As I understand the intended evaluation of the integral
$$\int_0^\infty \left(1+\frac{2}{\sqrt[x]{8}}-\frac{3}{\sqrt[x]{4}}\right)\,dx,\tag1$$
it should be transformed by substitution ($x\to1/x$) to
$$\int_0^\infty\frac{1+2\,e^{-x\ln8}-3\,e^{-x\ln4}}{x^2}\,dx,\tag2$$
first, and then by partial integration to
$$6\,\ln2\int_0^\infty\frac{e^{-x\ln4}-e^{-x\ln8}}{x}\,dx,\tag3$$
a Frullani integral.
An alternative way would be to use the definition of the improper integral (2):
$$\lim_{\varepsilon\to0}\int_\varepsilon^\infty\frac{1+2\,e^{-x\ln8}-3\,e^{-x\ln4}}{x^2}\,dx=\lim_{\varepsilon\to0}\left(\frac1\varepsilon+6\,\ln2\,\Gamma(-1,\varepsilon\ln8)-6\,\ln2\,\Gamma(-1,\varepsilon\ln4)\right)\tag4$$ with the upper incomplete gamma function $\Gamma(-1,z)$. Since
$$\Gamma(-1,z)=\frac{e^{-z}}z-\Gamma(0,z)$$ and
$$\Gamma(0,z)=-\gamma-\ln z-\sum^\infty_{k=0}\frac{(-z)^k}{k\cdot k!}$$
with the Euler-Mascheroni constant $\gamma$, it's not hard (though cumbersome) to calculate the limit in (4).
Best Answer
Do you know about the regularized lower incomplete gamma function $P(a,z)$ and upper $Q(a,z)$? Your integral then is:$$\int_0^\infty x^{k-1} P(x,t)dx= \mathcal M_x\{P(x,t)\}(k)$$
which is just The Mellin Transform. Let this expansion be used:
$$ \mathcal M_x\{P(x,t)\}(k) =\int_0^\infty \frac{x^{k-1}t^x}{Γ(x+1)}\sum_{n=0}^\infty \frac{(-1)^n x t^n}{(x+n)n!}dx=\sum_{n=0}^\infty \frac{(-t)^n}{n!}\int_0^\infty \frac{x^kt^x}{x!(x+n)}dx$$
Let’s try to use Ramanujan’s Master Theorem:
$$\int_0^\infty x^{k-1} P(x,t)dx=Γ(k)\varphi(-k)$$
Where $\varphi(m)$ is a type of generating function for $P(x,t)$:
$$P(x,t)=\sum_{m=0}^\infty \frac{(-1)^m x^m}{m!}\varphi(m)$$
Note that $$\varphi(m)\mathop=^\text{may}\varphi(m,t)$$
because of the $t$ parameter. This integral looks almost like a Mu function. Please correct me and give me feedback!