The solution to SDE $$dS(t)=\sigma S(t)dW_t$$ is $$S(t)=S(0)\exp(-\frac{1}{2}\sigma^2t+\sigma W_t)$$ the transition density for this martingale is $$p(S(t),t;S(0),0)=\frac{1}{S(t)\sigma \sqrt{2\pi t}}\exp{\left(-\frac{1}{2}\left[\frac{\log(S_t)-\log(S_0)-\sigma^2 t/2}{\sigma \sqrt{t}}\right]^2\right)}$$ But I can't get this density,here is my proof:
$$\mathbb{P}(S(t)\le u)=\mathbb{P}\left(S(0)\exp(-\frac{1}{2}\sigma^2t+\sigma W_t)\le u\right)$$ $$=\mathbb{P}\left(W_t\le \frac{\log\frac{u}{S(0)}+\sigma^2t/2}{\sigma}\right)=\frac{1}{\sqrt{2\pi}}\int_0^{\frac{\log\frac{u}{S(0)}+\sigma^2t/2}{\sigma}}e^{-\frac{1}{2}x^2}\,dx$$
$$p(S(t),t;S(0),0)=\frac{d}{du}\mathbb{P}(S(t)\le u)=\frac{1}{S(t)\sigma \sqrt{2\pi }}\exp{\left(-\frac{1}{2}\left[\frac{\log(S_t)-\log(S_0)+\sigma^2 t/2}{\sigma}\right]^2\right)}$$(take $u=S(t)$)
I got the wrong answer when it comes to $t$.Is there something wrong with my proof?