Complex analysis integration with contour

Question

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.

Answer 1

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 $\frac{\pi}{e^3}$.

Answer 2

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?

Answer 3

$$L=\lim_{R\to\infty}\int_{-R}^R\frac{\sin(x)}{x-3i}dx=\lim_{R\to\infty}\int_{-R}^R\frac{(x+3i)\sin(x)}{x^2+9}dx$$ $$L=\int_{-\infty}^\infty\frac{x}{x^2+9}\sin(x)dx+3i\int_{-\infty}^\infty\frac{\sin(x)}{x^2+9}dx$$ 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