I wrote the following integral for an integration bee earlier this year, with the intended solution being to manipulate the integrand into the form of a Frullani integral (https://en.wikipedia.org/wiki/Frullani_integral):
$$\int_0^\infty 1+\frac{2}{\sqrt[x]{8}}-\frac{3}{\sqrt[x]{4}}\,\mathrm dx$$
That being said, I know at least one person purportedly solved it using a non-Frullani method, in particular something involving the incomplete gamma function (https://en.wikipedia.org/wiki/Incomplete_gamma_function). I know nothing about the incomplete gamma function beyond a cursory skim of the Wikipedia page linked above, and I personally don't see how one would go about using it. Can the incomplete gamma function be used for this integral, and if so, how?
Solving a Frullani integral with the incomplete gamma function
calculusgamma functionintegration
Best Answer
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).