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.
Best Answer
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.