Complex analysis integration with contour Show that the next limit exists and finds its value
$$\lim_{R \to\infty}\int_{-R}^{R} \frac{\sin(x)}{x-3i}\ dx$$
My idea is to multiply by the conjugate in the denominator to obtain:
$$\int_{-R}^{R} \frac{\sin(x)}{x^2+9}\ dx$$
and then apply the residue theorem.
 A: Write your integral as
$$
\int_{-R}^R \frac{\exp(ix)-\exp(-ix)}{2i(x-3i)}\,\mathrm{d}x.
$$
So consider
$$
\int_{C_+}\frac{\exp(iz)}{(z-3i)}\,\mathrm{d}z\text{ and }\int_{C_-}\frac{\exp(-iz)}{(z-3i)}\,\mathrm{d}z
$$
where $C_+$ is an appropriate contour contour in the upper half plane, together with $[-R,R]$, and $C_-$ is an appropriate contour in the lower half plane, $0\leq t\leq\pi$ together with $[-R,R]$.  Can you see how to finish it from here?
A: First of all, note that
$$
\int_{-R}^R \frac{\sin x}{x-3i}dx = \Re \left(3i \int_{-R}^R \frac{e^{ix}}{x^2+9}dx \right)+ \Im\left(
 \int_{-R}^R \frac{x e^{ix}}{x^2+9} dx\right)
$$
Let us now look to each integral on the RHS. In what follows $C_R$ denotes the arc of the circumference centered in $0$ that connects $R$ and $-R$. In both cases we have integrals of the form $\int_{C_R} f(z) e^{imz} dz$ with $m>0$ and $|f(z)| < \frac{M}{R^k}$ for some $M,k>0$ and for all $z \in C_R$. In these conditions, it can be shown that 
$$
\lim_{R \to \infty} \int_{C_R} f(z) e^{imz} dz = 0.
$$
Hence, for $R>3$,
$$
\int_{-R}^R \frac{e^{ix}}{x^2+9} dx + \int_{C_R} \frac{e^{iz}}{z^2+9} dz = 2\pi i Res(\frac{e^{iz}}{z^2+9}, 3i)
$$
$$
\int_{-R}^R \frac{x e^{ix}}{x^2+9} dx + \int_{C_R} \frac{z e^{iz}}{z^2+9} dz = 2\pi i Res(\frac{z e^{iz}}{z^2+9}, 3i)
$$
Taking the limit as $R \to \infty$, since the integrals over $C_r$ will vanish, we will get
$$
\lim_{R\to \infty} \int_{-R}^R \frac{e^{ix}}{x^2+9} dx = 2\pi i Res(\frac{e^{iz}}{z^2+9}, 3i) = \frac{\pi}{3e^3}
$$
$$
\lim_{R\to \infty} \int_{-R}^R \frac{x e^{ix}}{x^2+9} dx = 2\pi i Res(\frac{z e^{iz}}{z^2+9}, 3i)=\frac{i \pi}{e^3}
$$
So, as a final result, we will obtain $\frac{\pi}{e^3}$. 
A: $$L=\lim_{R\to\infty}\int_{-R}^R\frac{\sin(x)}{x-3i}dx=\lim_{R\to\infty}\int_{-R}^R\frac{(x+3i)\sin(x)}{x^2+9}dx$$
$$L=\int_{-\infty}^\infty\frac{x}{x^2+9}\sin(x)dx+3i\int_{-\infty}^\infty\frac{\sin(x)}{x^2+9}dx$$
and we can say that:
$$\frac{x}{x^2+9}\le\frac{1}{x}\tag{1}$$
$$\frac{1}{x^2+9}\le\frac{1}{x^2}\tag{2}$$
If we apply this we get:
$$\int_{-\infty}^\infty\frac{x}{x^2+9}\sin(x)dx\le\int_{-\infty}^\infty\frac{\sin(x)}{x}=\pi\tag{3}$$
$$\int_{-\infty}^\infty\frac{\sin(x)}{x^2+9}=0\tag{4}$$
And so we can say:
$$L\le\pi$$
so the limit exists
