A nice double integral

Let $\mathcal{I}(\alpha)=\int_0^1\int_0^x\frac{x(t^\alpha-1)}{\ln t}\ dt\ dx.$ Differentiating $\mathcal{I}(\alpha)$ with respect to $\alpha$ yields $\begin{aligned} \frac{d\mathcal{I}}{d\alpha}&=\int_0^1x\int_0^x\frac{\partial}{\partial\alpha}\left[\frac{t^\alpha-1}{\ln t}\right]\ dt\ dx\\ &=\int_0^1x\int_0^xt^\alpha\ dt\ dx\\ &=\int_0^1\frac{x^{\alpha+2}}{\alpha+1}\ dt\ dx\\ &=\frac{1}{(\alpha+1)(\alpha+3)}\\ \mathcal{I}(\alpha)&=\frac12\int\left[\frac{1}{\alpha+1}-\frac{1}{\alpha+3}\right]\ d\alpha\\ &=\frac12\ln\left|\frac{\alpha+1}{\alpha+3}\right|+C. \end{aligned}$ For $\alpha=0$ implying $\mathcal{I}(0)=0$ then $C=\frac12\ln3$ . Hence $\mathcal{I}(\alpha)=\frac12\ln\left|\frac{3(\alpha+1)}{\alpha+3}\right|$ and $\mathcal{I}(3)=\int_0^1\int_0^x\frac{x(t^3-1)}{\ln t}\ dt\ dx=\large\color{#3D99F6}{\frac12\ln2}\approx\color{#D61F06}{0.346574}.$

$\displaystyle \begin{aligned} \int_0^1 \int_0^x \frac{x(t^3-1)}{\ln t}\,dt\,dx &= \int_0^1 \int_t^1 \frac{x(t^3-1)}{\ln t}\,dx\,dt \\ &=\frac{1}{2}\int_0^1 \frac{(1-t^2)(t^3-1)}{\ln t}\,dt \\ &=\frac{1}{2}\int_0^{\infty} \frac{(e^{-2y}-1)(e^{-3y}-1)}{y}e^{-y}\,dy\,\,\,\,\,\,(t=e^{-y}) \\ &=\frac{1}{2}\left(\int_0^{\infty} \frac{e^{-6y}-e^{-3y}}{y}\,dy+\int_0^{\infty} \frac{e^{-y}-e^{-4y}}{y}\,dy\right) \\ &= \frac{1}{2}\left(\ln\frac{1}{2}+\ln 4\right)\\ &=\boxed{\dfrac{\ln 2}{2}} \\ \end{aligned}$

The final two integrals can be evaluated using Frullani's Integral .