= Solution
{c}
Let $F(z)=e^{i\lambda z}/(z-i)$. For $\lambda>0$, close the interval $[-R,R]$ by the upper semicircle, counterclockwise. Its only enclosed pole is $z=i$, with <residue> $e^{-\lambda}$. The <residue theorem> gives
$$
\int_{-R}^RF(x)\,dx+\int_{\Gamma_R^+}F(z)\,dz=2\pi i e^{-\lambda}.
$$
On $z=Re^{i\theta}$, $|e^{i\lambda z}|=e^{-\lambda R\sin\theta}$ and $|z-i|\geq R-1$. Hence the arc is bounded by
$$
\frac R{R-1}\int_0^\pi e^{-\lambda R\sin\theta}\,d\theta\longrightarrow0
$$
by the supplied estimate, with parameter $\lambda R$.
For $\lambda<0$, close below, with clockwise orientation. The pole at $i$ lies outside, and the exponential decays there. Parametrizing the lower arc yields the same bound with $|\lambda|R$ in place of $\lambda R$. Its contribution again tends to zero, and no residue remains. Therefore
$$
\boxed{\lim_{R\to\infty}\int_{-R}^R\frac{e^{i\lambda x}}{x-i}\,dx
=\begin{cases}2\pi i e^{-\lambda},&\lambda>0,\\0,&\lambda<0.\end{cases}}
$$
\Image[/past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2008-fourier-contours.png]
{title=The exponential selects the half-plane in which the closing arc decays}
This is the contour-decay mechanism behind <Jordan's lemma>; the pole contributes only when the decaying closure encloses it.
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