Unable to get the approximated result

Question

I am stuck while trying to find the limiting solution of the given equation, which is $$ f(\omega)=B_o \frac{2 \mathcal{D}}{\omega^2 d^4} \int_0^{\infty} d \epsilon \frac{\epsilon^3 e^{-2 \epsilon}}{1+\frac{\mathcal{D}^2}{\omega^2 d^4} \epsilon^4} $$ Limit used is $d^2 \omega / \mathcal{D} \ll 1$ and the solution is $$ f(\omega)=B_0\left[\ln \left(\mathcal{D} / 4 \omega d^2\right)+2 \gamma\right] / \mathcal{D} $$ I tried various times but didn't get the given solution. Where $\gamma=0.577$, which is the EulerMascheroni constant.

Answer 1

Using the same notations as @Svyatoslav in comments, we have, using the Meijer-G function, $$g(N)=N\int_0^\infty\frac{x^3e^{-2x}}{1+Nx^4}dx=2 \sqrt{\frac{2}{\pi^3}}\,N\,G_{1,5}^{5,1}\left(\frac{1}{16 N}| \begin{array}{c} 1 \\ 1,1,\frac{5}{4},\frac{3}{2},\frac{7}{4} \\ \end{array} \right)$$

Expanding as series for large values of $N$, using $k=N^{\frac 14}$ $$g(N)= \log(k)-(\gamma +\log (2))+\frac{\pi }{k\sqrt{2} }\left(1-\frac{1}{\sqrt{2} k}+\frac{2}{3 k^2}+ O\left(\frac{1}{k^3}\right)\right)$$ which is very accurate : the relative error is less than $0.1$% if $k \geq 6.68$ (that is to say $N\geq 1990$).