Problem 11 - Biocalculus Calculus for the Life Sciences

Question

Determine whether each integral is convergent or divergent. Evaluate those that are convergent. \(\int_{-\infty}^{\infty} x e^{-x^{2}} d x\)

Step-by-Step Solution

Answer

The integral is convergent and evaluates to 0.

Step 1: Setting up the Integral

The given integral is \( \int_{-\infty}^{\infty} x e^{-x^{2}} \, dx \). We need to determine whether this integral is convergent or divergent.

Step 2: Analyze the Symmetry

Notice that the integrand \( x e^{-x^2} \) is an odd function, meaning that \( f(-x) = -f(x) \). This is because the term \( x \) changes sign while \( e^{-x^{2}} \) remains unchanged when \( x \) is replaced by \( -x \).

Step 3: Using Properties of Odd Functions and Integrals

Integrals over symmetric limits \([-a, a]\) for odd functions are zero. Thus,\[ \int_{-\infty}^{\infty} x e^{-x^{2}} \, dx = \int_{-a}^{a} x e^{-x^{2}} \, dx = 0 \] for all \( a \), and letting \( a \to \infty \) shows:\[ \int_{-\infty}^{\infty} x e^{-x^{2}} \, dx = 0. \]

Step 4: Conclusion on Convergence

Since the integral evaluates to 0, it is convergent. This is due to the symmetry and properties of odd functions over symmetric intervals.

Key Concepts

Understanding Convergence and Divergence

When we talk about convergence and divergence, we're exploring whether an integral or series "settles" to a specific value as you evaluate it over certain limits. For example, when examining a definite integral like \( \int_{-\infty}^{\infty} x e^{-x^{2}} \, dx \), we want to see if it gives a real number as a result.

If the integral sums to a finite value, it is said to be convergent. Conversely, if it tends to infinity or doesn't settle to a single value, it is divergent.
Here are a few properties to keep in mind:

• An integral is convergent if it results in a finite number when evaluated over infinite limits.
• Divergent integrals either grow without bound or oscillate without approaching a single value.

Thus, determining convergence is essential to ensure that the integral or series behaves nicely and yields meaningful results.

Integration of Odd Functions

Odd functions have a distinct property that makes them stand out in integrals, especially over symmetric intervals. These functions satisfy the condition \( f(-x) = -f(x) \), effectively mirroring about the origin on the graph.

The function \( x e^{-x^2} \) in our integral falls into the category of odd functions because the change in the sign of \( x \) results in the entire function changing sign.

• The significance of integrating odd functions over symmetric intervals, such as from \(-a\) to \(a\), is that their integrals become zero. This is because the positive and negative parts of the function cancel each other out perfectly.
• Therefore, if you integrate an odd function from \(-\infty\) to \(\infty\), you can conclude that the integral converges to zero without performing the actual integration steps.

Understanding this property significantly simplifies complex integrations and is a powerful tool in calculus.

Properties of Symmetric Intervals

Symmetric intervals are ranges that are equidistant from zero, such as \([-a, a]\). These intervals possess special traits when dealing with certain functions, particularly odd functions as seen in our example of \( x e^{-x^2} \).

Here's why symmetric intervals are essential:

• When integrating functions over symmetric intervals, values from the negative side can offset those from the positive side. This perfectly applies to odd functions.
• The mathematical beauty of symmetric intervals lies in their ability to simplify calculations by reducing problems of undefined behavior over infinite limits into simpler, often null results.

Using these properties, we were able to see that an otherwise complex integral like the one given can resolve quickly to zero without endless computations just by recognizing its symmetry and the odd nature of the function in question.