I stumbled on an interesting integral doing some physics exercise which did not require its closed form (if it has any). It has, however, sparked my interest and I tried my best to find it, but I couldn't. The integral is :
$$\int_0^R \frac{\text{d}x}{\sqrt{\ln(1+x)}}$$ with $$R>0$$
It looks like $$\int_0^R \frac{\text{d}x}{(x+1)\sqrt{\ln(1+x)}}$$ which is easy to integrate, but it seems like the first one is a lot harder. Wolfram Alpha does not display any results in the general case (for any positive real $R$), even if it does find some results for characteristic values like $R$=1. This makes me think there is no easy closed form for this integral. I am, however, open to any suggestions on how to tackle this more efficiently, or any solution, even if it is probably beyond my maths level.
Edit : This wasn't so hard indeed, I should have thought about this. I'm still puzzled about the complex substitution necessary to find tired's result (with a missing factor of $-i$ apparently), so I'll look for some confirmation before I accept anything.
Ok, this is not so difficult:
Setting $x+1=y$ our integral in question is:
$$ I=\int_1^{R+1}\frac{1}{\sqrt{\log(y)}}dy $$
next, using $y=e^q$ we obtain:
$$ I=\int_0^{\log(R+1)}\frac{e^q}{\sqrt{q}}dq $$
then, setting $q=p^2$ we get
$$ I=2\int_{0}^{\sqrt{\log(R+1)}}e^{p^2}dp=-i\sqrt{\pi}\text{Erf}(i\sqrt{\log(R+1)}) $$ if I’m not mistaken. $\text{Erf}(z)$ denotes the Error function. Note that it is legitimate to substitute $p\rightarrow ip$ because our integrand happens to be an entire function.