Question

$\int_{-\infty}^{\infty} \frac{\cos x}{4x^2 + 1} dx$

Answer 1

$\frac{\pi}{2\sqrt{e}}$

Answer 2

The integral $\int_{-\infty}^{\infty} \frac{\cos x}{4x^2 + 1} dx$ can be evaluated using the method of contour integration in complex analysis.

1. Define the Complex Function:
Consider the complex function $f(z) = \frac{e^{iz}}{4z^2 + 1}$. We are interested in the real part of $\int_{-\infty}^{\infty} f(x) dx$, since $\cos x = \text{Re}(e^{ix})$.

2. Identify the Poles:
The poles of $f(z)$ are the values of $z$ for which the denominator is zero:
$4z^2 + 1 = 0$
$4z^2 = -1$
$z^2 = -\frac{1}{4}$
$z = \pm \sqrt{-\frac{1}{4}} = \pm \frac{i}{2}$

3. Choose the Contour:
We use a semi-circular contour $C_R$ in the upper half-plane. This contour consists of two parts:

• The real axis from $-R$ to $R$, denoted as $L_R$.
• A semi-circle $\Gamma_R$ of radius $R$ in the upper half-plane, centered at the origin.

4. Identify Poles Inside the Contour:
For a sufficiently large $R$, only the pole $z_0 = \frac{i}{2}$ lies inside the contour $C_R$ (since it's in the upper half-plane).

5. Calculate the Residue:
The residue of $f(z)$ at a simple pole $z_0$ is given by $\text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0) f(z)$.
For $z_0 = \frac{i}{2}$:
$\text{Res}\left(f, \frac{i}{2}\right) = \lim_{z \to \frac{i}{2}} \left(z - \frac{i}{2}\right) \frac{e^{iz}}{4z^2 + 1}$
Factor the denominator: $4z^2 + 1 = 4(z^2 + \frac{1}{4}) = 4\left(z - \frac{i}{2}\right)\left(z + \frac{i}{2}\right)$.
So,
$\text{Res}\left(f, \frac{i}{2}\right) = \lim_{z \to \frac{i}{2}} \left(z - \frac{i}{2}\right) \frac{e^{iz}}{4\left(z - \frac{i}{2}\right)\left(z + \frac{i}{2}\right)}$
$\text{Res}\left(f, \frac{i}{2}\right) = \lim_{z \to \frac{i}{2}} \frac{e^{iz}}{4\left(z + \frac{i}{2}\right)}$
Substitute $z = \frac{i}{2}$:
$\text{Res}\left(f, \frac{i}{2}\right) = \frac{e^{i(i/2)}}{4\left(\frac{i}{2} + \frac{i}{2}\right)} = \frac{e^{-1/2}}{4i}$

6. Apply the Residue Theorem:
According to the residue theorem, $\oint_{C_R} f(z) dz = 2\pi i \sum \text{Res}(f, z_k)$, where the sum is over all poles inside $C_R$.
$\oint_{C_R} f(z) dz = 2\pi i \times \frac{e^{-1/2}}{4i} = \frac{2\pi e^{-1/2}}{4} = \frac{\pi e^{-1/2}}{2}$

7. Evaluate the Integral over the Contour as $R \to \infty$:
The integral over $C_R$ can be written as:
$\oint_{C_R} f(z) dz = \int_{L_R} f(z) dz + \int_{\Gamma_R} f(z) dz$
$\oint_{C_R} f(z) dz = \int_{-R}^{R} \frac{e^{ix}}{4x^2 + 1} dx + \int_{\Gamma_R} \frac{e^{iz}}{4z^2 + 1} dz$

As $R \to \infty$:
The integral over $L_R$ becomes the desired integral: $\lim_{R \to \infty} \int_{-R}^{R} \frac{e^{ix}}{4x^2 + 1} dx = \int_{-\infty}^{\infty} \frac{e^{ix}}{4x^2 + 1} dx$.
For the integral over $\Gamma_R$, we can apply Jordan's Lemma. Since $P(z) = 1$ and $Q(z) = 4z^2+1$, and $\text{deg}(Q) = 2 > \text{deg}(P) = 0$, Jordan's Lemma states that $\lim_{R \to \infty} \int_{\Gamma_R} \frac{e^{iz}}{4z^2 + 1} dz = 0$.

8. Equate the Results:
Therefore, as $R \to \infty$:
$\int_{-\infty}^{\infty} \frac{e^{ix}}{4x^2 + 1} dx = \frac{\pi e^{-1/2}}{2}$

9. Extract the Real Part:
We know that $e^{ix} = \cos x + i \sin x$.
So, $\int_{-\infty}^{\infty} \frac{\cos x + i \sin x}{4x^2 + 1} dx = \frac{\pi e^{-1/2}}{2}$
$\int_{-\infty}^{\infty} \frac{\cos x}{4x^2 + 1} dx + i \int_{-\infty}^{\infty} \frac{\sin x}{4x^2 + 1} dx = \frac{\pi e^{-1/2}}{2}$

The integrand $\frac{\cos x}{4x^2 + 1}$ is an even function, and $\frac{\sin x}{4x^2 + 1}$ is an odd function. The integral of an odd function over a symmetric interval $[-\infty, \infty]$ is zero.
Thus, $\int_{-\infty}^{\infty} \frac{\sin x}{4x^2 + 1} dx = 0$.

Therefore, by comparing the real parts:
$\int_{-\infty}^{\infty} \frac{\cos x}{4x^2 + 1} dx = \text{Re}\left(\frac{\pi e^{-1/2}}{2}\right) = \frac{\pi e^{-1/2}}{2} = \frac{\pi}{2\sqrt{e}}$