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binomial integral

jacks

Well-known member
Apr 5, 2012
226
for a non nagative integer $n$, If $\displaystyle I_{n}=\int_{0}^{1}\binom{x}{n}dx$, then $I_{n}=$

where $\displaystyle \binom{n}{r} = \frac{n!}{r!.(n-r)!}$
 

Mr Fantastic

Member
Jan 26, 2012
66
for a non nagative integer $n$, If $\displaystyle I_{n}=\int_{0}^{1}\binom{x}{n}dx$, then $I_{n}=$

where $\displaystyle \binom{n}{r} = \frac{n!}{r!.(n-r)!}$
This does not make sense because on the one hand I assume x is meant to be continuous over the interval [0, 1] and yet on the other hand the combinatorial as you have defined it is only valid for integer values of x.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
This does not make sense because on the one hand I assume x is meant to be continuous over the interval [0, 1] and yet on the other hand the combinatorial as you have defined it is only valid for integer values of x.
May be for real \(x\) and \(r\) a natural number he intends:
\[{x \choose r}=\frac{\Gamma(x+1)}{r!\;\Gamma(x-r+1)}\]

Though my money is on asking the wrong question.

CB
 

CaptainBlack

Well-known member
Jan 26, 2012
890
for a non nagative integer $n$, If $\displaystyle I_{n}=\int_{0}^{1}\binom{x}{n}dx$, then $I_{n}=$

where $\displaystyle \binom{n}{r} = \frac{n!}{r!.(n-r)!}$
Can you please post the original, or full question? As MrF points out as asked this makes no sense so we suspect this is not the full or actual question.

CB
 

chisigma

Well-known member
Feb 13, 2012
1,704
According to...

http://mathworld.wolfram.com/BinomialCoefficient.html

... the definition of the factorial function as...

$\displaystyle z!=\int_{0}^{\infty} t^{z}\ e^{-t}\ dt$ (1)

... allows the definition of binomial coefficient as...

$\displaystyle \binom {x}{y}= \frac{x!}{y!\ (x-y)!}$

... where x and y are, in most general case, complex numbers...

Kind regards

$\chi$ $\sigma$
 

CaptainBlack

Well-known member
Jan 26, 2012
890
According to...

http://mathworld.wolfram.com/BinomialCoefficient.html

... the definition of the factorial function as...

$\displaystyle z!=\int_{0}^{\infty} t^{z}\ e^{-t}\ dt$ (1)

... allows the definition of binomial coefficient as...

$\displaystyle \binom {x}{y}= \frac{x!}{y!\ (x-y)!}$

... where x and y are, in most general case, complex numbers...

Kind regards

$\chi$ $\sigma$
Which is the same thing as replacing them by gamma functions.

CB
 

awkward

Member
Feb 18, 2012
36
We can avoid reference to the gamma function if we regard \( \binom{x}{n} \) as a polynomial when n is non-negative integer:
\( \binom{x}{n} = \frac{x (x-1) (x-2) \cdots (x-n+1)} {n!} \)

See "Binomial coefficients as polynomials" in
http://en.wikipedia.org/wiki/Binomial_coefficient
 

chisigma

Well-known member
Feb 13, 2012
1,704
Very well!... now that is seems that $\displaystyle \binom{x}{n}$ is a polynomial of degree n in x, we can proceed to the computation of $\displaystyle I_{n}=\int_{0}^{1} \binom{x}{n}\ dx$. To do that let's start from the well known binomial series expansion...

$\displaystyle (1+t)^{x}= \sum_{n=0}^{\infty} \binom{x}{n}\ t^{n}$ (1)

... and then we use (1) to arrive to the identity...

$\displaystyle \int_{0}^{1} (1+t)^{x}\ dx= \frac{t}{\ln (1+t)} = \sum_{n=0}^{\infty} t^{n}\ \int_{0}^{1} \binom{x}{n}\ dx$ (2)

But the (2) is the McLaurin expansion of $\displaystyle \frac{t}{\ln (1+t)}$ so that is...

$\displaystyle I_{n}=\int_{0}^{1} \binom{x}{n}\ dx= \frac{1}{n!}\ \lim_{t \rightarrow 0} \frac{d^{n}}{d x^{n}}\ \frac{t}{\ln (1+t)}$ (3)

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
As seen in the previous post the $\displaystyle I_{n}=\int_{0}^{1} \binom{x}{n}\ dx$ are the coefficients of the McLaurin expansion of the function $\displaystyle f(t)=\frac{t}{\ln (1+t)}$. Because is...

$\displaystyle \frac{\ln (1+t)}{t} = 1 -\frac{t}{2}+\frac{t^{2}}{3}-...$ (1)

... a comfortable way to compute them is to impose the identity...

$\displaystyle (I_{0}+I_{1}\ t +I_{2}\ t^{2}+...)\ (1 -\frac{t}{2}+\frac{t^{2}}{3}-...)=1$ (2)

... and from (2) we derive recursively...

$\displaystyle I_{0}=1$

$\displaystyle I_{1}=\frac{I_{0}}{2}=\frac{1}{2}$

$\displaystyle I_{2}=\frac{I_{1}}{2}-\frac{I_{0}}{3}=-\frac{1}{12}$

$\displaystyle I_{3}=\frac{I_{2}}{2}-\frac{I_{1}}{3}+\frac{I_{0}}{4}=\frac{1}{24}$

... and so one...

Kind regards

$\chi$ $\sigma$