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I am not sure, how proof this integrate $\int_2^\infty\frac{dx}{x\log^s(x)}=\int_2^\infty\frac{\log'(x)}{\log^{s}(x)}=\frac1{1-s}\log^{1-s}(x)\Big|_2^\infty$ without strong intuition and showing the primary function.

Do you have any idea?

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    $\begingroup$ Let $$u=log(x)$$ $\endgroup$ – Zach Jul 19 at 19:01
  • $\begingroup$ What base is the logarithm? e ? $\endgroup$ – Mandelbrot Jul 19 at 19:22
  • $\begingroup$ Are you sure the answer isn't $=\frac{1}{1-s} \ln^{1-s}(x) \vert_{\ln 2}^{\infty}$ $\endgroup$ – Mandelbrot Jul 19 at 19:25
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To prove it, note that $(\log^a(x))' =a\log^{a-1}(x)(\log'(x)) =\dfrac{a\log^{a-1}(x)}{x} $.

Therefore $(\log^{1-s}(x))' =\dfrac{(1-s)\log^{(1-s)-1}(x)}{x} =\dfrac{(1-s)\log^{-s}(x)}{x} $ so $(\dfrac{\log^{1-s}(x)}{1-s})' =\dfrac{\log^{-s}(x)}{x} =\dfrac1{x\log^{s}(x)} $.

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