Using an integral to generate rational approximations of $\pi$

Question

Let $$f(r)=\int_0^1\frac{x^r(1-x)^r}{1+x^2}dx.$$ One surprising fact is that $f(4)=\frac{22}{7}-\pi.$ This got me thinking. Surely, it can't be a coincidence that a fairly accurate rational approximation for $\pi$ shows up in this way, right? Well, this is what I've found so far: $$\boxed{f(0)=\frac{\pi}{4}≈0.785}$$ $$\boxed{f(4)=\frac{22}{7}-\pi≈0.001}$$ $$\boxed{f(8)=4\pi-\frac{188684}{15015}≈3.64×10^{-6}}$$ $$\boxed{f(12)=\frac{431302721}{8580495}-16\pi≈1.18×10^{-8}}$$ $$\boxed{f(16)=64\pi-\frac{5930158704872}{29494189725}≈4.00×10^{-11}}$$ Let $k\in\{0,1,2,\ldots\}.$ It seems to me that $f(4k)=4^{k-1}((-1)^k\pi+(-1)^{k+1}R_k),$ where $R_k$ is some rational approximation of $\pi.$ We have: $$\boxed{R_0=0}$$ $$\boxed{R_1=\frac{22}{7}}$$ $$\boxed{R_2=\frac{188684}{15015×4}}$$ $$\boxed{R_3=\frac{431302721}{8580495×16}}$$ $$\boxed{R_4=\frac{5930158704872}{29494189725×64}}$$ I have three conjectures at this point: $$\textbf{C1:}\lim_{r\to\infty}f(r)=0.$$ $$\textbf{C2:}\forall k\in\{0,1,2,\ldots\},f(4k)=4^{k-1}((-1)^k\pi+(-1)^{k+1}R_k),R_k\in\mathbb{Q}.$$ $$\textbf{C3:}\lim_{k\to\infty}R_k=\pi.$$ Clearly, $(\textbf{C1}\wedge\textbf{C2})\implies\textbf{C3}.$ How do we go about proving these conjectures? It seems like $\textbf{C2}$ is the toughest of the three. $\textbf{C1}$ makes intuitive sense as the integrand is a decreasing function of $r.$

Answer 1

As mentioned in comments, C1 follows immediately from the dominated convergence theorem, and as you said, proving C2 proves C3.

For C2, one can check the proof of Wikipedia. It is stated that \begin{align} \frac1{2^{2n-2}} \int_0^1\frac{x^{4n}(1-x)^{4n}}{1+x^2}\,dx = \sum_{j=0}^{2n-1}\frac{(-1)^j}{2^{2n-j-2}(8n-j-1)\binom{8n-j-2}{4n+j}} +(-1)^n\left(\pi-4\sum_{j=0}^{3n-1}\frac{(-1)^j}{2j+1}\right) \end{align} from which we infer that $$R_k = 4\sum_{j=0}^{3k-1} \frac{(-1)^j}{2j+1} - \sum_{j=0}^{2k-1}\frac{(-1)^{k+j}}{2^{2k-j-2}(8k-j-1)\binom{8k-j-2}{4k+j}}$$ is a finite sum of rationals, hence rational.

Given that the other answers (at time of writing) somehow do not seem to answer the question, the only point of this answer is to have a copy of Wikipedia's proof on this website as well. (Apparently the page was nominated for deletion before...) Here it is -

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For all integers $k ≥ 0$ and $\ell ≥ 2$ we have \begin{align} x^k(1-x)^\ell&=(1-2x+x^2)x^k(1-x)^{\ell-2}\\ &=(1+x^2)\,x^k(1-x)^{\ell-2}-2x^{k+1}(1-x)^{\ell-2}. \end{align}

Applying this formula recursively $2n$ times yields $$ x^{4n}(1-x)^{4n} =\left(1+x^2\right)\sum_{j=0}^{2n-1}(-2)^jx^{4n+j}(1-x)^{4n-2(j+1)}+(-2)^{2n}x^{6n}. $$ Furthermore,

\begin{align} x^{6n}-(-1)^{3n} &=\sum_{j=1}^{3n}(-1)^{3n-j}x^{2j}-\sum_{j=0}^{3n-1}(-1)^{3n-j}x^{2j}\\ &=\sum_{j=0}^{3n-1}\left((-1)^{3n-(j+1)} x^{2(j+1)}-(-1)^{3n-j}x^{2j}\right)\\ &=-(1+x^2)\sum_{j=0}^{3n-1} (-1)^{3n-j}x^{2j}, \end{align}

where the first equality holds, because the terms for $1 ≤ j ≤ 3n – 1$ cancel, and the second equality arises from the index shift $j → j + 1$ in the first sum.

Application of these two results gives

\begin{align}\frac{x^{4n}(1-x)^{4n}}{2^{2n-2}(1+x^2)} =\sum_{j=0}^{2n-1} & \frac{(-1)^j}{2^{2n-j-2}}x^{4n+j}(1-x)^{4n-2j-2}\\ & {} -4\sum_{j=0}^{3n-1}(-1)^{3n-j}x^{2j}+(-1)^{3n}\frac4{1+x^2}.\tag{1} \end{align}

For integers $k, \ell ≥ 0$, using integration by parts $\ell$ times, we obtain

\begin{align} \int_0^1x^k(1-x)^\ell\,dx &=\frac \ell{k+1}\int_0^1x^{k+1}(1-x)^{\ell-1}\,dx\\ &\,\,\,\vdots\\ &=\frac \ell{k+1} \frac{\ell-1}{k+2}\cdots\frac1{k+\ell}\int_0^1x^{k+\ell}\,dx\\ &=\frac{1}{(k+\ell+1)\binom{k+\ell}{k}}.\tag{2} \end{align}

Setting $k = \ell = 4n$, we obtain

$$\int_0^1 x^{4n} (1-x)^{4n}\,dx = \frac{1}{(8n+1)\binom{8n}{4n}}.$$

Integrating equation (1) from 0 to 1 using equation (2) and $\arctan(1) = \frac{π}4$, we get the claimed equation involving $\pi$.

Answer 2

The integral can be evaluated with Laplaces method when $r\to\infty$. From the link, one has that an integral on the form according to below with a minimum of $p(x)$ at $x=a$ to leading order is given by

$$ \int_a^b \exp(-rp(x))q(x)dx \sim \frac{Q}{\mu} \Gamma\left(\frac{\lambda}{\mu}\right) \frac{e^{-rp(a)}}{(Pr)^{\lambda/\mu}}, \quad r\to \infty $$

where the constants are found from the Taylor expansions of $p(x)$ and $q(x)$ around $x=a$ respectively. In this case, one has

$$ f(r) = \int_0^1 \frac{x^r(1-x)^r}{1+x^2} dx=\int_0^1 \exp\left(-rp(x) \right) q(x)dx $$

with $p(x) = -\ln(x(1-x))$ and $q(x) = (1+x^2)^{-1}$. With a minimum at $x = 1/2$ one has that around $x=1/2$,

$$ p(x) = \ln(4)+4\left(x-\frac{1}{2}\right)^2 +... $$ $$ q(x) = \frac{4}{5}-\frac{16}{25}\left(x-\frac{1}{2}\right)+... $$

Splitting the integral into two parts, the first from $0$ to $1/2$ and the second from $1/2$ to $1$, one has that

$$ \int_{1/2}^1 \exp\left(-rp(x) \right) q(x)dx \sim \frac{4^{-r}}{5}\sqrt{\frac{\pi}{r}} $$

and letting $x\to x+1/2$ gives similarly that the integral from $0$ to $1/2$ is the same as the one above. Hence, to leading order, one has

$$ f(r) = \int_0^1 \frac{x^r(1-x)^r}{1+x^2} dx \sim \frac{2\cdot4^{-r}}{5}\sqrt{\frac{\pi}{r}}, \quad r \to \infty $$

Now, e.g. looking at $f(4k)$ with $k=8$ gives that $f(4k) = 6.712\cdot 10^{-21}$ whereas the asymptotic expression gives $f(4k) \sim \frac{4^{-4k}}{5} \sqrt{\frac{\pi}{k}} = 6.794\cdot 10^{-21}$.

If it is possible to get $\textbf{C}_2$ or $\textbf{C}_3$ from here, I'm not sure. I mean, you could take your assumption from $\textbf{C}_2$ and plug in the asymptotic expression, which would give $R_k \sim \pi +(-1)^{k-1}\frac{4^{-5k+1}}{5} \sqrt{\frac{\pi}{k}}$ but I don't really think it makes much sense that way.

Answer 3

$$f(r)=\int_0^1\frac{x^r(1-x)^r}{1+x^2}dx$$ $$f(r)=\sqrt{\pi }\,\,\frac{ \Gamma (r+1)}{2^{2 r+1}\,\,\Gamma \left(r+\frac{3}{2}\right)}\, _3F_2\left(1,\frac{r+1}{2},\frac{r+2}{2};r+1,\frac{2r+3}{ 2};-1\right)$$ In terms of summation $$f(r)=\Gamma (r+1)\sum_{n=0}^\infty(-1)^n\,\frac{\Gamma (2 n+r+1)}{\Gamma (2 (n+r+1))}$$ which will converge very fast since $$a_n=\frac{\Gamma (2 n+r+1)}{\Gamma (2 (n+r+1))}\quad \implies \quad \log\left(\frac{a_{n+1}}{a_n}\right)=-\frac{r+1}{n}+\frac{(r+1) (3 r+4)}{4 n^2}+O\left(\frac{1}{n^3}\right)$$

As you noticed, if $r=4k$, we obtained what you observed and conjectured.

Computing up to $r=100$ $$\log(f(r))=a + b\,r$$ with $R^2=0.9999985$ $$\begin{array}{|llll} \hline \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ a & -2.14647 & 0.04008 & \{-2.22601,-2.06693\} \\ b & -5.55932 & 0.00069 & \{-5.56069,-5.55795\} \\ \end{array}$$

Edit

@AxelT approach is very appealing and could be extended a little bit.

Writing $$q(x)=\sum_{n=0}^m a_n\,\left(x-\frac{1}{2}\right)^n$$ $$I_n=\int_{0}^1 \exp\left(-r\,p(x) \right) \left(x-\frac{1}{2}\right)^n\,dx $$ $$I_n=2^{-(n+2 r+1)} r^{-\frac{n+1}{2}} \left(\Gamma \left(\frac{n+1}{2}\right)-\Gamma \left(\frac{n+1}{2},r\right)\right)$$ $$I_n \sim \frac 1{2^{n+2 r+1} }\left(r^{-\frac{n+1}{2}} \Gamma \left(\frac{n+1}{2}\right) -\frac{e^{-r} (n+2 r-1)}{2 r^2}\right)$$

With $m=2$ $$f(r)\sim\frac 1{2^r}\left(\frac{2}{5} \sqrt{\frac{\pi}{r}}-\frac{4}{25 r}-\frac{\sqrt{\pi }}{125 r^{3/2}}-e^{-r}\frac{ 28 r-26}{125 r^2}\right)=g(r)$$ which is a decent approximation even for small values of $r$ $$\left( \begin{array}{ccc} r & \log(g(r)) & \log(f(r)) \\ 1 & -2.02322 & -2.02517 \\ 2 & -3.66861 & -3.63135 \\ 3 & -5.20670 & -5.17007 \\ 4 & -6.71015 & -6.67309 \\ 5 & -8.19183 & -8.15398 \\ 6 & -9.65820 & -9.61981 \\ 7 & -11.1133 & -11.0747 \\ 8 & -12.5599 & -12.5213 \\ 9 & -13.9998 & -13.9615 \\ 10 & -15.4344 & -15.3963 \\ \end{array} \right)$$

Answer 4

Lazy way to prove C1, C2, C3, without evaluating $R_k$:

We have an isomorphism of fields: $$\mathbb{Q}[i]\to\mathbb{Q}[x]/(1+x^2),$$ identifying $i\mapsto x$.

Thus to find the remainder left after dividing a polynomial $p$ by $1+x^2$, we may simply evaluate $p(i)=a+bi$ (with $a,b\in\mathbb{Q}$) and take $a+bx$ as the remainder.

In particular $x^{4k}(1-x)^{4k}$ evaluated at $i$ is just $((1-i)^4)^k=(-4)^k$. Thus $$ \frac{x^{4k}(1-x)^{4k}}{1+x^2} = p(x)+\frac{(-4)^k}{1+x^2}, $$ where $p(x)\in \mathbb{Q}[x]$.

Thus $$\int_0^1 \frac{x^{4k}(1-x)^{4k}}{1+x^2} dx = \int_0^1 p(x) dx +\int_0^1 \frac{(-4)^k}{1+x^2} dx$$ $$ =R_k'+(-1)^k 4^{k-1}\pi, $$ with $R_k'\in \mathbb{Q}$.

This is just C2 with $R_k=(-4)^{1-k} R_k'$.

For C1 just note that ${\rm sup}_{x\in [0,1]}\{x(1-x)\}=\frac14$ and ${\rm inf}_{x\in [0,1]}\{1+x^2\}=1$, so $$0\leq \int_0^1 \frac{x^{r}(1-x)^{r}}{1+x^2} dx \leq \frac1{4^r}.$$

C3 follows immediately.

Answer 5

Recurrence

Define $$\newcommand{\rmod}[1]{\quad\left(\text{mod}\ \ {#1}\right)} f(n)=\int_0^1\frac{x^n(1-x)^n}{1+x^2}\,\mathrm{d}x\tag1 $$ We would like to find a recurrence relation for $f(n)$. To do this, we will find $a,b,c$ so that $$ ax^2(1-x)^2+bx(1-x)+c\equiv0\rmod{1+x^2}\tag2 $$ Using $\{1,x\}$ as a basis for $\left.\mathbb{R}[x]\middle/\left(1+x^2\right)\right.$, $$ \begin{align} 1&\equiv1&\rmod{1+x^2}\tag{3a}\\ x(1-x)&\equiv1+x&\rmod{1+x^2}\tag{3b}\\ x^2(1-x)^2&\equiv2x&\rmod{1+x^2}\tag{3c} \end{align} $$ Working from $(3)$, we can see that $$ x^2(1-x)^2-2x(1-x)+2\equiv0\rmod{1+x^2}\tag4 $$ Noting that $\left(u^2-2u+2\right)\left(u^2+2u+2\right)=u^4+4$ and setting $u=x(1-x)$, multiply $(4)$ by $x^2(1-x)^2+2x(1-x)+2$ to get $$ x^4(1-x)^4+4\equiv0\rmod{1+x^2}\tag5 $$ In fact, $$ x^4(1-x)^4+4 =\left(x^6-4x^5+5x^4-4x^2+4\right)\left(1+x^2\right)\tag6 $$ Thus, using the Beta Function Integral, we get $$ \begin{align} \hspace{-24pt}f(n+4)+4f(n) &=\int_0^1x^n(1-x)^n\frac{x^4(1-x)^4+4}{1+x^2}\mathrm{d}x\tag{7a}\\ &=\int_0^1x^n(1-x)^n\left(x^6-4x^5+5x^4-4x^2+4\right)\mathrm{d}x\tag{7b}\\ &=\frac1{n+1}\left(\frac1{\binom{2n+7}{n+1}}-\frac4{\binom{2n+6}{n+1}}+\frac5{\binom{2n+5}{n+1}}-\frac4{\binom{2n+3}{n+1}}+\frac4{\binom{2n+1}{n+1}}\right)\tag{7c}\\[6pt] &\in\mathbb{Q}\tag{7d} \end{align} $$ Since $f(0)=\frac\pi4$, we get that $$ f(4n)\equiv(-4)^n\frac\pi4\rmod{\mathbb{Q}}\tag8 $$ Furthermore, since $x(1-x)\le\frac14$, we get that $$ f(4n)\le\frac1{256^n}\tag9 $$

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Conjectures

The conjectures in the question can be verified using $(8)$ and $(9)$.

$\textbf{C1}$ follows directly from $(9)$.

$\textbf{C2}$ follows from $(8)$, which says that $\frac{f(4n)}{4^{n-1}}=(-1)^n\pi+(-1)^{n-1}R_n$, for some $R_n\in\mathbb{Q}$.

$\textbf{C3}$ follows from $\textbf{C2}$ and $(9)$: $|\pi-R_n|=\frac{f(n)}{4^{n-1}}\le\frac4{1024^n}$

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Computation of the Integral

We can use $\text{(7c)}$ to compute $f(4n)$. $$ \begin{array}{r|l} n&f(4n)&\approx f(4n)&\approx\frac1{256^n}\\\hline 0&\frac\pi4&0.7853981634&1.0000000000\\ 1&\frac{22}7-\pi&0.0012644893&0.0039062500\\ 2&4\pi-\frac{188684}{15015}&3.6479922066\times10^{-6}&1.5258789063\times10^{-5}\\ 3&\frac{431302721}{8580495}-16\pi&1.1814790762\times10^{-8}&5.9604644775\times10^{-8}\\ 4&64\pi-\frac{5930158704872}{29494189725}&4.0282588889\times10^{-11}&2.3283064365\times10^{-10}\\ 5&\frac{26856502742629699}{33393321606645}-256\pi&1.4141199610\times10^{-13}&9.0949470177\times10^{-13}\\ 6&1024\pi-\frac{423877461668007447086}{131762096268962445}&5.0587290930\times10^{-16}&3.5527136788\times10^{-15} \end{array} $$

Answer 6

Some thoughts.

$$f(p)=\int_0^1 \frac{x^p(1-x)^p}{1+x^2}\mathrm dx$$

Now, recall that $$\pi=4\int_0^1 \frac{1}{1+x^2}\mathrm dx$$

So $$f(4k)-4^{k-1}(-1)^{k}\pi \\ =\int_0^1 \frac{x^{4k}(1-x)^{4k}}{1+x^2}\mathrm dx~+(-1)^{k+1}4^k\int_0^1\frac{1}{1+x^2}\mathrm dx \\ =\int_0^1 \frac{x^{4k}(1-x)^{4k}+(-1)^{k+1}4^k}{1+x^2}\mathrm dx$$

Maybe there is some way to show that this is rational?

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Another interesting finding - I accidentally typed $(1-x^{4k})$ instead of $(1-x)^{4k}$ when trying to evaluate $f(4k)$, but strangely "typo" integral,

$$g(p)=\int_0^1\frac{x^p(1-x^p)}{1+x^2}\mathrm dx$$

Seems to have the strange property that $g(4k)\in\mathbb Q$ when $k\in\mathbb N$. For example, Mathematica is telling me that $$g(4\cdot 1)=\frac{2}{35} \\ g(4\cdot 2)=\frac{196}{6~435} \\ g(4\cdot 3)=\frac{208~786}{10~140~585} \\ g(4\cdot 4)=\frac{489~772~744}{31~556~720~475} \\ \text{etc}.$$

There must be something deeper going on.