Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
3  
Note: $\log_x e = \frac{1}{\log_e x}=\frac{1}{\ln x}$ –  Thomas Andrews Nov 28 '12 at 21:26
1  
Do you mean $\log_x e$? If so note that $\log_x(e)=\frac{\ln e}{\ln x}$ –  Stefan Nov 28 '12 at 21:27
1  
Not in terms of elementary functions. It is the logarithmic integral $\text{li}(x)$. Important in several places, notably the distribution of prime numbers. –  André Nicolas Nov 28 '12 at 21:30
2  
This is known as $Li(x)$. –  Raskolnikov Nov 28 '12 at 21:30
2  
@Hooman: Are you sure you don't mean $\log x$ with base $e$? I think that would be much more likely... –  Jesse Madnick Nov 29 '12 at 0:25

2 Answers 2

up vote 3 down vote accepted

$$ \int\log_x(e)\,\mathrm{d}x=\int\frac1{\log(x)}\,\mathrm{d}x $$ 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.

share|improve this answer
    
@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, $\log_x(e)=\frac{\log(e)}{\log(x)}=\frac1{\log(x)}$. –  robjohn Nov 29 '12 at 2:10
    
Thanks a lot for the proof,,, –  Hooman Nov 29 '12 at 2:18

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 $\log_w e$.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.