Evaluate infinite integral using complex analysis

Question

Can anyone please give me a detailed solution of $\int_{-\infty}^{\infty}\frac{1}{1+3x+5x^2} dx$. I have tried looking at the solutions of other problems like these but I am unable to work this out. A simple and detailed solution that works for a lot of problems like these would be very helpful.

Answer 1

Let $\gamma$ be a contour with two parts. $\gamma_1$ is the portion of real line between $[-R,R]$ And $\gamma_2$ is the semi-circular path of radius $R$ through the portion of the complex plain with positive real part.

$$\int_{-R}^R f(x) \, dx = \oint_\gamma f(z) \, dz - \int_{\gamma_2} f(z) \, dz $$

To evaluate $\oint_\gamma f(z) \, dz$ we just need to evaluate the residuals with positive imaginary part.

$$(5x^2 + 3x + 1) = 5\left(x + \frac {3 + i\sqrt {11}}{10}\right) \left(x + \frac {3 - i\sqrt {11}}{10}\right)$$

We have one residual in the region at $x = \frac {-3 + i\sqrt {11}}{10}$

Which will evaluate to $2\pi \frac {1}{\sqrt {11}}$

And what remains is to show that $\int_{\gamma_2} f(z) dz = 0$

$$z = Re^{it}\\ dz = Rie^{it}\\ \int_{\gamma_2} f(z) \, dz = \int_0^\pi \frac {Rie^{it}}{5R^2 e^{2it} + 3R e^{it} + 1} < \pi \frac {R}{5R^2}$$

Taking the limit a R goes to infinity, we see that this indeed equals 0.