Applying the Feynman-Kac theorem on the PDE $$ F_t + \frac12 \sigma^2 X_t^2 F_{xx} = 1 $$ with the terminal boundary condition $$F(T,x) = (ln (x))^4 $$ I arrived at the following equation: $$ F= -(T-t) + \mathbb{E}\{(ln X_0)^4 + (-\frac12 \sigma (T-t) + \sigma B_{T-t})^4 | X_T=x\} $$
Is it acceptable to leave the solution at that? If not, how do I compute the conditional expectation given above?
The underlying stochastic process of interest is a geometric Brownian motion with no drift: $$dX_t=\sigma X_tdW_t$$ So the terminal probability density given $X_t=x$ is $$p(y,T-t|x,t)=\frac{1}{y}\frac{1}{\sqrt{2\pi\sigma^2(T-t)}}\exp\bigg\{-\frac{1}{2\sigma^2(T-t)}\left(\ln\bigg(\frac{y}{x}\bigg)+\frac{1}{2}\sigma^2(T-t)\right)^2\bigg\}$$ and the solution to the PDE (without considering $1$, which can be added after as it is deterministic) is given by the expectation $$E[\ln(X_T)^4|X_t=x]=\int_{(0,\infty)}\ln(y)^4p(y,T-t|x,t)dy$$ However, since $X_t$ is lognormally distributed, then $\ln(X_t)$ is Gaussian.Therefore $$E[\ln(X_T)^4|X_t=x]=\int_{(-\infty,\infty)}z^4\frac{1}{\sqrt{2\pi\sigma^2(T-t)}}\exp\bigg\{-\frac{(z-(\ln(x)-\frac{1}{2}\sigma^2(T-t)))^2}{2\sigma^2(T-t)}\bigg\}dz$$ Which is just the fourth moment of a Gaussian with mean $\mu=\ln(x)-\frac{1}{2}\sigma^2(T-t)$ and variance $\eta^2=\sigma^2(T-t)$. Therefore $$E[\ln(X_T)^4|X_t=x]=\mu^4+6\mu^2\eta^2+3\eta^4$$ and ultimately, by Feynman-Kac $$F(x,t)=-(T-t)+(\mu^4+6\mu^2\eta^2+3\eta^4)$$