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A while ago I started reading these notes on special functions, but I got stuck trying to find all positive numbers $T$ satisfying $$\int_T^\infty x^{-\log{x}} dx = \int_0^Tx^{-\log{x}}dx$$

I noticed the integrand is $e^{-(\log{x})^2}$, but this isn't quite a Gaussian because the former is missing a factor of $x^{-1}$. Based on the straightforwardness of the other problems, I suspect there is a clever way of doing this, but unfortunately I'm not seeing it at the moment.

Hints would be greatly appreciated! Complete solutions are alright too, but in that case would you please use spoiler formatting? Thanks!

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2 Answers 2

up vote 1 down vote accepted

$\displaystyle T=\sqrt{\mathrm e}$

Hint: Starting from your remark, use the change of variable $t=\sqrt2\log x$ to transform everything into gaussian integrals $\Phi(t)=\frac1{\sqrt{2\pi}}\int\limits_{-\infty}^t\mathrm e^{-s^2/2}\mathrm ds$ for some suitable values of $t$. Use identities such as $\Phi(+\infty)=1$ and a second change of variables $s=t-\frac1{\sqrt2}$ to reach an equation solved by $\sqrt2\log T$. Inverting this equation should yield something similar to the spoiler above.

If nothing works, post a comment to complain.

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Got it, thank you! –  sourisse Nov 1 '12 at 21:30
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First of all, there is only one such number, because the integrand in positive, bounded from above and converges at $\infty$. The integrand $x^{-\log x} dx$ can be rewritten as $ x^{1-\log x}dx/x = e^{\log x - (\log x)^2} d(\log x) = e^{u-u^2}du $ where $u=\log x$, and this can be rewritten as $e^{1/4-v^2}dv$ where $v=u-1/2$. Finally, $x=0$ corresponds to $v=-\infty$ and $x=\infty$ corresponds to $v=\infty$. The $T$ you are looking for corresponds to $v=0$, which gives you $\log T -1/2 =0$, or $T=\sqrt e$.

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