# Evaluating the integral $\frac{1}{2^{2n-2}}\int_0^1\frac{x^{4n}\left(1-x\right)^{4n}}{1+x^2} dx$

Prove that :

$$\frac{1}{2^{2n-2}}\int \limits_{0}^{1} \dfrac{x^{4n}\left(1-x\right)^{4n}}{1+x^2} dx =$$$$\sum \limits_{j=0}^{2n-1}\dfrac{(-1)^j}{2^{2n-j-2}\left(8n-j-1\right)\binom{8n-j-2}{4n+j}} + (-1)^n\left(\pi-4\sum \limits_{j=0}^{3n-1}\dfrac{(-1)^j}{2j+1}\right)\,\,\,\,(♣)$$ where $\binom{a}{b}= {}_aC_b=\frac{a!}{b!(a-b)!}$.

I was reading an article on "$\frac{22}{7}$ exceeds $\pi$ ", when came across the generalized form $(♣)$.

Putting the value of $n=1$ in $(♣)$, we get the famous Putnam Problem :

$$0<\int \limits_0^1\dfrac{x^4(1-x)^4}{1+x^2}dx=\dfrac{22}{7}-\pi.\,\,\,\,(♠)$$

I know how to evaluate $(♠)$ by using expansion and polynomial long-division and then term-wise integration. The other integrals involving $n>1$ provide better approximations of $\pi$. But I am unable to get even close to evaluating $(♣)$ because of the '$n$'s. I tried to do polynomial division but got badly stuck. The $1+x^2$ in the denominator reminds me of $\arctan(x)$, but what can I do?

Can anyone provide a bit of help as to how to evaluate this lovely integral ? Also I would like to know if it is okay to treat $n$ as a real number here instead of a natural number.

• If I properly remember, I think I saw a solution using the regularized generalized hypergeometric function. – Claude Leibovici Feb 24 '17 at 7:47
• @ClaudeLeibovici So it is not that simple as its special case involving n=1 ??? Also, can $n$ be a floating point number ? Is there a "simple"/elementary way of evaluating the integral ? – user399078 Feb 24 '17 at 7:50
• Please avoid using \large for no real reason. – Did Feb 24 '17 at 8:39
• @ClaudeLeibovici it is indeed not too difficult to reformulate this integral into a form $\sim {_3F_2}(..,..,..;..,..;-1)$ but how to show that this series terminates after finite terms? – tired Feb 24 '17 at 19:00
• my try would be to use parity and integrate around a dogbone contour in the complex plane. take into account the residues at $\pm i$ and $\infty$ – tired Feb 24 '17 at 19:03

Let us do some preliminary proofs.

$$\color{red}{\displaystyle 4\int\limits_0^1 \dfrac{x^{6n}}{1+x^2}\; dx = (-1)^n \left(\pi-4\sum\limits_{j=0}^{3n-1}\dfrac{(-1)^j}{2j+1}\right)}\tag 1 \\$$

Proof : Denote the integral by $\mathfrak{A}$ and so,

$\mathfrak{A}\\ =\displaystyle (-1)^n \left(4\int\limits_0^1\dfrac{1-1+(-1)^n x^{6n}}{1+x^2}\; dx\right) \\ \displaystyle =(-1)^n \left(4\int\limits_0^1 \dfrac{1}{1+x^2}\; dx - 4\int\limits_0^1\dfrac{1-(-1)^n x^{6n}}{1+x^2}\; dx\right)\\ \displaystyle =(-1)^n \left(\pi -4\int\limits_0^1\sum\limits_{j=0}^{3n-1}(-x^2)^j\right)\\ \displaystyle =(-1)^n \left(\pi-\sum_{j=0}^{3n-1}\dfrac{(-1)^j}{2j+1}\right)$

$$\color{blue}{\displaystyle \sum\limits_{j=0}^{2n-1} \left(\dfrac{-2x}{(1-x)^2}\right)^j = \dfrac{(1-x)^2}{1+x^2}\left(1-\left(\dfrac{2x}{(1-x)^2}\right)^{2n}\right)}\tag 2$$

Proof : It follows directly from the GP formula that,

$\displaystyle \sum\limits_{j=0}^{2n-1} \left(\dfrac{-2x}{(1-x)^2}\right)^j = \dfrac{1-\left(\dfrac{-2x}{(1-x)^2}\right)^{2n}}{1+\left(\dfrac{2x}{(1-x)^2}\right)}$

Just simplifying the denominator we get,

$\displaystyle \sum\limits_{j=0}^{2n-1} \left(\dfrac{-2x}{(1-x)^2}\right)^j = \dfrac{(1-x)^2}{1+x^2}\left(1-\left(\dfrac{2x}{(1-x)^2}\right)^{2n}\right)$

Coming back to the problem and denote the Integral by I,

$\displaystyle \dfrac{1}{2^{2n-2}}\int\limits_0^1 \dfrac{x^{4n}(1-x)^{4n}}{1+x^2}\; dx \\ = \displaystyle \dfrac{1}{2^{2n-2}}\int\limits_0^1 \dfrac{x^{4n}(1-x)^{4n}-2^{2n}(x^{6n}-x^{6n})}{1+x^2}\; dx \\= \displaystyle \dfrac{1}{2^{2n-2}}\int\limits_0^1 \dfrac{x^{4n}(1-x)^{4n}-2^{2n}x^{6n}}{1+x^2}\; dx+4\int\limits_0^1 \dfrac{x^{6n}}{1+x^2}\; dx = \displaystyle \dfrac{1}{2^{2n-2}}\int\limits_0^1 x^{4n}(1-x)^{4n}\dfrac{(1-x)^2}{1+x^2}\left(1-\left(\dfrac{-2x}{(1-x)^2}\right)^{2n}\right)\; dx + \underbrace{(-1)^n \left(\pi-4\sum\limits_{j=0}^{3n-1}\dfrac{(-1)^j}{2j+1}\right)}_{\text{From (1)}} \\ \displaystyle = \dfrac{1}{2^{2n-2}}\int\limits_0^1 x^{4n}(1-x)^{4n} \underbrace{\left(\displaystyle \sum\limits_{j=0}^{2n-1} \left(\dfrac{-2x}{(1-x)^2}\right)^j\right)}_{\text{From (2)}}\; dx + (-1)^n \left(\pi-4\sum\limits_{j=0}^{3n-1}\dfrac{(-1)^j}{2j+1}\right) \\ \displaystyle = \dfrac{1}{2^{2n-2}}\sum\limits_{j=0}^{2n-1} (-2)^j \beta(4n+j+1,4n-2j-1)+(-1)^n \left(\pi-4\sum\limits_{j=0}^{3n-1}\dfrac{(-1)^j}{2j+1}\right) \\ \displaystyle = \sum \limits_{j=0}^{2n-1}\dfrac{(-1)^j}{2^{2n-j-2}\left(8n-j-1\right){{8n-j-2}\choose{4n+j}}} + (-1)^n\left(\pi-4\sum \limits_{j=0}^{3n-1}\dfrac{(-1)^j}{2j+1}\right)\\$

$$\color{red}{\boxed{\mathfrak{Proved}}}$$

• Very nice derivation. I have seen mathfrak ($\mathfrak{abc}$) used for decoration for the first time here. +1 – Paramanand Singh Feb 27 '17 at 13:27
• @ParamanandSingh Thanks ! I am still learning to be more decorative with my latex. – Aditya Narayan Sharma Feb 27 '17 at 14:55
• Thanks for such a nice and fancy answer. (+unity) + (50 Reputation). I could have offered a bigger bounty, but as you might see I myself have very low rep. Thanks again !!! ^_^ ;-) – user399078 Feb 27 '17 at 16:42
• Welcome :-) , Don't worry +50 is perfect for this one, a bigger bounty some other time. – Aditya Narayan Sharma Feb 27 '17 at 17:09