That looks even nastier than the original.
$endgroup$
– J. M. is a poor mathematician
Aug 31 '10 at 7:27


















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