Is it possible to calculate the integral of $\log e$ with base of $x$?
Calculating the integral of $\log e$ with base of $x$
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3Note: $\log_x e = \frac{1}{\log_e x}=\frac{1}{\ln x}$ – 2012-11-28
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1Do you mean $\log_x e$? If so note that $\log_x(e)=\frac{\ln e}{\ln x}$ – 2012-11-28
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0Ok , How to calculate integral of ln e/ln x – 2012-11-28
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1Not 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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2[This is known as $Li(x)$.](http://en.wikipedia.org/wiki/Logarithmic_integral_function) – 2012-11-28
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0Can 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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0@Hooman There is no (known?) closed form for the integral in terms of the elementary functions. – 2012-11-28
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0So it doesn't have answer . – 2012-11-28
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0Only if your test was about numeric integration. – 2012-11-28
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0Either the test gave you a problem you can't solve, or you mis-entered the problem here. – 2012-11-28
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0Question 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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1Really? The question sheet actually said, "log $e$ with base of $x$"? In those exact words? – 2012-11-28
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0Occasionally 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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2@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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0@GerryMyerson this is what happens when someone doesn't take advice of friends and transfer from MIT to Caltech (stupid university ...) – 2012-11-29
2 Answers
$$ \int\log_x(e)\,\mathrm{d}x=\int\frac1{\log(x)}\,\mathrm{d}x $$ 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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0@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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0Thanks a lot for the proof,,, – 2012-11-29
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 $\log_w e$.)