For which $k>0$ process $X=(e^{kW_s^2})_{s \ge 0}$ belong to $\mathcal{L}^2_{\infty }(W)$ and for which belong to $\Lambda ^2_{\infty }(W)$. Set one localization sequence $(\tau_n)_{n \ge 0}$ for local martingale $ \int X dW$
So Ive started from the top and I stacked at the begginig. I mean: I recall the definition of $\mathcal{L}^2_{\infty }(W)$ and to set the $k$ I need to calculate: $\mathbb{E}\int_0^{\infty}e^{2W_s^2}ds$ and it is $\int_0^{\infty}\mathbb{E}e^{2W_s^2}ds$ I have problem with that expectation value. I tried use definition of it because we know that $W_s$~$N(0,s)$.
Can you help me?
chi-squared random variable with k degrees of freedom is given as $$\chi^2=\sum\limits_{i=1}^{k}{{{X}_{i}}^{2}} $$ where the $X_i$'s are all independent and have $N(0, 1)$ distributions. Also recall that I claimed that $\chi^2$ has a gamma distribution with parameters $r=k=2$ and $\alpha=\frac{1}{2}$, thus $$M_{\chi^{\,2}}(t)=\left(\frac{1}{1-2t}\right)^{\frac{k}{2}},\,\,\,\,|t|<\frac{1}{2}$$ on the other hand we know $$\frac{W_s-0}{\sqrt{s}}\sim N(0,1) $$ then $$\frac{{W_s}^2}{s}\sim \chi^2$$ as a result $$E\left[e^{2{W_s}^2}\right]=E\left[e^{2s\,\frac{{W_s}^2}{s}}\right]=M_{\chi^{\,2}}\left(2s\right)=\frac{1}{\sqrt{1-4s}}$$