$\int_1^e \! \frac{\mathrm{d}x}{x\,\sqrt{-\ln \ln x}}$ $$\int\limits_1^e \! \frac{\mathrm{d}x}{x\,\sqrt{-\ln \ln x}}$$
I can't find any antiderivative, is it possible to calculate the definite integral?
 A: Indeed

$$I:=\int\limits_1^e \frac{dx}{x\sqrt{-\log \log x}} = \sqrt{\pi}.$$

When one makes the change of variables $x=e^u$, one has that
$$I = \int\limits_0^1 \frac{du}{\sqrt{-\log u}} = \int \limits_0^1 \left(\log \frac{1}{u}\right)^{-1/2} \; du.$$
One should now make the change of variables $u = e^t$. This yields the desired result
$$I = - i \int\limits_{-\infty}^0 \frac{e^t}{\sqrt{t}} = \sqrt{\pi}.$$

One need only show that
$$\int\limits \frac{e^x}{\sqrt{x}} \; dx = \sqrt{\pi} \, \mathrm{erfi}(\sqrt{x}),$$
where $\mathrm{erfi}(x)$ denotes the imaginary error function.
A: Thanks!
With knowing the following I was able to solve the problem.

$$\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$$

Back to the definite integral:
$$\int\limits_1^e \! \frac{\mathrm{d}x}{x\,\sqrt{-\ln \ln x}}$$
substitute $u = ln(x)$
$$= \int\limits_0^1 \! \frac{\mathrm{d}u}{\sqrt{-\ln(u)}}$$
substitute $r = ln(u)$
$$= \int\limits_{-\infty}^0 \! \frac{e^r}{\sqrt{-r}}\mathrm{d}r = \int\limits_{0}^{\infty} \! \frac{e^{-r}}{\sqrt{r}}\mathrm{d}r$$
substitute $r = z^2$
$$= 2 \cdot \int\limits_{0}^{\infty} \! e^{-z^2}\mathrm{d}z = \int\limits_{-\infty}^{\infty} \! e^{-z^2}\mathrm{d}z = \sqrt{\pi}$$
Hence 

$$\int\limits_1^e \! \frac{\mathrm{d}x}{x\,\sqrt{-\ln \ln x}} = \sqrt{\pi}$$

