An interesting double definite integral

First, to make it simpler to write, define:

$\displaystyle \text{li}(x)=\int_0^x \frac{1}{\ln t}\,dt$

$\text{li}(x)$ (Logarithmic Integral ) is termed as a special function but the following solution will not use any special properties related to this function.

$\displaystyle \int_0^1 \text{li}(x) \frac{x^{p-1}}{\sqrt{\ln\frac{1}{x}}}\,dx$

Use the substitution $x=e^{-u^2} \Rightarrow dx=-2ue^{-u^2}\,du$.

$\displaystyle \Rightarrow I(p)=2\int_0^{\infty} \text{li}(e^{-u^2})e^{-pu^2}\,du$

Divide both the sides by two and use integration by parts to obtain:

$\displaystyle \frac{I(p)}{2}=\left(u\text{li}(e^{-u^2})e^{-pu^2}\right|_0^{\infty}-\int_0^{\infty} u\left(\text{li}(e^{-u^2})(-2pue^{-pu^2})+\frac{e^{-pu^2}(-2ue^{-u^2})}{\ln(e^{-u^2})}\right)\,du$

$\displaystyle \Rightarrow \frac{I(p)}{4}=p\int_0^{\infty} u^2\text{li}(e^{-u^2})e^{-pu^2}\,du-\int_0^{\infty} e^{-(p+1)u^2}\,du$

$\displaystyle \Rightarrow \frac{I(p)}{4}=-p\frac{I'(p)}{2}-\frac{\sqrt{\pi}}{2\sqrt{p+1}}$

$\displaystyle \Rightarrow I'(p)+\frac{I(p)}{2p}=-\frac{\sqrt{\pi}}{2p\sqrt{p+1}}$

The above is a linear differential equation and its solution is:

$\displaystyle I(p)=-2\sqrt{\frac{\pi}{p}}\ln\left(\sqrt{p}+\sqrt{p+1}\right)+C$

As $p$ tends to infinity, $I(p)$ tends to zero, hence $C=0$.

$\displaystyle \Rightarrow I(p)=-2\sqrt{\frac{\pi}{p}}\ln\left(\sqrt{p}+\sqrt{p+1}\right)$

Please feel free to ask me any doubts regarding the solution. :)

$I(p)= \int_0^1 \frac{x^{p-1} \rm li(x)}{\sqrt{-\ln x}} \ \mathrm{d}x\overset{x=e^{-u^2}} = 2\int_0^{\infty} e^{-pu^2} \rm li(e^{- u^2})\mathrm{d}u=\frac{2}{\sqrt{p}} \int_0^{\infty} e^{-u^2} \rm li(e^{-\frac{u^2}{p}}) \ \mathrm{d}u$ Use the series described in the logarithmic integral wiki page to get : $I(p) = \frac{2}{\sqrt{p}} \int_0^{\infty} e^{- u^2} \left(\gamma + \ln \frac{u^2}{p} +\sum_{n\geq 1 } \frac{(-1)^n p^{-n}u^{2n}}{n\cdot n!}\right) \ \mathrm{d}u$ Now, it is just expand and Taylor series to finish (too lazy to finish it) but here's the main idea : $\int_0^{\infty} e^{-u^2}(\gamma +\ln \frac{u^2}{p} ) \mathrm{d}u = -\frac{-\sqrt{\pi}\ln(4p)}{2}$ and $\int_0^{\infty} u^{2n} e^{u^2} \ \mathrm{d}u =\frac{(2n+1)!!}{2^n}$

Fun Fact: This can also be written as $-2\sqrt{\frac\pi p}\operatorname{arcsinh}(\sqrt p)$.