 Solution

If $g^{\prime }$ never vanishes on $[a,b]$, integration by parts gives $I(\lambda )=[f(t)e^{i\lambda g(t)}/(i\lambda g^{\prime }(t)){]}_a^b-(i\lambda {)}^{-1}{\int }_a^b(f/g^{\prime }{)}^{\prime }{e}^{i\lambda g(t)}dt$, hence $\boxed{I(\lambda )=O({\lambda }^{-1})}$ for smooth $f,g$. Endpoint terms usually attain this order; special cancellation can give faster decay.

For the final integral, take the imaginary part of ${\int }_0^1{e}^{i\lambda (t^3-t)}dt$. Its sole stationary point is $t_0=1/\sqrt{3}$, with $g(t_0)=-2/(3\sqrt{3})$ and $g^{\prime \prime }(t_0)=2\sqrt{3}$. The stationary phase formula in part (a) therefore gives