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What is the value of

$$\int\frac{1}{\log x}dx$$ I have tried many times, but failed everytime.

Can anyone help me out in solving this question.

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2 Answers 2

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If you don't like special functions symbols:

$\int\dfrac{1}{\log x}dx=\int\dfrac{\ln10}{\ln x}dx$

Let $u=\ln x$ ,

Then $x=e^u$

$dx=e^u~du$

$\therefore\int\dfrac{\ln10}{\ln x}dx$

$=\int\dfrac{e^u\ln10}{u}du$

$=\int\sum\limits_{n=0}^\infty\dfrac{u^n\ln10}{n!u}du$

$=\int\sum\limits_{n=0}^\infty\dfrac{u^{n-1}\ln10}{n!}du$

$=\int\left(\dfrac{\ln10}{u}+\sum\limits_{n=1}^\infty\dfrac{u^{n-1}\ln10}{n!}\right)du$

$=\ln10\ln u+\sum\limits_{n=1}^\infty\dfrac{u^n\ln10}{n!n}+C$

$=\ln10\ln\ln x+\sum\limits_{n=1}^\infty\dfrac{(\ln x)^n\ln10}{n!n}+C$

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    $\begingroup$ What is the valor of n in this context? $\endgroup$
    – Dinesh
    Apr 18, 2014 at 10:59
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    $\begingroup$ $\log$ probably denotes the natural logarithm here. $\endgroup$
    – user65203
    Sep 18, 2016 at 10:04
  • $\begingroup$ I think that it could be interesting to justify the change in the order of summation (for the readers). $\endgroup$ Oct 19, 2016 at 13:07
  • $\begingroup$ Correct, but not useful if one wishes to integrate over $[0,1-\delta)$ for $0<\delta<1$. $\endgroup$
    – Mark Viola
    Nov 16, 2021 at 14:26
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I assume that $\log x$ denotes the natural logarithm.

The antiderivative you are analyzing has no expression in terms of “elementary function”, just like the case of the antiderivative of $e^{-x^2}$.

You can transform it in various ways; with the substitution $t=-\log x$, we have $x=e^{-t}$ and $dx=-e^{-t}\,dt$, so we obtain $$ \int\frac{e^{-t}}{t}\,dt $$ but this doesn't help much.

See http://mathworld.wolfram.com/LogarithmicIntegral.html and http://mathworld.wolfram.com/ExponentialIntegral.html for those two closely related integrals.

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  • $\begingroup$ You can Taylor expand the $e^{-t}$ to get a series expansion. $\endgroup$ Jul 9, 2017 at 11:19

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