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I am having quite a bit of difficulty integrating,

$$ \int \frac{\mathrm{d}x}{\ln x } $$

I believe a u-substitution will not work since if $ u = \ln(x) $ then $ \mathrm{d}u = \frac{\mathrm{d}x}{x} $ and I don't believe I can eliminate the variable $ x $.

Also I am not sure how to apply integration by parts to the problem. I'd imagine since the difficulty is integrating $ \frac{1}{\ln x} $ you would set $ u = \frac{1}{\ln x}$ and so $ u' = - \frac{1}{x( \ln x)^2} $, $ v' = 1 $ and $ v = x $. From the method of integration by parts,

$$ \int \frac{\mathrm{d} x}{\ln x} = \frac{x}{\ln x} + \int \frac{\mathrm{d} x}{(\ln x)^2 }$$

Which appears more complicated than the initial problem unless this suggests that the solution is an infinite series. Applying a few more integrations by parts it seems to be expressible as,

$$ \sum^\infty_{n=0} \frac{x (n!)}{(\ln x)^{n+1}} $$

However this series does not appear to converge and anyways I might be completely off track here. I've just gotten to Taylor Series and have a some idea as to how they can be used to approximate functions, perhaps one centered at $ \mathrm{e} $ could be used?

user38770
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2 Answers2

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$\frac{1}{\ln x}$ has no elementary antiderivative. If you are so inclined you could express $\ln x$ as a series.

$\ln (1+x) = x - x^2/2 + x^3/3 - ....$

$\to \ln (1+x) = (-1+1+x) - (-1+1+x)^2/2 + (-1+1+x)^3/3 - ....$

$\to \ln (x) = (-1+x) - (-1+x)^2/2 + (-1+x)^3/3 - ....$

http://en.wikipedia.org/wiki/Mercator_series

Or better yet: https://www.physicsforums.com/threads/integral-of-1-ln-x.224473/

And the fifth one: http://en.wikipedia.org/wiki/List_of_integrals_of_logarithmic_functions

BCLC
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  • Thank you! I recall now that $ \ln (1 + x) $ can be expressed as an infinite series, since it is the integral of the power series expansion of the geometric series with $ r = -x $ where $ |x| < 1$? I think I can figure out the rest from your answer, I just need a little time go back through all the calculus I have poor grasp of! – user38770 Oct 04 '14 at 22:21
  • @user38770 see other links too... – BCLC Oct 05 '14 at 00:10
  • Nice series at the end of your question btw – BCLC Oct 05 '14 at 00:16
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$x=e^{z} \implies dx=e^z \ dz$

$\therefore\displaystyle\int\dfrac{dx}{\ln x}=\displaystyle\int\dfrac{e^z}{z}dz=\displaystyle\int{\displaystyle\sum_{i=0}^n\dfrac{z^{i-1}}{i!}}dz=\displaystyle\sum_{i=0}^n\displaystyle\int\dfrac{z^{i-1}}{i!}dz=\displaystyle\sum_{i=0}^n\dfrac{z^{i-1}}{i\cdot i!}$