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I am currently trying to deduce that $$ \mathbb{E}[e^{i \lambda W_t}-1] = -\frac{\lambda^2}{2} \mathbb{E}\left[ \int_0^te^{i\lambda W_s}ds\right] \hspace{10mm} (*) $$ where $W_t$ is a Brownian motion process.

This is stated in my lecture notes, shortly after it is said that for any twice differentiable complex function $f$ $$ f(W_t) - f(0) = \int_0^t f'(W_s) dW_s + \int_0^t f''(W_s) ds $$ So I am assuming that this result should be used to deduce $(*)$. I have attempted to do this as follows:

  1. Define $f(x) = e^{i \lambda x}$
  2. Then $f(W_t) - f(0) = e^{i \lambda W_t} - 1$, which resembles the LHS of $(*)$.
  3. $f'(x) = i \lambda e^{i \lambda x}$ and $f''(x) = - \lambda^2 e^{i \lambda x}$, so the RHS of the given integral formula may be written as $$ i \lambda \int_0^t e^{i \lambda W_s} dW_s - \lambda^2 \int_0^t e^{i \lambda W_s} ds $$

So I suppose that I have reduced this problem to showing that $$ -\frac{\lambda^2}{2} \mathbb{E}\left[ \int_0^te^{i\lambda W_s}ds\right] = \mathbb{E} \left[ i \lambda \int_0^t e^{i \lambda W_s} dW_s - \lambda^2 \int_0^t e^{i \lambda W_s} ds \right] $$

Can anyone see how I might do this (assuming that this is the correct approach to this problem)?

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Your approach is correct but there's a typo in your formula for $f(W_t)$ (which is a special case of Ito's formula). It should read $$f(W_t) - f(0) = \int_0^t f'(W_s) dW_s + \frac12 \int_0^t f''(W_s) ds$$ Then taking $f(x) = e^{i \lambda x}$ and following your approach, we get that the problem is equivalent to showing that $$-\frac{\lambda^2}{2} \mathbb{E}\left[ \int_0^te^{i\lambda W_s}ds\right] = \mathbb{E} \left[ i \lambda \int_0^t e^{i \lambda W_s} dW_s - \frac{\lambda^2}{2} \int_0^t e^{i \lambda W_s} ds \right]$$ which is the same as showing $$\mathbb{E} \left[ \int_0^t e^{i \lambda W_s} dW_s \right ] = 0$$ This is a standard result that will follow from showing that $\int_0^t e^{i \lambda W_s} dW_s$ is a martingale.

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