Finding $\int \ln\ln x\ dx$

Question

Find $\int \ln\ln x\,\mathrm dx$.

I tried$$\int \ln\ln x\,\mathrm dx=x\ln\ln x -\int \frac{1}{\ln x}\,\mathrm dx\\=x\ln x\ln x-\sqrt{x}{\ln x}-\int \frac{1}{(\ln x)^2}\,\mathrm dx$$ It seems more and more difficult. And I tried substitution, i.e. substituting the $\ln \ln x$ to $t$, then $=\int t\ e^{e^t}e^t\,\mathrm dt$ and then I can not integral it.

There is no simple expression for your integral. Moreover, the first line of analysis appears incorrect: the last term should be $ \int 1 / ln x dx $ — WA Don, Apr 30 '20 at 12:51
Did you mean it can not be represented as the elementary function?@GerryMyerson — fractal, Apr 30 '20 at 12:53

score 1 · Answer 1 · answered Apr 30 '20 at 12:50

1

$$ \int\ln\ln(x)dx=x\ln\ln(x)-\int\frac{dx}{\ln x}=x\ln\ln(x)-\text{Li}(x)+c $$

answered Apr 30 '20 at 12:50

Tuvasbien

8,907

Could you provide more details abour what Li(x) is? – fractal Apr 30 '20 at 12:58
1

https://en.wikipedia.org/wiki/Logarithmic_integral_function – Henry Apr 30 '20 at 12:58
Thank you! I got it. – fractal Apr 30 '20 at 13:22

Finding $\int \ln\ln x\ dx$

1 Answers1