Using integer relation algorithms to guess closed forms

Source

In this post, we'll see about guessing closed forms of the answers obtained by numerical methods. In particular, we'll use the excellent math toolkit by David Bailey et. al. aimed at experimental mathematics -- arprec. Compile and run mathtool.

The following problems are taken from Brilliant

  1. \begin{equation*} \displaystyle \int_0^{\frac{\pi}{3}} x \left(\ln{\left(2 \sin{\frac{x}{2}}\right)}\right)^2 \, dx = \frac{c\, \pi^a}{b} \end{equation*}

    Find $a, b,$ and \(c\).

    Start mathtool, and enter the commands in sequence: (Only the relevent output is shown after command executions)

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    integrate[x*log[2*sin[x/2]]^2,{x,0,pi/3}]
-- snip --
> 0.25554854129290762855238976168333131037737175253636607542005616591624
epsilon=-50
pslq[0.25554854129290762855238976168333131037737175253636607542005616591624, table[pi^i,{i,2,4}]]
-- snip --
> Relation:  0 =
> +  6480.* pslq001
> +     0.* pslq002
> +     0.* pslq003
> +   -17.* pslq004
> Result[ 37] through Result[ 40] =
>      6480.00000000000000000000000000000000000000000000000000000000000000000000
>       0.00000000000000000000000000000000000000000000000000000000000000000000e0
>       0.00000000000000000000000000000000000000000000000000000000000000000000e0
>       -17.00000000000000000000000000000000000000000000000000000000000000000000

    So, the above output means, \(0= 6480\times 0.255548541292907628552389761683331310377371752536366075420056165916 - 17\times \pi^4\), and hence \(c=17, a=4, b=6480\)

    When doing such computations, it's good to have more digits when computing integral or sums, and reduce the epsilon value when using pslq, so that it checks for fewer decimal places when trying for an integer relation. Otherwise, it is likely to miss the relation when the numerical accuracy is kept high. digits=100 and epsilon=-50 worked well for me in most cases.

  2. \begin{equation*} \displaystyle \int_0^{2\log{\phi}} \log{\left(2\, \sinh{\frac{x}{2}}\right)} = -\frac{\pi^a}{b} \end{equation*} 
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    epsilon=-100
integrate[log[2*(exp[x/2]-exp[-x/2])/2],{x,1e-100,2*log[(1+sqrt[5])/2]}]
> -0.98696044010893586188344909998761511353136994072407906264133493762200
epsilon=-50
pslq[-0.98696044010893586188344909998761511353136994072407906264133493762200, table[pi^i,{i,1,4}]]
> Relation:  0 =
> +  10.* pslq001
> +   0.* pslq002
> +   1.* pslq003
> +   0.* pslq004
> +   0.* pslq005
> Result[ 29] through Result[ 33] =
>        10.00000000000000000000000000000000000000000000000000000000000000000000
>       0.00000000000000000000000000000000000000000000000000000000000000000000e0
>         1.00000000000000000000000000000000000000000000000000000000000000000000
>       0.00000000000000000000000000000000000000000000000000000000000000000000e0
>       0.00000000000000000000000000000000000000000000000000000000000000000000e0

    In this example, mathtool chokes when lower limit is 0 saying argument is too large, so keep it close to 0. Then using pslq, we see that a=2 and b=10.

  3. \begin{equation*} \displaystyle \int_{-\infty}^{\infty} \dfrac{\log{\left(1 + e^{2x}\right)}}{1 + e^{3\, x}} = \frac{a \pi^b}{c} \end{equation*} 
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    epsilon=-100
Integrate[Log[1 + Exp[2*x]]/(1 + Exp[3* x]), {x, -Infinity, Infinity}]

    If we try to execute it directly, it complains "argument too large". So, we transform it by substituting \(e^x=y\)

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    Integrate[Log[1 + y^2]/(1 + y^3)/y, {y, 0, Infinity}]
> 0.59400396858408176872614992128884242944017635321356610251561824949472
epsilon=-50
pslq[0.59400396858408176872614992128884242944017635321356610251561824949472,table[pi^i,{i,1,5}]]
> Relation:  0 =
> +  216.* pslq001
> +    0.* pslq002
> +  -13.* pslq003
> +    0.* pslq004
> +    0.* pslq005
> +    0.* pslq006
> Result[  7] through Result[ 12] =
>       216.00000000000000000000000000000000000000000000000000000000000000000000
>       0.00000000000000000000000000000000000000000000000000000000000000000000e0
>       -13.00000000000000000000000000000000000000000000000000000000000000000000
>       0.00000000000000000000000000000000000000000000000000000000000000000000e0
>       0.00000000000000000000000000000000000000000000000000000000000000000000e0
>       0.00000000000000000000000000000000000000000000000000000000000000000000e0

    Hence, \(a=13, b=2, c=216\)

  4. \begin{equation*} \displaystyle \int_0^1 \log{x}\log{\left(1-x\right)} dx = \frac{a}{b}-\frac{\pi^c}{d} \end{equation*} 
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    epsilon=-100
integrate[log[x]*log[1-x],{x,0,1}]
> 0.35506593315177356352758483335397481078105009879320156226444177062999
epsilon=-50
pslq[0.35506593315177356352758483335397481078105009879320156226444177062999, 1,table[pi^i,{i,1,5}]]
> Relation:  0 =
> +  -6.* pslq001
> +  12.* pslq002
> +   0.* pslq003
> +  -1.* pslq004
> +   0.* pslq005
> +   0.* pslq006
> +   0.* pslq007
> Result[ 27] through Result[ 33] =
>        -6.00000000000000000000000000000000000000000000000000000000000000000000
>        12.00000000000000000000000000000000000000000000000000000000000000000000
>       0.00000000000000000000000000000000000000000000000000000000000000000000e0
>        -1.00000000000000000000000000000000000000000000000000000000000000000000
>       0.00000000000000000000000000000000000000000000000000000000000000000000e0
>       0.00000000000000000000000000000000000000000000000000000000000000000000e0
>       0.00000000000000000000000000000000000000000000000000000000000000000000e0

    Therefore, \(a=2, b=1, c=2, d=6\)