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Is it possible to calculate the integral of $\log e$ with base of $x$?

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    @GerryMyerson this is what happens when someone doesn't take advice of friends and transfer from MIT to Caltech (stupid university ...)2012-11-29

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$ \int\log_x(e)\,\mathrm{d}x=\int\frac1{\log(x)}\,\mathrm{d}x $ This is known as the Log-Integral. $ \begin{align} \mathrm{li}(x) &=\int_0^x\frac1{\log(t)}\,\mathrm{d}t\\[6pt] &=\lim_{a\to0^+}\int_0^{1-a}\frac1{\log(t)}\,\mathrm{d}t +\int_{1+a}^x\frac1{\log(t)}\,\mathrm{d}t\\[6pt] &=\lim_{a\to0^+}\int_{-\infty}^{\log(1-a)}e^s\frac{\mathrm{d}s}{s} +\int_{\log(1+a)}^{\log(x)}e^s\frac{\mathrm{d}s}{s}\\[6pt] &=\lim_{a\to0^+}\log|\log(1-a)|\,e^{\log(1-a)}-\int_{-\infty}^{\log(1-a)}\log|s|\,e^s\,\mathrm{d}s\\ &\hphantom{\lim_{a\to0^+}}+\int_{\log(1+a)}^{\log(x)}\frac{\mathrm{d}s}{s} +\int_{\log(1+a)}^{\log(x)}(e^s-1)\frac{\mathrm{d}s}{s}\\[6pt] &=\lim_{a\to0^+}\log|\log(1-a)|\,(1-a)-(-\gamma)+\log|\log(x)|-\log|\log(1+a)|\\ &\hphantom{\lim_{a\to0^+}}+\int_0^{\log(x)}(e^s-1)\frac{\mathrm{d}s}{s}\\[6pt] &=\gamma+\log|\log(x)|+\int_0^{\log(x)}(e^s-1)\frac{\mathrm{d}s}{s}\\[6pt] &=\gamma+\log|\log(x)|+\sum_{k=1}^\infty\frac{\log(x)^k}{k\,k!} \end{align} $ where $\gamma$ is the Euler-Mascheroni Constant.

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    Thanks a lot for the proof,,,2012-11-29
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A guess: Since this is reported in comments below the question to have been in a "question sheet" in a course, is it possible that something like the following happened?

The question sheet says "Find the derivative $f'(w)$ if $f(w)=$

#1 etc. etc. etc.

#2 etc. etc. etc.

#3 etc. etc. etc.

#4 ${}\qquad\displaystyle \int_1^w (\log_x e)\,dx$

#5 etc. etc. etc."

Often students lose sight of the words at the beginning and mistakenly think they're being asked to find the integral.

postscript: ($f'(w)$ would of course be $\log_w e$.)