Problem 47 - Calculus and its applications

Question

Determine whether each improper integral is convergent or divergent, and find its value if it is convergent. $$ \int_{0}^{\infty} x e^{-x^{2}} d x $$

Step-by-Step Solution

Answer

The integral converges to \( \frac{1}{2} \).

Step 1: Identify the Integral Type

The integral given is \( \int_{0}^{\infty} x e^{-x^{2}} \, dx \). Since the upper limit is infinity, it indicates an improper integral.

Step 2: Set Up the Limit

For improper integrals with infinite limits, we use limit notation: \[ \int_{0}^{\infty} x e^{-x^{2}} \, dx = \lim_{b \to \infty} \int_{0}^{b} x e^{-x^{2}} \, dx \] Evaluate the definite integral first, then take the limit as \( b \to \infty \).

Step 3: Perform Substitution

To solve the integral, use substitution. Let \( u = -x^2 \). Then \( du = -2x \, dx \), which gives \( x \, dx = -\frac{1}{2} du \).

Step 4: Change Limits of Integration

When \( x = 0 \), \( u = 0 \). When \( x = b \), \( u = -b^2 \). So, the integral becomes: \[ \lim_{b \to \infty} \int_{0}^{-b^2} -\frac{1}{2} e^{u} \, du \]

Step 5: Evaluate the Integral

The integral \( \int e^{u} \, du = e^u \). Thus, \[ \lim_{b \to \infty} \left[ -\frac{1}{2} e^u \right]_{0}^{-b^2} = \lim_{b \to \infty} \left( -\frac{1}{2} e^{-b^2} + \frac{1}{2}\right)\]

Step 6: Calculate the Limit

Evaluate the limit: \[ \lim_{b \to \infty} \left( -\frac{1}{2} e^{-b^2} + \frac{1}{2} \right) \] As \( b \to \infty \), \( e^{-b^2} \to 0 \), hence \[ \lim_{b \to \infty} \left( 0 + \frac{1}{2} \right) = \frac{1}{2} \]

Step 7: Conclusion

The integral \( \int_{0}^{\infty} x e^{-x^{2}} \, dx \) converges to \( \frac{1}{2} \). Thus, it is convergent and its value is \( \frac{1}{2} \).

Key Concepts

Convergence of Integrals

Understanding the convergence of integrals is crucial for solving problems involving improper integrals. Improper integrals occur when an integral has at least one infinite limit of integration or an integrand that approaches infinity within the interval.

In the exercise, the integral \( \int_{0}^{\infty} x e^{-x^{2}} \, dx \) is improper because the upper limit is infinity. To determine if such an integral converges, we use limits to examine the behavior as the variable approaches infinity or a point of discontinuity.

In this case, by setting up the limit \( \lim_{b \to \infty} \int_{0}^{b} x e^{-x^{2}} \, dx \), we focus on whether the limit exists and is finite. If it is, the integral is convergent, and its value can be determined. Conversely, if the limit does not exist or is infinite, the integral diverges.

Substitution in Integration

Substitution is a powerful tool in integration, simplifying complex integrals into more manageable forms. It's closely related to the chain rule in differentiation and involves replacing a part of the integrand with a new variable to make the integral easier to solve.

In our exercise, we use substitution to tackle the integral \( \int x e^{-x^{2}} \, dx \). We set \( u = -x^2 \), thus transforming the integral into a simpler form.

• Set \( du = -2x \, dx \), giving us \( x \, dx = -\frac{1}{2} du \).
• By substitution, change the limits of the integral: when \( x = 0 \), \( u = 0 \), and when \( x = b \), \( u = -b^2 \).

This step converts a potentially complex integral into one involving \( e^u \), which is straightforward to evaluate. Substitution streamlines the process, enabling us to focus on the core elements of the integral.

Limit Process in Calculus

The limit process is vital in calculus, especially when dealing with improper integrals and infinite bounds. Limits allow us to rigorously define and compute integrals or functions as they approach infinity or points of discontinuity.

In our exercise, once substitution is applied and the integral is transformed, we arrive at \( \lim_{b \to \infty} \int_{0}^{-b^2} -\frac{1}{2} e^{u} \, du \).

• The integration yields \( -\frac{1}{2} e^u \), evaluated from \( 0 \) to \( -b^2 \).
• As \( b \to \infty \), the term \( e^{-b^2} \to 0 \) due to the exponential function's rapid decay.

The limit therefore simplifies to \( \frac{1}{2} \).

This calculation showcases how limits help us find the actual value of an improper integral, ensuring a clear understanding of its behavior at infinity.