# Evaluating a real definite integral using residue theorem

I'm trying to evaluate the following integral: $$I = \int_{-\infty}^{\infty} \frac{\sin x}{9x^2 + 4} dx$$

From what I understood(probably misunderstood) from reading the textbook, the approach should be applying something like this:

$$I = \oint \frac{\sin z}{{9(z + 2i/3)}{(z-2i/3)}} dz = 2 \pi i \sum{\text{residues (upper half-plane)}}$$

But from inspection, the function is odd, so $I = 0$. However, since only the pole $z = +2i/3$ is enclosed by the countour, and $Res( z=+2i/3) \neq 0$, my approach wouldn't result in the correct value. I also don't understand why this seems to be recommended, why can't I use the Residue Theorem to evaluate the integral, using the entire unit circle as countour? So something like $I = 2\pi i (\text{Sum of enclosed residues)}$, and since it seems $Res( z=+2i/3) = - Res( z=-2i/3)$, this would give the correct result. What should be the correct approach to evaluate this integral using residue theorem?

• Come on, the integral of an odd integrable function over $\mathbb{R}$ is zero. – Jack D'Aurizio Jul 3 '17 at 20:07
• @JackD'Aurizio Yes, I know, but I'm supposed to solve this using complex integrations techniques – Vinícius Aguiar Jul 3 '17 at 20:08
• $$I = \text{Im}\int_{-\infty}^{+\infty}\frac{e^{iz}\,dz}{9z^2+4} = -2\pi\, \text{Re}\,\text{Res}\left(\frac{e^{iz}}{9z^2+4},z=\frac{2i}{3}\right)=0.$$ – Jack D'Aurizio Jul 3 '17 at 20:12
• Thank you! Will this always give me the same value as $I = 2\pi i (\text{Sum of enclosed residues)}$? ps: Since this comment answers my question I'd be glad to accept it if posted as an answer – Vinícius Aguiar Jul 3 '17 at 20:19

We have that $$J=\int_{-\infty}^{+\infty}\frac{e^{iz}}{9z^2+4}\,dz = \lim_{R\to +\infty}\oint_{\gamma_R}\frac{e^{iz}}{9z^2+4}\,dz \tag{1}$$ where $\gamma_R$ is the boundary of a semicircle in the upper half-plane, counter-clockwise oriented, with the diameter having endpoints $\pm R$. By the residue theorem $$J = 2\pi i\cdot\text{Res}\left(\frac{e^{iz}}{9z^2+4},z=\frac{2i}{3}\right)=2\pi i\lim_{z\to\frac{2i}{3}}\frac{e^{iz}\left(z-\frac{2i}{3}\right)}{9z^2+4}=\frac{\pi}{6 e^{2/3}}\tag{2}$$ hence $I = \text{Im } J = \color{red}{0}$. The pole at $-\frac{2i}{3}$ does not matter since it is not enclosed by $\gamma_R$.
Of course, this approach only works for functions where we can say the circular part drops to zero. In your case, the function will not drop to zero over the circular part, and we get nothing useful about the real integral. However, exchange sin with the imaginary part of $e^{iz}$ and it works. So, the problem was that the real integral is not equal to the residue of the upper plane.