definite integral. closed form? Does this definite integral have a closed form $?$:
$$
\int_{0}^{1}
\frac{\mathrm{e}^{1/\log\left(x\right)}}
{x^{1/5}\,
\left[\log\left(x\right)\right]^{1/5}}
\,\mathrm{d}x
$$
Although it may not, I'm hopeful for a closed form. The troublesome part is the fractional exponents.

*

*For example if the denominator were $x^{1}$ times $\log^{n}\left(x\right)$ then the integral would turn out to be related to $\Gamma\left(n\right).$

*However when both exponents go fractional, it becomes much more diffiucult for me.

*I have a hunch that the closed form is in terms of a modified Bessel function of the second kind due to having evaluated similar integrals.

Although this integral looks daunting I think it can be solved $!$.
Thanks so much.
 A: With CAS help and Mellin Transform:
$$\int_0^1 \frac{\exp \left(\frac{1}{\log (x)}\right)}{\sqrt[5]{x} \sqrt[5]{\log (x)}} \, dx=\\\mathcal{M}_a^{-1}\left[\int_0^1 \mathcal{M}_a\left[\frac{\exp \left(\frac{a}{\log (x)}\right)}{\sqrt[5]{x}
   \sqrt[5]{\log (x)}}\right](s) \, dx\right](1)=\\\mathcal{M}_a^{-1}\left[\int_0^1 \frac{(-1)^{-s} \Gamma (s) \log ^{-\frac{1}{5}+s}(x)}{\sqrt[5]{x}} \, dx\right](1)=\\\mathcal{M}_s^{-1}\left[-(-1)^{4/5}
   \left(\frac{4}{5}\right)^{-\frac{4}{5}-s} \Gamma (s) \Gamma \left(\frac{4}{5}+s\right)\right](1)=\\-(-1)^{4/5} *\sqrt[5]{2}* 5^{2/5} *K_{\frac{4}{5}}\left(\frac{4}{\sqrt{5}}\right)\approx0.302595\, -0.219848 i$$
Where:
$\mathcal{M}_a[f(a)](s)$ is Mellin Transform,
$\mathcal{M}_s^{-1}[f(s)](a)$ is Inverse Mellin Transform,
$K_{\frac{4}{5}}\left(\frac{4}{\sqrt{5}}\right)$ is modified Bessel function of the second kind.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{m \in
\braces{\pm 1,\,\pm 3,\,\pm 5,\ldots}}$:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{1}
{\expo{1/\ln\pars{x}} \over x^{1/5}\,\,\bracks{\vphantom{\large A}\ln\pars{x}}^{1/5}}\,\dd x} =
\int_{0}^{1}
{\expo{1/\ln\pars{x}} \over x^{1/5}\,\,\bracks{\vphantom{\large A}\expo{m\pi\ic}\verts{\ln\pars{x}}}^{1/5}}\,\dd x
\end{align}
Lets $\ds{x \equiv \expo{-1/t} \iff t = -\,{1 \over \ln\pars{x}}}$:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{1}
{\expo{1/\ln\pars{x}} \over x^{1/5}\,\,\bracks{\vphantom{\large A}\ln\pars{x}}^{1/5}}\,\dd x}
\\[5mm] = &\
\expo{-m\pi\ic/5}\int_{0}^{\infty}
{\expo{-t} \over \expo{-1/\pars{5t}}\,\,t^{-1/5}}\,
\pars{\expo{-1/t}\,{1 \over t^{2}}\,\dd t}
\\[5mm] = &\
\expo{-m\pi\ic/5}\int_{0}^{\infty}t^{-9/5}\,\,
\exp\pars{-t - {4 \over 5}\,t}\,\dd t
\\[5mm] = &
\expo{-m\pi\ic/5}
\bracks{50^{1/5}\,\on{K}_{4/5}\pars{4\root{5} \over 5}}
\label{1}\tag{1}
\end{align}
where $\ds{\on{K}_{\nu}}$ is a Modified Bessel Function. The
last result is from DLMF.

Since $\ds{\quad 50^{1/5}\,\on{K}_{4/5}\pars{4\root{5} \over 5} \approx 0.3740}$, the $\ds{\underline{\mbox{initial integral value}}}$ is
$$
\approx 0.302595 + 0.219848\ic
$$
when $\ds{m = \pm 1}$ besides, of course, other values which arise from the $\ds{\expo{-m\pi\ic/5}}$ periodicity.

The final result, which agrees with
$\ds{\tt Mathematica}$, is given by
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{1}
{\expo{1/\ln\pars{x}} \over x^{1/5}\,\,\bracks{\vphantom{\large A}\ln\pars{x}}^{1/5}}\,\dd x} =
\expo{\pi\ic/5}
\bracks{50^{1/5}\,\on{K}_{4/5}\pars{4\root{5} \over 5}}
\\[5mm] = &
\pars{{1 + \root{5} \over 4} +
\root{5 - \root{5} \over 8}\ic}50^{1/5}
\,\on{K}_{4/5}\pars{4\root{5} \over 5}
\\[5mm] \approx &\ \bbx{0.302595 + 0.219848\,\ic} \\ &
\end{align}
