How to integrate $\int \ln(\ln(x))dx$?

Question

How to integrate

$$\int \ln(\ln(x))dx?$$

I am trying this question by substituting $\ln(x)=u$. Therefore $\frac{1}{x}dx=du$. Therefore $dx=e^{u}du$. So, the integral will become

$$\int (\ln(u))e^{u}du.$$

Now I am thinking of applying Integration by Parts. So, here according to the rule of ILATE, $u=(\ln(u))$ and $v=e^{u}$. Let $I=(\ln(u))e^{u}$. Now after applying IBP we will get

$$I= (\ln(u))e^{u}-\int \frac{e^{u}}{u}du.$$

[Because $\int e^{u}du=e^{u}$]. But after this step, I can't proceed further. Please help me out. Because, I am facing problem in integrating $$\int \frac{e^{u}}{u}du.$$

Answer 1

$$\int \ln(\ln x)dx = \int (\ln(\ln x) + \frac{1}{\ln x} - \frac{1}{\ln x})dx$$

$$= \int d(x\cdot\ln(\ln x)) - \int \frac{1}{\ln x}dx$$

$$ = x\cdot\ln(\ln x) + c - \int \frac{1}{\ln x}dx$$

The right hand side integral is also known as "logarithmic integral" and is denoted by $\operatorname{li}(x)$. Thus:

$$\int \ln(\ln x)dx = x\cdot\ln x(\ln x) - \operatorname{li}(x) +c$$

Edit: About the question owner's result $e^u.\ln u - \int \frac{e^u}{u}du$ is same result with I found. Just reverse the substitutions $u = \ln x$ :

$$e^{\ln x} . \ln(\ln x) - \int \frac{e^{\ln x}}{\ln x} \frac{dx}{x}$$ $$x. \ln(\ln x) - \int \frac{1}{\ln x} dx$$ $$x. \ln(\ln x) - \operatorname{li}(x)$$

I would like to say that the integral $\int \frac{e^u}{u}du$ is also known as "exponential integral".

Both exponential and logarithmic integrals can't be represented with standart mathematical functions. Actually you found the correct result.