A closed form for $\int_0^\infty\left(\frac{2^{-x}-3^{-x}}x\right)^adx,\ a\notin\mathbb{Z}^+$

Let $$I(a)=\int_0^\infty\left(\frac{2^{-x}-3^{-x}}x\right)^adx.$$ $I(a)$ has closed form representations for all $a\in\mathbb{Z}^+$.

• Is there any algebraic (or at least period) $a\notin\mathbb{Z}^+$ such that $I(a)$ has a closed form representation?

• In particular, does $\displaystyle I\left(\frac12\right)=\int_0^\infty\sqrt{\frac{2^{-x}-3^{-x}\vphantom|}{x}}\ dx\ $ have a closed form representation?

For $a=1$ we have: $$ \int_{0}^{+\infty}\frac{2^{-x}-3^{-x}}{x}\,dx = \log\left(\frac{\log 3}{\log 2}\right)$$ by Frullani's theorem. For $a=2$ we can apply integration by parts: $$ \int_{0}^{+\infty}\left(\frac{2^{-x}-3^{-x}}{x}\right)^2\,dx = 2\int_{0}^{+\infty}\frac{2^{-x}-3^{-x}}{x}(3^{-x}\log 3-2^{-x}\log 2)\,dx$$ so: $$ I(2)=2\left(\log 3\cdot\log\left(\frac{\log 9}{\log 6}\right)-\log2\cdot\log\left(\frac{\log 6}{\log 4}\right)\right).$$ by Frullani's theorem again. To give an extension of the technique for non-integer $a$, we just need to compute some fractional derivatives/integrals. At least in theory. For $a=\frac{1}{2}$, we just need to compute the semiderivative of $\frac{1}{\sqrt{x}}$ (easy) and the semi-integral of $\sqrt{2^{-x}-3^{-x}}$ (hard).