Integral ${\large\int}_0^1\left(-\frac{\operatorname{li} x}x\right)^adx$

Let $\operatorname{li} x$ denote the logarithmic integral $$\operatorname{li} x=\int_0^x\frac{dt}{\ln t}.$$ Consider the following parameterized integral: $$I(a)=\int_0^1\left(-\frac{\operatorname{li} x}x\right)^adx.$$

Can we find a closed form for this integral?

We can find some special values of this integral: $$I(0)=1,\,\,I(1)=1,\,\,I(2)=\frac{\pi^2}6,\,\,I(3)\stackrel?=\frac{7\zeta(3)}2$$ The last value was suggested by numeric computations, and I do not yet have a proof for it.

Can we prove the conjectured value of $I(3)$?

One could expect that $I(4)$ might be a simple rational (or at least algebraic) multiple of $\pi^4$ but I could not find such a form.

Can we find closed forms for $I(4),I(5)$ and other small integer arguments?

$I(5)$ doesn't seem to have a closed form that I could find. On the other hand, I found numerically that $$ \begin{eqnarray}I(4) &=& -\tfrac{32}{3} \text{Li}_4(\tfrac{1}{2})+\tfrac{52}{3} \zeta(4)+\tfrac{8}{3} \zeta (2) \log^22-\tfrac{4}{9}\log^4 2. \end{eqnarray}$$

Case 3. The way to do the integral $I(3)$ is to write $$ \mathrm{li}(x) = -E_1(\log\tfrac1x), $$ and make the change of variable $t=1/u$, so that $$ I(a) = \int_1^\infty u^{a-2}E_1(\log u)^a\,du. $$ This way, the exponential integral takes the form $$ E_1(x) = \int_1^\infty \frac{dt}{t} e^{-x t}, $$ so that the integral $I(a)$ is equal to $$ I(a) = \int_1^\infty \frac{u^{a-2-\sum t}}{t_1\cdots t_a}\,du\,dt_1\cdots dt_a. $$ The integral over $u$ can be done easily enough: $$ I(a) = \int_1^\infty \left(\prod \frac{dt}{t}\right) \frac{1}{\sum t+1-a}. $$

For $a=3$, Mathematica can do this three-dimensional integral: $$ I(3) = \int_1^\infty\frac{ds\,du\,dv}{suv}\frac{1}{s+u+v-2} = \tfrac72\zeta(3). $$

In reference to Kirill's answer, I will show that indeed $$ I(3) = \int_{1}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{xyz} \frac{dx \, dy \, dz}{x+y+z-2} = \frac{7}{2}\zeta(3).$$

I will make the same change of variables I made in my answer to your other more recent question.

$$ \begin{align} I(3) &= \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{yz} \int_{1}^{\infty} \frac{1}{y+z-2}\left( \frac{1}{x} - \frac{1}{x+y+z-2} \right) \, dx \, dy \, dz \\ &= \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{yz} \frac{\log (y+z-1)}{y+z-2} \, dy \, dz \\ &= 2 \int_{1}^{\infty} \int_{1}^{z} \frac{1}{yz} \frac{\log (y+z-1)}{y+z-2} \, dy \, dz \\ &= 2 \int_{2}^{\infty} \int_{u-1}^{u^{2}/4} \frac{1}{v} \frac{\log(u-1)}{u-2} \frac{dv \, du}{\sqrt{u^{2}-4v}} \tag{1} \\ & = 4 \int_{2}^{\infty} \frac{\log(u-1)}{u-2}\int_{0}^{u-2} \frac{1}{u^{2}-t^{2}} \, dt \, du \tag{2} \\ &= 2 \int_{2}^{\infty} \frac{\log^{2}(u-1)}{u(u-2)} \, du \\ &= 2 \int_{1}^{\infty} \frac{\log^{2}(w)}{w^{2}-1} \, dw \\ &= 2 \int_{0}^{1} \frac{\log^{2}(s)}{1-s^{2}} \, ds \tag{3} \\ &= 2 \int_{0}^{1} \log^{2}(s) \sum_{n=0}^{\infty} s^{2n} \, ds \\ &= 2 \sum_{n=0}^{\infty} \int_{0}^{1} \log^{2}(s) s^{2n} \, ds \\ &= 4 \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{3}} \\ &= 4 \left(\sum_{n=1}^{\infty} \frac{1}{n^{3}} - \sum_{n=1}^{\infty} \frac{1}{(2n)^{3}} \right) \\ &= 4 \left( \zeta(3) - \frac{\zeta(3)}{8}\right) \\ &= \frac{7}{2} \zeta(3) \end{align}$$

$(1)$ Make the change of variables $u=y+z, v=yz$.

$(2)$ Make the substitution $t^2 = u^{2}-4v$.

$(3)$ Make the substitution $s = \frac{1}{w}$.