I've attempted as follows
We want to know if the stochastic process is square-integrable or not by evaluting if $E[\int_{0}^{T} X_t^2] < \infty $ or no, my solution is:
$ \begin{align} E\bigg[\int_{0}^{T} X_t^2\bigg] &= E\bigg[\int_{0}^{T} (e^{(W_t)^4})^2dt\bigg]\\ &= \int_{0}^{T} E\bigg[e^{2W_t^4}\bigg]dt\quad\text{(Fubini's Theorem)}\\ &= \int_{0}^{T}\big(\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi t}}e^{2W_t^4}e^{-\frac{Wt^2}{2t}}dW_t\big)dt\\ &= \int_{0}^{T}\big(\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi t}}e^{\frac{4tW_{t}^4 - W_{t}^2}{2t}}dW_t\big)dt\\ \end{align} $
Then I got stuck and don't know how to proceed, can someone please point out how I can evaluate this integral?
Many thanks!
By Tonelli-Fubini we can claim $E[\int_{[0,t]}X_s^2ds]=\int_{[0,t]}E[X_s^2]ds$. Then $$E[X_t^2]=E[e^{2W_t^4}]=\int_{\mathbb{R}}\frac{1}{\sqrt{2\pi t}}e^{2x^4-\frac{x^2}{2t}}dx=\int_{\mathbb{R}}\frac{1}{\sqrt{2\pi t}}e^{\frac{x^2(4tx^2-1)}{2t}}dx=\infty,\,\forall t>0$$ because for $|x|\geq M(t)$ for some $M(t)$ the exponential term shoots up to infinity as the exponentiated term becomes positive (graph it).