Intuition behind using Ito's lemma Even though I am already familiar with applying Ito's lemma in stochastic differential equations (SDEs), I fail to understand the idea behind this technique.
In words, why is it not possible to perform the following integration over time for the SDE below? Using Ito's lemma, the straightforward (and faulty) integration over time fails to capture the term $-\frac{1}{2}t$ in the exponent of $X(t)$, but what exactly goes wrong?
Let $W(t)$ be a Brownian motion.
$$dX(t) = X(t) dW(t) \implies \frac{dX(t)}{X(t)} = dW(t)$$
$$\int^{t}_{0} \frac{dX(t)}{X(t)} = \int^{t}_{0} dW(t) \implies \ln(X(t)) - \ln(X(0)) = W(t) - W(0)$$
$$\ln(X(t)) = \ln(X(0)) + W(t) \implies X(t) = X(0)\exp(W(t))$$
Thank you in advance. 
 A: Consider a time step $\Delta t\ll 1$. Then subdivide into a large number $N$ of smaller intervals $N\delta t=\Delta t$. Then the Euler-discretized equation leads to a discrete solution 
$$
X_{t+Δt}=X_t\prod_{k=0}^{N-1}(1+δW_{t+kδt})
$$
With sufficiently high probability the increments $δW_{t+kδt}=W_{t+(k+1)δt}-W_{t+kδt}$ are small, so that the product can be converted into the exponential of a sum
\begin{align}
\prod_{k=0}^{N-1}(1+δW_{t+kδt})
&=\exp\left(\sum_{k=0}^{N-1}\ln(1+δW_{t+kδt})\right)
\\
&=\exp\left(\sum_{k=0}^{N-1}δW_{t+kδt} -\frac12\sum_{k=0}^{N-1}δW_{t+kδt}^2+\frac13\sum_{k=0}^{N-1}δW_{t+kδt}^3+... \right)
\end{align}
The first term is just $ΔW_t=W_{t+Δt}-W_t$, the second is with a probability close to 1 just $-\frac12Δt$ and the third has size $Δt\maxδW_{t+kδt}$ which by assumption is close to zero. All following terms are smaller.
So yes, the Ito term comes from the quadratic terms of the Taylor expansion and the informal rule that $(dW_t)^2=dt$.
A: Because $X_t$ is not differential in the strong sense. Suppos what you did is correct. Applying Itô formula on $f(x)=x_0e^{x}$ yield $$\mathrm d (x_0e^{W_t}) =\mathrm d f(X_t)=x_0e^{W_t}\,\mathrm d W_t+\frac{1}{2}x_0e^{W_t}\,\mathrm d t\neq x_0e^{W_t}\,\mathrm d W_t.$$
