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I've been struggling a while now trying to find an alternate form for $$G=\int_0^{\infty}{x^{1/x-x}dx}$$ in function of known constants, (like $e$,$\pi$,etc.)

Having looked at the problem, I don't think I have the knowledge to solve this problem.

Worth noting:

$$ G \approx 1.32073040087 $$

$$ G = \int_0^1{x^{1/x-x}d\left(x-\frac{1}{x}\right)},\tiny \left(\forall f(x)=f(1/x):\int_0^{\infty}{f(x)dx}=\int_0^1{f(x)d\left(x-\frac{1}{x}\right)}\right ) $$

$$ G = \int_{-\infty}^{\infty}{e^{u(1-2\sinh(u))}du} , \small x=e^u $$

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    $\begingroup$ I highly doubt it has any closed form, in fact the original expression is likely the most simple representation $\endgroup$ – Yuriy S Dec 27 '17 at 16:01
  • $\begingroup$ It may help to rewrite the original as $$ \int_0^\infty e^{\ln(x)/x - x\ln(x)}\,dx $$ $\endgroup$ – Omnomnomnom Dec 27 '17 at 16:02
  • $\begingroup$ Is there a representation for $\int_{0}^{1}x^{\frac{1}{x}}\,dx$ in terms of Sophomore-dream-like series? Maybe we can get something interesting in terms of the inverse function of $t\sinh t$ over $\mathbb{R}^+$ (is it related to the Lambert $W$ function in a simple way?), since $$ G= \int_{0}^{+\infty}2\cosh(t)\,e^{-2t\sinh t}\,dt.$$ $\endgroup$ – Jack D'Aurizio Dec 27 '17 at 17:55

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