Estimate the value of the integral: integral from 0

Estimate the value of the integral: $$ \int_0^{\infty} \frac{\cos ax}{4+x^2} \, dx $$

We will estimate this integral using bounds and then compute the exact value.

Method 1: Simple bounds

Since $|\cos ax| \leq 1$ for all $x$:

$$\left|\frac{\cos ax}{4+x^2}\right| \leq \frac{1}{4+x^2}$$

Therefore:

$$\left|\int_0^{\infty} \frac{\cos ax}{4+x^2} \, dx\right| \leq \int_0^{\infty} \frac{1}{4+x^2} \, dx$$

Computing the right side:

$$\int_0^{\infty} \frac{dx}{4+x^2} = \int_0^{\infty} \frac{dx}{4(1+x^2/4)} = \frac{1}{4}\int_0^{\infty} \frac{dx}{1+(x/2)^2}$$

Let $u = x/2$, so $du = dx/2$:

$$= \frac{1}{4} \cdot 2\int_0^{\infty} \frac{du}{1+u^2} = \frac{1}{2}[\arctan u]_0^{\infty} = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$$

Therefore:

$$\boxed{\left|I\right| \leq \frac{\pi}{4} \approx 0.785}$$

Method 2: Behavior for different values of $a$

The value of the integral depends on the parameter $a$.

Case 1: $a = 0$

$$I = \int_0^{\infty} \frac{1}{4+x^2} \, dx = \frac{\pi}{4}$$

Case 2: $a > 0$ (large)

For large $|a|$, $\cos ax$ oscillates rapidly, causing cancellation. The integral becomes smaller.

Case 3: $a < 0$

By symmetry of cosine: $\cos(-ax) = \cos(ax)$, so the integral is the same for $a$ and $-a$.

Method 3: Exact calculation using residue theorem or Laplace transform

The exact value can be computed using complex analysis or looking up tables.

The integral is related to the Laplace transform:

$$\mathcal{L}\{\cos(at)\}(s) = \frac{s}{s^2+a^2}$$

However, our integral has the form:

$$\int_0^{\infty} \frac{\cos ax}{4+x^2} \, dx$$

Using the formula for Fourier cosine transforms and properties of the Laplace transform, or using contour integration in the complex plane, the exact result is:

$$\boxed{I = \frac{\pi}{4}e^{-2|a|}}$$

Step-by-step verification of the formula

For $a = 0$:

$$I = \frac{\pi}{4}e^{0} = \frac{\pi}{4}$$

This matches our direct calculation. ✓

For $a > 0$, the integral decreases exponentially.

Numerical estimates for different values of $a$

Using $I = \frac{\pi}{4}e^{-2|a|}$:

Alternative derivation using symmetry

Since $\cos ax$ is even, we can write:

$$\int_0^{\infty} \frac{\cos ax}{4+x^2} \, dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{\cos ax}{4+x^2} \, dx$$

This integral can be evaluated using the residue theorem from complex analysis.

The function has poles at $x = \pm 2i$ in the complex plane.

For $a > 0$, closing the contour in the upper half-plane and using the residue at $x = 2i$:

$$\text{Residue at } x = 2i: \lim_{x \to 2i}(x-2i)\frac{e^{iax}}{(x-2i)(x+2i)} = \frac{e^{ia(2i)}}{4i} = \frac{e^{-2a}}{4i}$$

By the residue theorem:

$$\int_{-\infty}^{\infty} \frac{\cos ax}{4+x^2} \, dx = 2\pi i \cdot \frac{e^{-2a}}{4i} = \frac{\pi e^{-2a}}{2}$$

Therefore:

$$\int_0^{\infty} \frac{\cos ax}{4+x^2} \, dx = \frac{1}{2} \cdot \frac{\pi e^{-2a}}{2} = \frac{\pi e^{-2a}}{4}$$

For $a < 0$, using $\cos(-ax) = \cos(ax)$, we get $e^{-2|a|}$.

Final answer

$$\boxed{\int_0^{\infty} \frac{\cos ax}{4+x^2} \, dx = \frac{\pi}{4}e^{-2|a|}}$$

Bounds:

$$0 \leq I \leq \frac{\pi}{4} \approx 0.785$$

The maximum occurs at $a = 0$, and the integral decays exponentially as $|a|$ increases.

Summary

1. For $a = 0$: $I = \frac{\pi}{4} \approx 0.785$
2. For general $a$: $I = \frac{\pi}{4}e^{-2|a|}$
3. The integral decreases exponentially with $|a|$
4. Upper bound: $0 \leq I \leq \frac{\pi}{4}$

Key insights:

• The integral depends exponentially on the parameter $a$
• Maximum value is $\frac{\pi}{4}$ at $a = 0$
• Rapid oscillation of $\cos ax$ causes cancellation for large $|a|$
• The formula $I = \frac{\pi}{4}e^{-2|a|}$ is exact
• This is a standard result in Fourier analysis and signal processing