Calculate the integral: integral from - infinity

Calculate the integral: $$ \int_{-\infty}^{\infty} \frac{1}{1+x^2} \, dx $$

This is an improper integral because the limits of integration are infinite. We need to evaluate it as a limit.

Step 1: Express as limits

Since both limits are infinite, we split the integral at a convenient point (typically 0) and evaluate each part separately:

$$\int_{-\infty}^{\infty} \frac{1}{1+x^2} \, dx = \int_{-\infty}^0 \frac{1}{1+x^2} \, dx + \int_0^{\infty} \frac{1}{1+x^2} \, dx$$

Each part is defined as a limit:

$$= \lim_{a \to -\infty} \int_a^0 \frac{1}{1+x^2} \, dx + \lim_{b \to \infty} \int_0^b \frac{1}{1+x^2} \, dx$$

Step 2: Find the antiderivative

The antiderivative of $\frac{1}{1+x^2}$ is a standard result:

$$\int \frac{1}{1+x^2} \, dx = \arctan x + C$$

Verification:

$$\frac{d}{dx}[\arctan x] = \frac{1}{1+x^2}$$

This confirms our antiderivative is correct. ✓

Step 3: Evaluate the first limit (from $-\infty$ to 0)

$$\lim_{a \to -\infty} \int_a^0 \frac{1}{1+x^2} \, dx = \lim_{a \to -\infty} [\arctan x]_a^0$$

$$= \lim_{a \to -\infty} (\arctan 0 - \arctan a) = 0 - \lim_{a \to -\infty} \arctan a$$

Since $\lim_{x \to -\infty} \arctan x = -\frac{\pi}{2}$:

$$= 0 - \left(-\frac{\pi}{2}\right) = \frac{\pi}{2}$$

Step 4: Evaluate the second limit (from 0 to $\infty$)

$$\lim_{b \to \infty} \int_0^b \frac{1}{1+x^2} \, dx = \lim_{b \to \infty} [\arctan x]_0^b$$

$$= \lim_{b \to \infty} (\arctan b - \arctan 0) = \lim_{b \to \infty} \arctan b - 0$$

Since $\lim_{x \to \infty} \arctan x = \frac{\pi}{2}$:

$$= \frac{\pi}{2} - 0 = \frac{\pi}{2}$$

Step 5: Add the results

$$\int_{-\infty}^{\infty} \frac{1}{1+x^2} \, dx = \frac{\pi}{2} + \frac{\pi}{2} = \pi$$

Final Answer

$$\boxed{\int_{-\infty}^{\infty} \frac{1}{1+x^2} \, dx = \pi}$$

Alternative approach: Direct evaluation

We can also evaluate this in one step:

$$\int_{-\infty}^{\infty} \frac{1}{1+x^2} \, dx = \lim_{R \to \infty} \int_{-R}^R \frac{1}{1+x^2} \, dx$$

$$= \lim_{R \to \infty} [\arctan x]_{-R}^R = \lim_{R \to \infty} (\arctan R - \arctan(-R))$$

Using $\arctan(-R) = -\arctan R$:

$$= \lim_{R \to \infty} (\arctan R - (-\arctan R)) = \lim_{R \to \infty} 2\arctan R$$

$$= 2 \cdot \frac{\pi}{2} = \pi$$

This approach works because the integrand is an even function: $f(-x) = f(x)$.

Verification using symmetry

Since $\frac{1}{1+x^2}$ is even:

$$\int_{-\infty}^{\infty} \frac{1}{1+x^2} \, dx = 2\int_0^{\infty} \frac{1}{1+x^2} \, dx = 2 \cdot \frac{\pi}{2} = \pi$$

Geometric interpretation

The integral $\int_{-\infty}^{\infty} \frac{1}{1+x^2} \, dx$ represents the total area under the curve $y = \frac{1}{1+x^2}$ over the entire real line.

Properties of the function $f(x) = \frac{1}{1+x^2}$:

• Always positive: $f(x) > 0$ for all $x$
• Even function: $f(-x) = f(x)$ (symmetric about the $y$-axis)
• Maximum at $x = 0$: $f(0) = 1$
• Approaches 0 as $x \to \pm\infty$: $\lim_{|x| \to \infty} f(x) = 0$
• Bell-shaped curve (similar to a normal distribution but with heavier tails)

Despite extending to infinity in both directions, the total area is finite (equals $\pi$).

Why the integral converges

The integral converges because the function $\frac{1}{1+x^2}$ decays fast enough as $x \to \pm\infty$.

For large $|x|$:

$$\frac{1}{1+x^2} \approx \frac{1}{x^2}$$

And we know that:

$$\int_1^\infty \frac{1}{x^2} \, dx = 1 \quad \text{(converges)}$$

More generally, $\int_1^\infty \frac{1}{x^p} \, dx$ converges if and only if $p > 1$.

Since our function behaves like $\frac{1}{x^2}$ for large $|x|$, and $p = 2 > 1$, the integral converges.

Connection to the Cauchy distribution

The function $\frac{1}{\pi(1+x^2)}$ is the probability density function of the standard Cauchy distribution in probability theory.

The integral we computed shows that:

$$\int_{-\infty}^{\infty} \frac{1}{\pi(1+x^2)} \, dx = \frac{1}{\pi} \cdot \pi = 1$$

This confirms that it's properly normalized as a probability density function (total probability = 1).

Related integrals

General form:

$$\int_{-\infty}^{\infty} \frac{1}{a^2+x^2} \, dx = \frac{1}{a} \int_{-\infty}^{\infty} \frac{1}{1+\left(\frac{x}{a}\right)^2} \, dx$$

Using the substitution $u = \frac{x}{a}$:

$$= \frac{1}{a} \cdot a \int_{-\infty}^{\infty} \frac{1}{1+u^2} \, du = \int_{-\infty}^{\infty} \frac{1}{1+u^2} \, du = \pi$$

Or more directly:

$$\int_{-\infty}^{\infty} \frac{1}{a^2+x^2} \, dx = \frac{1}{a}[\arctan(x/a)]_{-\infty}^{\infty} = \frac{1}{a}\left(\frac{\pi}{2} - \left(-\frac{\pi}{2}\right)\right) = \frac{\pi}{a}$$

Example:

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

Comparison with the Gaussian integral

This integral is often compared with the famous Gaussian integral:

$$\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}$$

Both integrals:

• Converge to finite values involving $\pi$
• Represent areas under bell-shaped curves
• Are fundamental in probability theory
• Have even integrands (symmetric about the $y$-axis)

However, the Cauchy distribution has "heavier tails" (decays like $\frac{1}{x^2}$) compared to the Gaussian (decays like $e^{-x^2}$).

Historical note

This integral and the arctangent function are intimately connected with $\pi$. In fact, one way to compute $\pi$ numerically is through the arctangent:

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

This is known as Leibniz's formula (when expressed as a series):

$$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots$$

Summary

1. The integral $\int_{-\infty}^{\infty} \frac{1}{1+x^2} \, dx$ is improper with infinite limits
2. We evaluate it by splitting at 0 or using symmetry
3. The antiderivative is $\arctan x$
4. Using $\lim_{x \to \infty} \arctan x = \frac{\pi}{2}$ and $\lim_{x \to -\infty} \arctan x = -\frac{\pi}{2}$
5. The integral converges to $\pi$
6. The convergence is due to the $\frac{1}{x^2}$ decay at infinity

Key lessons:

• Improper integrals over infinite intervals can converge to finite values
• The arctangent function naturally connects to $\pi$ through integration
• Symmetry (even functions) can simplify calculations
• The decay rate at infinity determines convergence ($\frac{1}{x^p}$ converges for $p > 1$)
• This integral appears throughout mathematics and physics (Cauchy distribution, Lorentzian, etc.)