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$\begingroup$

I did a bit of work on this, but I'm not so sure about the parts towards the end. Starting with$$\int\frac{1}{\ln x}\,dx$$$$u=\ln x,1=\frac{dx}{du}\frac{1}{x},dx=x\,du,dx=e^{\ln x}du,dx=e^u\,du$$$$\int\frac{e^u}{u}\,du=\int\frac{1}{u}e^u\,du=\int\frac1u\sum_{n=1}^\infty\frac{u^n}{n!}\,du=\int\sum_{n=1}^\infty\frac{u^{n-1}}{n!}\,du$$$$\sum_{n=1}^\infty\frac{u^n}{n\cdot n!}=\sum_{n=1}^\infty\frac{\ln(x)^n}{n\cdot n!}$$which WolframAlpha tells me converges to $-\ln\left(-\ln x\right)-\Gamma\left(0,-\ln x\right)-\gamma$. I'm not sure what happened in this last step.

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    This is the [logarithmic integral](http://mathworld.wolfram.com/LogarithmicIntegral.html) that you are considering, which is nonelementary...2012-04-20
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    The $\gamma$ is Euler's gamma constant, concerning which there is much information in texts and on websites. I don't know what WA means by $\Gamma(x,y)$ but surely some documentation is available?2012-04-21
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    have you tried integration by parts?2012-06-26
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    @user1434040: That won't work. The integral is non-elementary.2012-06-26
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    @GerryMyerson $\Gamma(x,y)$ is the incomplete gamma function2012-08-31

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