Basic contour integral of $\frac{\sin z}{z - i}$

Question

I'm trying to evaluate a basic contour integral, but am getting an incorrect factor of 2 in my answer. Evaluating $$\int_{-\infty}^{\infty} \frac{\sin z}{z - i} = \Im \left( \int_{-\infty}^{\infty} \frac{e^{i z}}{z - i} \right) \, ,$$

I express this as a closed semi-circular contour of radius $R$ in the upper-half plane and take $\lim R \to \infty$. By Jordan's lemma, the integral over the arc contour vanishes and we can evaluate the integral by the residue theorem

$$\int_{-\infty}^{\infty} \frac{\sin z}{z - i} = \Im \left( 2 \pi i \, \left. \mathrm{Res}\left( \frac{e^{i z}}{z - i} \right) \right|_{z = i} \right) = \frac{2 \pi}{e} \, ,$$

while the correct answer is $\pi / e$. I've stared at this for a bit and still cannot see what I've done wrong. Can anyone quickly point out how the factor of 2 cancels?

Answer 1

Note that we have

$$\begin{align} \int_{-\infty}^\infty \frac{\sin(x)}{x-i}\,dx&=\int_{-\infty}^\infty \frac{(x+i)\sin(x)}{x^2+1}\,dx\\\\ &=\int_{-\infty}^\infty \frac{x\sin(x)}{x^2+1}\,dx\\\\ &=-i\int_{-\infty}^\infty \frac{xe^{ix}}{x^2+1}\,dx \end{align}$$

Now, we can close the contour in the upper-half plane and write

$$\begin{align} \int_{-\infty}^\infty \frac{\sin(x)}{x-i}\,dx&=2\pi i \text{Res}\left(-i\frac{xe^{ix}}{x^2+1}, z=i\right)\\\\ &=\frac{\pi}{e} \end{align}$$

as expected!