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I am interested to evaluate integral of the form $\displaystyle\int {g(x)}^{g'(x)}\,\mathrm dx$. I have got this simple example: $\displaystyle\int_{0}^{1} \left(\frac{1}{x}\right) ^{\log x}\,\mathrm dx$. Wolfram Alpha gives a nice closed form which is defined as shown below: $$\int_0^1\left(\dfrac1x\right)^{\log(x)}\,\mathrm dx = -\dfrac12\sqrt[4]{e}\sqrt\pi\left(\mathrm{erf}\left(\dfrac12\right) - 1\right) \approx 0.545641\tag{1}$$

Now my question here is: Is there any mathematical basis that gives us rule(s) to evaluate integrals of the form $\displaystyle\int {g(x)}^{g'(x)}\,\mathrm dx$? And at the same time how can we arrive at the solution in $(1)$?

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    $\begingroup$ Change $t=\log x$ works for the identity. Not clear what $g^{g'}$ is. For the example, it is $(g')^g$ if $g'$ is the derivative. $\endgroup$ – A.Γ. May 27 '18 at 22:08
  • $\begingroup$ Very easy, just change of variables and completing the square. $\endgroup$ – Szeto May 27 '18 at 22:33
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My Attempt to answer my question :

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    $\begingroup$ what is "Posons" :-)? $\endgroup$ – tired May 27 '18 at 22:47
  • $\begingroup$ Sorry i solved it in french , it means : assume $\endgroup$ – zeraoulia rafik May 27 '18 at 22:48
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    $\begingroup$ @tired. "Posons" = let $\endgroup$ – Claude Leibovici May 28 '18 at 4:24

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