Area under the graph of $r\mapsto\binom nr$

The question:

Given $$n$$ is a natural number and $$r$$ is varying from $$0$$ to $$n$$, find the area under the graph of $$r\mapsto\binom nr$$, taking the $$\Gamma$$-function definition of factorial.

Background: I'm just a high school student and I don't know enough calculus as to how to even approach a question like this. The question arose when my high school Maths teacher told us that there exists a function known as the $$\Gamma$$-function which is an extension of the factorial function to all real numbers. I immediately had this question in mind since binomial theorem is all about factorials. I've taught myself some of the properties of the $$\Gamma$$-function and learnt how to solve some definite integrals using the $$\Gamma$$-function since then, but I just don't know how to approach this one. It has been nagging me ever since then. Even letting me know whether this is solvable or not (by hand and not computers) would be very helpful.

• This is a good question ! Dec 8, 2018 at 8:59
• Working with high accuracy $I_5=31.3749$ and $I_{10}=1023.4546$. Quite close to $2 ^5=32$ and $2^{10}=1024$ Dec 8, 2018 at 9:09

If I properly understand, you would like to compute $$I_n=\int_0^n \binom{n}{r}\,dr=n!\, \int_0^n \frac {dr} {(n-r)!\,\, r!}=\Gamma (n+1)\, \int_0^n \frac{dr}{\Gamma (r+1)\, \Gamma (n-r+1)}$$ Unfortunately, there is no closed form even for $$\int_1^a {\Gamma (r)}\,dr\qquad \text{or} \qquad \int_1^a \frac{dr}{\Gamma (r)}$$ and you will be facing numerical integration.
For you curiosity, I give you below some values of $$\log_{10}(I_n)$$ since $$I_n$$ varies extremely fast $$\left( \begin{array}{cc} n & \log_{10}(I_n) \\ 10 & 3.01007 \\ 20 & 6.02060 \\ 30 & 9.03090 \\ 40 & 12.0412 \\ 50 & 15.0515 \\ 60 & 18.0618 \\ 70 & 21.0721 \\ 80 & 24.0824 \\ 90 & 27.0927 \\ 100 & 30.1030 \end{array} \right)$$ If you plot them, you could see that this is almost $$\log_{10}(I_n)= n\log_{10}(2)\implies I_n \sim 2^n$$ which is normal since $$I_n \sim \sum_{r=0}^n \binom{n}{r}=2^n$$