Calculate Integral $\int_{0}^{1}x^{2014}\cdot \left(1-x\right)^{2014}dx$

[juantheron]

Calculation of Integral $\displaystyle \int_{0}^{1}x^{2014}\cdot \left(1-x\right)^{2014}dx$

 

[magneto1]

Spoiler

If we use the beta function:
\[
B(a+1,b+1) = \int_0^1 x^a (1-x)^b dx,
\]
and the property of
\[
B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}
\]
and the fact that for integer $n$, $\Gamma(n) = (n-1)!$, we have:
\[
B(a,b) = \frac{(a-1)!(b-1)!}{(a+b-1)!}.
\]

For this problem $a = b = 2013$, so $B(2013,2013) = \frac{(2014!)^2}{4029!}$.

 

[chisigma]

Calculation of Integral $\displaystyle \int_{0}^{1}x^{2014}\cdot \left(1-x\right)^{2014}dx$

[sp]Is...

$\displaystyle \int_{0}^{1} x^{2014}\ (1-x)^{2014}\ dx = \frac{1}{2015}\ (|x^{2015}\ (1-x)^{2014}|_{0}^{1} + 2014\ \int_{0}^{1} x^{2015}\ (1-x)^{2013}\ d x ) = \frac{2014}{2015} \int_{0}^{1} x^{2015}\ (1-x)^{2013}\ dx\ (1) $

Proceeding in this way we obtain...

$\displaystyle \int_{0}^{1} x^{2014}\ (1-x)^{2014}\ dx = \frac{2014 \cdot 2013 ... 2 \cdot 1}{2015 \cdot 2016 ... 4028 \cdot 4029} = \frac{(2014!)^{2}}{4029!}\ (2)$

[/sp]

Kind regards

$\chi$ $\sigma$

 

[juantheron]

Thanks http://mathhelpboards.com/members/magneto/ and chisigma Nice solution.

My Solution is same as Chisigma(Integration by parts):

Spoiler

Let $\displaystyle I(m,n) = \int_0^1 \! x^m(1-x)^n \, dx$.

We have, for $n \ge 1$,

\[ \begin{eqnarray}
I(m,n) &=& \left. -\frac{x^{m+1}(1-x)^n}{m+1}\right]_0^1 + \int_0^1 \frac{n}{m+1}x^{m+1}(1-x)^{n-1} \, dx \nonumber \\
&=& \frac{n}{m+1}I(m+1,n-1) \nonumber
\end{eqnarray}\]
Repeating this gives us
\[ \begin{eqnarray}
I(m,n) &=& \frac{n}{m+1}\cdot\frac{n-1}{m+2}\cdot\cdots\cdot\frac{1}{m+n}I(m+n,0) \nonumber \\
&=& \frac{n!}{\frac{(m+n)!}{m!}}\cdot\frac{1}{m+n+1} \nonumber \\
&=& \frac{1}{m+n+1}\cdot\frac{1}{\binom{m+n}{m}} \nonumber
\end{eqnarray} \]

Now Put $m=2014$ and $n=2014$

So $\displaystyle I(2014,2014) = \int_{0}^{1}x^{2014}\cdot (1-x)^{2014}dx = \frac{1}{2014+2014+1}\cdot \frac{1}{\binom{2014+2014}{2014}} = \frac{1}{4029}\cdot \frac{1}{\binom{4028}{2014}}$

 

[CellCoree]

I would approach this integral by first recognizing that it is a special case of the Beta function, which is defined as $\displaystyle B(x,y) = \int_{0}^{1}t^{x-1}(1-t)^{y-1}dt$. In this case, we have $x = 2015$ and $y = 2015$, so the integral can be rewritten as $\displaystyle \int_{0}^{1}x^{2014}(1-x)^{2014}dx = B(2015,2015)$.

Using the properties of the Beta function, we can rewrite this as $\displaystyle B(2015,2015) = \frac{\Gamma(2015)\Gamma(2015)}{\Gamma(4030)}$, where $\Gamma(x)$ is the Gamma function. This simplifies to $\displaystyle \frac{(2014!)^2}{4029!}$.

Since this integral is a special case of the Beta function, we can also use the Beta function's connection to the binomial coefficients. This connection states that $\displaystyle B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} = \binom{x+y-1}{x}$.

Applying this to our integral, we get $\displaystyle \binom{4029}{2014} = \frac{4029!}{2014! \cdot 2015!}$, which is the same result obtained using the Beta function.

Therefore, the value of the integral is $\displaystyle \frac{(2014!)^2}{4029!} = \binom{4029}{2014}$.