# binomial integral

#### jacks

##### Well-known member
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
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
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

• jacks

#### CaptainBlack

##### Well-known member
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

• Mr Fantastic

#### chisigma

##### Well-known member
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
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
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
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

$\chi$ $\sigma$

• sbhatnagar and jacks

#### chisigma

##### Well-known member
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

$\chi$ $\sigma$

• sbhatnagar