Replace $x$ by $1-x$ , right

A much simpler and short method to solve this is considering that

$\int _{ 0 }^{ 1 }{ { x }^{ k }{ (1-x) }^{ n-k }dx\quad =\quad } \int _{ 0 }^{ 1 }{ { x }^{ n-k }{ (1-x) }^{ k } } dx$

Therefore, Denominator :-

$2\sum _{ k=0 }^{ 1007 }{ \int _{ 0 }^{ 1 }{ { x }^{ k }{ (1-x) }^{ n-k } } }$

and Numerator :-

$2014\sum _{ k=0 }^{ 1007 }{ \int _{ 0 }^{ 1 }{ { x }^{ k }{ (1-x) }^{ n-k } } }$

Hence Numerator / Denominator = 1007

P.S. This question is from JEE

Looking at the exponents in the integral, it hints us to consider a binomial expansion. So let $\displaystyle J = \int \limits_0^1 (tx + 1 - x)^n \text{d}x$ , where $t$ is a parameter. Let us evaluate $J$ by two different methods.

Firstly, $\begin{array}{c}& \\ J & = \int \limits_0^1 ((t-1)x +1)^n \text{d}x \\ & = \frac{((t-1)x + 1)^{n+1}}{(n+1)(t-1)} \bigg|_0^1 \\ & = \frac{t^{n+1} - 1}{(n+1)(t-1)} \\ & = \frac{1}{n+1} \big\{ \sum_{k=0}^n t^k \big\} \\ \end{array}$

Now, we will approach in a different manner so as to create the required expression. $\begin{array}{c}& \\ J & = \int \limits_0^1 (tx +1-x)^n \text{d}x \\ & = \sum_{k=0}^n \binom{n}{k} \int \limits_0^1 (tx)^k (1-x)^{n-k} \text{d}x \\ & = \sum_{k=0}^n \binom{n}{k} t^k \int \limits_0^1 (x)^k (1-x)^{n-k} \text{d}x \\ & = \sum_{k=0}^n \binom{n}{k} t^k I_n (k) \\ \end{array}$

Now, we can equate the final expression obtained in both the cases as they are ultimately $J$ . Therefore,

$\displaystyle \frac{1}{n+1} \big\{ \sum_{k=0}^n t^k \big\} = \sum_{k=0}^n \binom{n}{k} t^k I_n (k)$

Equating the coefficients of $t^k$ on both the sides, we obtain,

$\displaystyle I_n (k) = \frac{1}{(n+1) \binom{n}{k} }$

As we have obtained an expression for $I_n (k)$ , we can go and evaluate the sums.

Using the property that, $\displaystyle \binom{n}{k} = \binom{n}{n-k}$ , we can write,
$\displaystyle \sum_{k=0}^n \frac{k}{\binom{n}{k}} = \sum_{k=0}^n \frac{n-k}{\binom{n}{k}}$
(The RHS is the actually the terms summed in the reverse order. ) Thus, $\begin{array}{c}& \\ 2 \times \sum_{k=0}^n \frac{k}{\binom{n}{k}} & = \sum_{k=0}^n \frac{n}{\binom{n}{k}} \\ \frac{2}{n+1} \sum_{k=0}^n \frac{k}{\binom{n}{k}} & = \frac{n}{n+1} \sum_{k=0}^n \frac{1}{\binom{n}{k}} \\ \Rightarrow 2 \sum_{k=0}^{n} k I_{n}(k) & = n \sum_{k=0}^{n} I_{n}(k) \\ \Rightarrow \frac{\sum_{k=0}^{n} k I_{n}(k) }{\sum_{k=0}^{n} I_{n}(k) } & = \frac{n}{2} \\ \end{array}$

Therefore,
$\frac{\sum_{k=0}^{2014} k I_{2014}(k) }{\sum_{k=0}^{2014} k I_{2014}(k) } = \frac{2014}{2} = \boxed{1007}$

First let us consider the integral:

$\displaystyle\int_{0}^{1} x^{p}(1-x)^{q}\text{d}x$

Let us make the substitution $x=\sin^{2} \theta$

$\therefore \text{d}x=2\sin \theta\cos \theta\text{d}\theta$

Thus our integral becomes:

$2\displaystyle\int_{0}^{\pi/2} \sin^{2p+1} \theta\cos^{2q+1} \theta\text{d}\theta$

By Walli's formula we can show that the value of this integral is:

$2\dfrac{\bigg( (2p)(2p-2)(2p-4)...\bigg)\bigg( (2q)(2q-2)(2q-4)...\bigg)}{\bigg( (2p+2q+2)(2p+2q)(2p+2q-2)...\bigg)}$

$=\dfrac{(p)!(q)!}{(p+q+1)!}$

Now just substitute $p=k$ and $q=n-k$ to obtain:

$I_{n}(k)=\dfrac{(k)!(n-k)!}{(n+1)!}=\dfrac{1}{(n+1)\dbinom{n}{k}}$

Now let us consider the sum:

$\displaystyle\sum_{k=0}^{n} kI_{n}(k)=\mathfrak{S}(n)\text{(say)}$

$\mathfrak{S}(n)=\displaystyle\sum_{k=0}^{n} \dfrac{k}{(n+1)\dbinom{n}{k}}$

Replacing $k$ by $n-k$ :

$\mathfrak{S}(n)=\displaystyle\sum_{k=0}^{n} \dfrac{n-k}{(n+1)\dbinom{n}{k}}$

Adding the two equations above and dividing both sides by $2$ we get:

$\mathfrak{S}(n)=\dfrac{n}{2}\displaystyle\sum_{k=0}^{n} \dfrac{1}{(n+1)\dbinom{n}{k}}$

Thus the required answer is:

$\dfrac{\mathfrak{S}(2014)}{\displaystyle\sum_{k=0}^{2014} \dfrac{1}{(2015)\dbinom{2014}{k}}}=\dfrac{2014}{2}=\boxed{1007}$