Integration of logarithm

Question

$\int \ln(\ln \sqrt{x})^{\ln (x)}dx$ how should I integrate this? I think it can't be integrated. I don't know.

Answer 1

The integral can be transformed into some easy-integrate parts and the integral $\int \frac{dx}{\ln x}$ by integration by parts. But the latter has no closed form. For details see http://en.m.wikipedia.org/wiki/Li(x). Hope this is useful.

Answer 2

$$I = \int \ln(\ln \sqrt{x})^{\ln x} dx$$ $$ = \int \ln \left(\ln \left(x^{1/2}\right) \right)^{\ln x} dx$$ $$ = \int \ln \left(\frac{1}{2}\ln x \right)^{\ln x} dx$$ $$= \int (\ln x) \cdot \ln \left(\frac{1}{2}\ln x \right) dx$$ $$ = \int (\ln x) \cdot \left(\ln \frac{1}{2} + \ln x \right) dx$$

Let $$u = \ln x$$, $$x = e^u$$, $$\frac{du}{dx} = \frac{1}{x}$$, $$x = x du = e^u du$$

Therefore,

$$I = \int u \left( \ln \frac{1}{2} + u \right) e^u du$$ $$ = \int \left( \ln \frac{1}{2} \cdot u e^u + u^2 e^u \right) du$$ which can be integrated by parts.