logarithmic trickery

Consider that since \log(a)\log(b)=\log(a^{\log b})=\log(b^{\log{a}}, and therefore, somewhat unexpectly: a^{\log b} = b^{\log a}
we can use it to construct new identities thusly

Let A = \prod_{n=1}^{\infty} \sqrt[2]{1+\frac{1}{n^{2}}}

Let B = \prod_{m=1}^{\infty} \sqrt[e^{m}]{1+\frac{1}{m^{3}}}

Therefore, since we know that A^{\log(B)}=B^{\log(A)} we can use that
to explicitly calculate a new identity thusly:

1. Compute the logarithm of A, \log(A):
\log(A) = \frac{1}{2} \sum_{v=1}^{\infty} \frac{(-1)^{v+1}\zeta(2v)}{v}

2. Compute the logarithm of B, \log(B):
\log(B) = \sum_{s=1}^{\infty} \frac{(-1)^{s+1}\mathrm{Li}_{(3s)}(1/e)}{s}

3. Combine them:

\prod_{n=1}^{\infty} \prod_{s=1}^{\infty} \sqrt[s]{\left(1+\frac{1}{n^{2}}\right)^{(-1)^{s+1}\mathrm{Li}_{(3s)}(1/e)}} = \prod_{m=1}^{\infty} \prod_{v=1}^{\infty} \sqrt[ve^{m}]{\left(1+\frac{1}{m^{3}}\right)^{(-1)^{v+1}\zeta(2v)}}

>>> A = fp.nprod(lambda n: sqrt(1+1/(n*n)), [1,inf])
>>> A
>>> B = fp.nprod(lambda m: (1+1/(m**3))**(1/exp(m)), [1,inf])
>>> B
>>> A**log(B)
>>> B**log(A)


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