Problem 8 - Calculus for Biology and Medicine

Question

All the integrals in problem are improper and converge. Explain in each case why the integral is improper, and evaluate each integral. $$ \int_{-\infty}^{\infty} x e^{-x^{2} / 2} d x $$

Step-by-Step Solution

Answer

The integral is improper due to infinite limits and evaluates to zero using the symmetry property of odd functions.

Step 1: Identify Why the Integral is Improper

The given integral \[ \int_{-\infty}^{\infty} x e^{-x^2 / 2} \, dx \]is improper because the limits of integration include infinity, specifically from \(-\infty\) to \(\infty\). This makes the integral improper due to the presence of infinite limits of integration.

Step 2: Break Down the Integral

To solve an improper integral with infinite limits, we first express it in a way that allows calculation:\[ \int_{-\infty}^{\infty} x e^{-x^2 / 2} \, dx = \lim_{a \to -\infty, b \to \infty} \int_{a}^{b} x e^{-x^2 / 2} \, dx.\]Next, recognize that the function \(x e^{-x^2 / 2}\) is an odd function, which helps in solving the integral over symmetric intervals.

Step 3: Recognize Symmetry

The function \(f(x) = x e^{-x^2 / 2}\) is odd because \(f(-x) = -f(x)\). The integral of an odd function over a symmetric interval from \(-a\) to \(a\) is zero, that is:\[ \int_{-a}^{a} f(x) \, dx = 0.\]This property is especially useful for integrals over the entire real line.

Step 4: Apply Symmetry Property

Applying the property of an odd function, the integral \[ \int_{-\infty}^{\infty} x e^{-x^2 / 2} \, dx = 0 \]directly evaluates to zero, as the contributions from \(-\infty \text{ to } 0\) and from \(0 \text{ to } \infty\) cancel each other out.

Key Concepts

Infinite Limits

When tackling improper integrals, it's crucial to understand the notion of infinite limits. In the context of integrals, infinite limits occur when the integration bounds extend to positive or negative infinity. The given integral \( \int_{-\infty}^{\infty} x e^{-x^2 / 2} \, dx \) is a classic example since it spans from \(-\infty\) to \(\infty\).

This type of integral is labelled 'improper' because traditional calculus methods can't directly apply – you can't just plug infinity into a formula. Instead, these integrals require special techniques, usually involving limit processes. Here's how we deal with them:

• Recast the integral using limits: \( \lim_{a \to -\infty, b \to \infty} \int_{a}^{b} x e^{-x^2 / 2} \, dx \).
• Treat the evaluation as two separate limits if necessary to ensure convergence.

Recognizing the need for this approach helps manage and correctly evaluate integrals despite their initially daunting infinite boundaries.

Symmetric Intervals

Many integral problems can be simplified by spotting symmetry in the function and the limits of integration. Symmetric intervals are intervals that evenly surround the origin, typically taking the form \([-a, a]\). With our integral, consider how it spans from \(-\infty\) to \(\infty\), which is perfectly symmetrical.

Identifying symmetric intervals allows us to leverage specific mathematical properties, particularly those related to odd functions. For symmetric intervals:

• They help make predictions about the integral's value, often simplifying calculations.
• They serve as a powerful tool when dealing with functions and their symmetrical behavior.

For the integral \( \int_{-\infty}^{\infty} x e^{-x^2 / 2} \, dx \), symmetry suggests that the areas under the curve from \(-\infty\) to 0 and 0 to \(\infty\) might balance each other, leading to a more straightforward evaluation.

Odd Functions

Odd functions are an interesting concept with properties that can drastically simplify solving integrals. A function \(f(x)\) is called odd if \(f(-x) = -f(x)\). The function in this exercise, \(x e^{-x^2 / 2} \), meets this criterion since \(f(-x) = -x e^{-(-x)^2 / 2} = -x e^{-x^2 / 2} = -f(x)\).

Odd functions have a unique trait when integrated over symmetric intervals:

• The integral over any symmetric interval \([-a, a]\) is zero: \(\int_{-a}^{a} f(x) \, dx = 0\).
• This is because the area under the curve on the negative side of the y-axis cancels out the area on the positive side.

Applying this to the integral \(\int_{-\infty}^{\infty} x e^{-x^2 / 2} \, dx \) reveals that the total area from \(-\infty\) to 0 negates the area from 0 to \(\infty\), simplifying the evaluation to zero.

Grasping the characteristic of odd functions reduces complex-looking integrals into more straightforward calculations, showing the elegance of mathematical symmetry.