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The nonnegative series in part (b) has independent squared-standard-normal terms. If $G$ is standard normal, direct Gaussian integration gives $\mathbb{E}{e}^{-a{G}^2}=(1+2a{)}^{-1/2}$ for $a\ge 0$. Consequently, by independence and dominated convergence applied to the exponentials of increasing partial sums,Use the permitted product identity with $x=\sqrt{2\lambda }$. The answer isThis Laplace transform of the integrated square of Brownian motion equals $1$ at zero. Its first derivative there gives mean $1/2$, in agreement with ${\int }_0^1\mathbb{E}{W}_t^{2}dt$.