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

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    Note: $\log_x e = \frac{1}{\log_e x}=\frac{1}{\ln x}$2012-11-28
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    Do you mean $\log_x e$? If so note that $\log_x(e)=\frac{\ln e}{\ln x}$2012-11-28
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    Ok , How to calculate integral of ln e/ln x2012-11-28
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    Not in terms of elementary functions. It is the logarithmic integral $\text{li}(x)$. Important in several places, notably the distribution of prime numbers.2012-11-28
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    [This is known as $Li(x)$.](http://en.wikipedia.org/wiki/Logarithmic_integral_function)2012-11-28
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    Can you Please Explain How to Calcualte this integral , it was an extra credit questions in my today's exam ,2012-11-28
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    @Hooman There is no (known?) closed form for the integral in terms of the elementary functions.2012-11-28
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    So it doesn't have answer .2012-11-28
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    Only if your test was about numeric integration.2012-11-28
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    Either the test gave you a problem you can't solve, or you mis-entered the problem here.2012-11-28
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    Question is correct I have question sheet and I asked instructor to make sure about correctness of this problem,2012-11-28
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    Really? The question sheet actually said, "log $e$ with base of $x$"? In those exact words?2012-11-28
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    Occasionally students are asked to find the derivative of something like $\displaystyle w\mapsto\int_1^w (\log_x e)\,dx$, and they mistakenly think they need to find the integral. Could that be what happened here?2012-11-28
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    @Hooman: Are you sure you don't mean $\log x$ with base $e$? I think that would be much more likely...2012-11-29
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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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    @Hooman: $\log_b(a)=\frac{\log(a)}{\log(b)}$, where the $\log$s are in the same base (usually base $e$ for math and base 10 for engineering). Thus, $\log_x(e)=\frac{\log(e)}{\log(x)}=\frac1{\log(x)}$.2012-11-29
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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 $\log_w e$.)