Finding $ {\int e^\sqrt[k]{-a \ln(x)+b} \:dx} $ in terms of special functions

Question

I have come across this distribution in my work, and I was wondering if it were possible to represent its integral in some way.

$${\int e^{\sqrt[k]{-a \ln{x}+b} }\:dx}$$

I'm pretty sure there's no closed form solution to this, and attempting to take the Maclaurin series of this and integrating that is tedious to say the least.

Can this integral be expressed in terms of a special function? Even the hypergeometric function will do at a stretch.

Answer 1

Try plugging the exponent into the maclaurin series for the exponential function and integrating term by term.

Answer 2

Expanding with infinite sum:

$\int \exp \left((-a \ln (x)+b)^{1/k}\right) \, dx=\int \left(\sum _{n=0}^{\infty } \frac{(b-a \ln (x))^{n/k}}{n!}\right) \, dx=\sum _{n=0}^{\infty } \int \frac{(b-a \ln (x))^{n/k}}{n!} \, dx=\sum _{n=0}^{\infty } \frac{e^{b/a} \Gamma \left(\frac{k+n}{k},\frac{b}{a}-\ln (x)\right) \left(\frac{b}{a}-\ln (x)\right)^{-\frac{n}{k}} (b-a \ln (x))^{n/k}}{n!}=\sum _{n=0}^{\infty } \frac{e^{b/a} E_{-\frac{n}{k}}\left(\frac{b}{a}-\ln (x)\right) (b-a \ln (x))^{1+\frac{n}{k}}}{a n!}$

where: $\Gamma \left(\frac{k+n}{k},\frac{b}{a}-\ln (x)\right)$ is incomplete gamma function, $E_{-\frac{n}{k}}\left(\frac{b}{a}-\ln (x)\right)$ is exponential integral function.

Edited:

only for $k=2$: $$\int \exp \left(\sqrt{-a \ln (x)+b}\right) \, dx=\sum _{n=0}^{\infty } \frac{e^{b/a} E_{-\frac{n}{2}}\left(\frac{b}{a}-\ln (x)\right) (b-a \ln (x))^{1+\frac{n}{2}}}{a n!}=\sum _{n=0}^{\infty } \frac{\left(e^{b/a} (b-a \ln (x))^{1+\frac{n}{2}}\right) \int_1^{\infty } \frac{\exp \left(-t \left(\frac{b}{a}-\ln (x)\right)\right)}{t^{-\frac{n}{2}}} \, dt}{a n!}=\int_1^{\infty } \left(\sum _{n=0}^{\infty } \frac{\left(e^{b/a} (b-a \ln (x))^{1+\frac{n}{2}}\right) \exp \left(-t \left(\frac{b}{a}-\ln (x)\right)\right)}{(a n!) t^{-\frac{n}{2}}}\right) \, dt=\int_1^{\infty } \frac{e^{\frac{b}{a}-\frac{b t}{a}+\sqrt{t} \sqrt{b-a \ln (x)}} x^t (b-a \log (x))}{a} \, dt=e^{\sqrt{b-a \ln (x)}} x+\frac{e^{\frac{b}{a}+\frac{a b}{4 b-4 a \ln (x)}} \sqrt{\pi } x^{\frac{a^2}{-4 b+4 a \ln (x)}} \sqrt{b-a \ln (x)}}{2 \sqrt{\frac{b}{a}-\ln (x)}}+\frac{e^{\frac{b}{a}+\frac{a b}{4 b-4 a \ln (x)}} \sqrt{\pi } x^{\frac{a^2}{-4 b+4 a \ln (x)}} \text{erf}\left(\frac{-\frac{2 b}{a}+2 \ln (x)+\sqrt{b-a \ln (x)}}{2 \sqrt{\frac{b}{a}-\ln (x)}}\right) \sqrt{b-a \ln (x)}}{2 \sqrt{\frac{b}{a}-\ln (x)}}$$