I have tried many times to play with the positive real $k$ for the below convergent integral over $\mathbb{R}$ to get it's values , I have got the below nice identity but no way to prove it.
$$\int_{-\infty}^{+\infty} {\exp^{kx^2}(-x^2)}=\frac{2\Gamma(\frac 5 4)}{k^{\frac 1 4}}$$ .
Then my question here is : Any simple way to show that ?
The notation is a bit confusing, but I think what you need is
$$ \int_{-\infty}^{+\infty} \left[\exp(-x^2) \right]^{k x^2}{\rm d}x = \int_{-\infty}^{+\infty} \exp(- kx^4) {\rm d}x = 2 \int_{0}^{+\infty} \exp(- kx^4) {\rm d}x $$
To solve that one call $u = k x^4$, so that ${\rm d}u = x^{3}{\rm d}x = k [(u/k)^{1/4}]^3{\rm d}x = k u^{3/4} k^{-3/4}{\rm d}x = k^{1/4}u^{3/4}{\rm d}x $, so the integral becomes
$$ \int_{0}^{+\infty} \exp(- kx^4) {\rm d}x = k^{-1/4}\int_{0}^{+\infty}u^{-3/4}e^{-u}{\rm d}u $$
Can you take it from here?