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In Ramanujan's Notebooks, Vol 4, p.48 (and a related one in Quarterly Journal of Mathematics, XLV, 1914) there are various approximations, including the close (by just $10^{-7}$),

$$\pi^4 \approx 2^4+3^4+\frac{1}{2+\Big(\frac{2}{3}\Big)^2} = \frac{2143}{22}$$

We can also easily calculate $\pi^3 \approx 31.0062$ and know that,

$$ \int_0^1 \frac{x^4(1-x)^4}{1+x^2} \mathrm{d}x = \frac{22}{7} - \pi $$

$$ \int_0^1 \frac{x^8(1-x)^8(25+816x^2)}{28\cdot113(1+x^2)} \mathrm{d}x = \frac{355}{113} - \pi $$

Q: Are there then analogous integrals for these early approximations of $\pi^k$,

$$ \int_0^1 f(x)\, \mathrm{d}x = \pi^2 -\frac{1+2\times113}{23} $$

$$ \int_0^1 f(x)\, \mathrm{d}x = \pi^3 - 31 $$

$$ \int_0^1 f(x)\, \mathrm{d}x = \pi^4 -\frac{2143}{22} $$

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    $\begingroup$ Silly answer, if you let $f(x) = \pi^4 - \tfrac{2143}{22}$, then $\int_0^1 f(x) ~ dx $ would be equal to what you want. $\endgroup$ Aug 12, 2015 at 23:38
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    $\begingroup$ Presumably you want an integral of P/Q from A to B where P and Q are polynomials with integer coefficients and A,B are rational, I know that the square of pi comes up in some such integrals that need complex residue theory to evaluate,. e.g. P=1, Q=1/(1+x^4) for x from 0 to infinity.You might try looking at some of these. $\endgroup$ Aug 14, 2015 at 18:42
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    $\begingroup$ I know that $\displaystyle\int_0^\infty\,\frac{t}{\exp(t)-1}\,\text{d}t=\frac{\pi^2}{6}$ and $\displaystyle\int_0^\infty\,\frac{t^3}{\exp(t)-1}\,\text{d}t=\frac{\pi^4}{15}$. It is easy to manipulate these two integrals to get something of the forms $\displaystyle\int_0^1\,f(x)\,\text{d}x=\pi^2-\frac{227}{23}$ and $\displaystyle\int_0^1\,f(x)\,\text{d}x=\pi^4-\frac{2143}{22}$. However, frankly, I don't think that these integrals will be meaningful. $\endgroup$ Aug 15, 2015 at 3:26
  • $\begingroup$ @Batominovski: Perhaps something more difficult, like $\displaystyle\int_0^1 \frac{P(x)}{Q(x)} \mathrm{d}x$ where $P(x), Q(x)$ are polynomials with integer coefficients analogous to the examples. Just to see if it can be done. $\endgroup$ Aug 15, 2015 at 3:37
  • $\begingroup$ Let $\mathbb{A}$ be the field of algebraic numbers. You question reduces to finding $s_1,s_2,\ldots,s_n\in \mathbb{A}$ and $t_1,t_2,\ldots,t_n\in\mathbb{A}$ such that $\sum_{i=1}^n\,s_i\ln\left(t_i\right)=R(\pi)$, where $R(x)\in\mathbb{A}[x]$ is of degree at least $2$. I conjecture that this is impossible. $\endgroup$ Aug 15, 2015 at 4:48

1 Answer 1

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The answer is yes. The method is detailed in the paper by S. K. Lucas. The approximate fractions are obtained from truncating the exact continued fraction of the respected numbers, so the signs are alternating. We first give a few examples.

Results

For $\pi$

The continued fractions are $3, 22/7, 333/106, 355/113, 103993/33102, \dots$.

\begin{align} \pi - \frac{333}{106} &= \int_0^1 \frac{x^4 \, (1-x)^5 \, \left(74 \, x^2-53 \, x+21\right)}{106 \left(x^2+1\right)} \, dx. \\ \frac{355}{113} - \pi &= \int_0^1 \frac{x^{10} \, (1-x)^8 \, \left(886+95\,x^2\right)}{3164 \left(x^2+1\right)} \, dx. \end{align}

For $\pi^2$

The truncated continued fractions are $9, 10, 69/7, 79/8, 227/23, 10748/1089, \dots$,

\begin{align} \pi^2-\frac{69}{7} &= \int_0^1 \frac{4 \, x^{4} \, (1 - x)^3 \left(64 x^2 -39 x + 25\right)} {13 \, (1 + x^2) } \log(x^{-1}) \, dx \\ &= \int_0^1 \frac{24 \, x^{6} \, (1 - x)^2 \left(119 - 72 \, x^2\right)} {191 \, (1 + x^2) } \log(x^{-1}) \, dx. \end{align}

\begin{align} \frac{79}{8} - \pi^2 &= \int_0^1 \frac{4 \, x^6 \, (1-x)^3 (49 - 51 x + 100 x^2)} {17 \, (1 + x^2) } \log(x^{-1}) \, dx \\ &= \int_0^1 \frac{4 \, x^3 \, (1-x)^4 (25 + 2254 x^2)} {743 \, (1 + x^2) } \log(x^{-1}) \, dx \\ &= \int_0^1 \frac{ 24 \, x^5 \, (1-x)^2 \left[37 \, (x^2 + 1) - 73 \, x\right] } { 73 \, (1 + x^2) } \log(x^{-1}) \, dx \end{align}

\begin{align} \pi^2-\frac{227}{23} &= \int_0^1 \frac{4 \, x^{19} \, (1 - x)^4 \left(61847 x^2+87524\right)} {8559 \, (1 + x^2) } \log(x^{-1}) \, dx. \end{align}

For $\pi^3$

The truncated continued fractions are $31, 4930/159, 14821/478, \dots$.

\begin{align} \pi^3-31 &= \int_0^1 \frac{8 \, x^5 \, (1-x)^2 \, \left(324889-120736 \, x^2\right)} {445625 \, (1 + x^2) } \log^2 x \, dx\\ \frac{4930}{159}-\pi^3 &= \int_0^1 \frac{4 \, x^{10} \, (1-x)^4 \, \left(695774836+470936528857 \, x^2\right)} {470240754021 \, (1 + x^2) } \log^2 x \, dx. \end{align}

For $\pi^4$

The truncated continued fractions are $97, 195/2, 487/5, 1656/17, 2143/22, \dots$.

\begin{align} \pi^4-97 &= \int_0^1 \frac{240 \, x^{4} \, (1 - x)^{2} \,\left(3522267 x^2+1681375\right) } {3221561 \, (1 + x^2) } \log^3(x^{-1}) \, dx \\ \frac{195}{2}-\pi^4 &= \int_0^1 \frac{192 \, x^{6} \, (1 - x)^{2} \, \left(5657688 x^2+3056473\right) } {3641701 \, (1 + x^2) } \log^3(x^{-1}) \, dx. \\ \pi^4-\frac{487}{5} &= \int_0^1 \frac{15 \, x^{8} \, (1 - x)^{2} \, \left(3293858975 x^2+746556831\right) } {278611172 \, (1 + x^2) } \log^3(x^{-1}) \, dx. \\ \frac{1656}{17}-\pi^4 &= \int_0^1 \frac{480 \, x^{7} \, (1 - x)^{4} \, \left(8555775811 x^2+2883779820\right) } {39703971937 \, (1 + x^2) } \log^3(x^{-1}) \, dx. \\ \pi^4-\frac{2143}{22} &= \int_0^1 \frac{480 \, x^{31} \, (1 - x)^{4} \, \left(4071997316165706379 x^2+175446796437023645180\right) } {1199623593846005571607 \, (1 + x^2) } \log^3(x^{-1}) \, dx. \end{align}

Method

The idea is simple. We basically combine the following identities.

(1) \begin{align} \int_0^1 \log^{s-1}\left( x^{-1} \right) x^k \, dx = \frac{(s-1)!}{(k+1)^s}. \end{align}

For an even $s$

(2a) \begin{align} \int_0^1 \frac{ \log^{s-1}(x^{-1}) \, x }{1+x^2} \, dx &=2^{-s} \int_0^\infty \frac{ t^{s-1} }{ e^t + 1 } \, dt \\ &=2^{-s} \int_0^\infty t^{s-1} \left( e^{-t} - e^{-2\,t} + e^{-3t} - \cdots \right) \, dt \\ &= \frac{ (s-1)! \, (2^s - 2)}{4^s} \zeta(s) \\ &= \frac{ (1-2^{1-s}) \, |B_s| }{2 \, s} \, \pi^s. \end{align} where $\zeta(s)$ is the Riemann zeta function, $B_s$ is the Bernoulli number, which is rational. The last step is well known.

Similarly, for an odd $s$,

(2b) \begin{align} \int_0^1 \frac{ \log^{s-1}(x^{-1}) }{1+x^2} \, dx &=\int_0^\infty \frac{ t^{s-1} \, e^{-t} }{ e^{-2t} + 1 } \, dt \\ &=\int_0^\infty t^{s-1} \left( e^{-t} - e^{-3\,t} + e^{-5\,t} -\cdots \right) \, dt \\ &= (s-1)! \, \left(1-\frac{1}{3^s}+\frac{1}{5^s}-\cdots \right) \\ &= (s-1)! \, \beta(s) = \frac{|E_{s-1}|}{2^{s+1}} \, \pi^s, \end{align} where $E_s$ is the Euler number, which is also rational.

This means \begin{align} \frac{\pi}{4} &= \int_0^1 \frac{ 1 }{1+x^2} \, dx, \\ \frac{\pi^2}{48} &= \int_0^1 \frac{\log\left( x^{-1} \right) x}{1+x^2} \, dx\\ \frac{\pi^3}{16} &= \int_0^1 \frac{\log^2\left( x^{-1} \right) }{1+x^2} \, dx\\ \frac{7\,\pi^4}{192} &= \int_0^1 \frac{\log^3\left( x^{-1} \right) x}{1+x^2} \, dx. \end{align}

Now suppose we have a polynomial $P(x) = Q(x)(1 + x^2) + R(x)$, and we want $$ \int_0^1 \frac{ \log^s(x^{-1}) \, P(x) } { 1 + x^2 } \, dx = \pi^s - A, $$ where $A$ is an approximation of $\pi^s$ (we have assumed the possible sign, the case of negative sign is similar). To satisfy this equation, we demand,

\begin{align} \int_0^1 \frac{ \log^s(x^{-1}) \, R(x) } { 1 + x^2 } \, dx &= \pi^s \\ \int_0^1 \log^s(x^{-1}) \, Q(x) \, dx &= - A, \end{align}

This requires \begin{align} R(x) &= \begin{cases} \dfrac{ 2^{s+1} } { |E_{s-1}|} & \mathrm{for\; odd\;} s \\ \dfrac{ 2 \, s } { (1-2^{1-s}) \, |B_s| } x & \mathrm{for\; even\;} s \end{cases} \\ \sum_{n=0} \frac{ (s-1)! }{(n+1)^s} q_n &= -A, \end{align} where $Q(x) = \sum_{n=0} q_n \, x^n$.

Using these to rules to design $P(x)$, we get the above formulas. Particularly, we studied the form \begin{align} P(x) = x^u \,(1-x)^v (a \, x^2 + b \, x + c). \end{align}

For a particular set of $u$ and $v$, the constraint for $R(x)$ determines two parameters, say $b$ and $c$. The constraint for $A$ determines $a$. We then check if $a \, x^2 + b \, x + c$ is nonnegative definite. We then vary $u$ and $v$ to seek a simple combination of $a$, $b$ and $c$.

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  • $\begingroup$ Maybe you'd be interested in this related question? $\endgroup$ Feb 11, 2016 at 1:51
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    $\begingroup$ Mmm...has someone thought about using these ideas to give an alternative proof $\pi^s$ is irrational for positive $s$? It would be quite pointless since we already know $\pi$ is transcendental. That said, I think another proof using Beukers'-like methods would be a fun endeavor. $\endgroup$
    – Brian
    Jun 10, 2016 at 1:31

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