definite integral. closed form?

Question

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.

Answer 1

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.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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.

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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}