Problem 30 - Calculus Single Variable

Question

Determine whether the given improper integral converges or diverges. If it converges, then evaluate it. $$ \int_{-\infty}^{0} x \exp \left(-x^{2}\right) d x $$

Step-by-Step Solution

Answer

The integral converges and its value is \(-\frac{1}{2}\).

Step 1: Recognize the Type of Integral

The given integral \( \int_{-\infty}^{0} x \exp(-x^2) \, dx \) is an improper integral because it has an infinite limit at the lower bound, \(-\infty\). We need to examine whether it converges or diverges.

Step 2: Set Up the Limit for the Integral

Rewrite the integral as a limit: \( \lim_{a \to -\infty} \int_{a}^{0} x \exp(-x^2) \, dx \). This allows us to evaluate the integral over a finite interval and explore its behavior as we extend towards \(-\infty\).

Step 3: Check the Odd Function Condition

The integrand \( x \exp(-x^2) \) is an odd function, which means that \( f(-x) = -f(x) \). Since the interval of integration is symmetric about zero and the limits are from \(-\infty\) to \(0\), the integral over the entire symmetric interval should converge to zero if the function is integrable.

Step 4: Evaluate the Integral from \(-a\) to \(0\)

Compute \( \int_{a}^{0} x \exp(-x^2) \, dx \) where the antiderivative of \( x \exp(-x^2) \) can be found using substitution. Let \( u = -x^2 \), then \( du = -2x \, dx \) or \( x \, dx = -\frac{1}{2} du \). This leads to the integral becoming \( -\frac{1}{2} \int e^{u} \, du = -\frac{1}{2} e^{u} \). Substitute back, \( -\frac{1}{2} e^{-x^2} \).

Step 5: Apply Limits to the Antiderivative

Now apply the limits to the antiderivative: \[ \lim_{a \to -\infty} \left[ -\frac{1}{2} e^{-x^2} \right]_{a}^{0} = \lim_{a \to -\infty} \left[ -\frac{1}{2} e^{-0^2} - \left(-\frac{1}{2} e^{-a^2}\right)\right]. \]This simplifies to \( -\frac{1}{2} + \lim_{a \to -\infty} -\frac{1}{2} e^{-a^2} \).

Step 6: Evaluate the Limit at Infinity

Since \( e^{-a^2} \to 0 \) as \( a \to -\infty \), the term \( -\frac{1}{2} e^{-a^2} \to 0 \). Thus, the entire expression becomes \( -\frac{1}{2} + 0 = -\frac{1}{2} \).

Step 7: Conclude Convergence and Evaluate

The result from Step 6 shows the integral converges and its value is \(-\frac{1}{2}\). This indicates that despite the infinite interval, the improper integral has a finite value.

Key Concepts

Convergence and Divergence

When dealing with improper integrals like \( \int_{-\infty}^{0} x \exp(-x^2) \, dx \), it's important to check if they converge or diverge. Convergence means that as you take the limit of the integral towards infinity or negative infinity, the integral approaches a specific, finite number. Divergence, on the other hand, means that the integral does not approach any finite number, and instead it could tend to infinity, negative infinity, or oscillate without settling down.
To determine convergence for the integral at hand, we rewrite it as a limit: \( \lim_{a \to -\infty} \int_{a}^{0} x \exp(-x^2) \, dx \). This approach helps to deal with the infinite limits by instead working over a variable, finite interval. Assessing whether this limit results in a finite number will tell us if the integral converges. In this instance, after all calculations, it shows convergence with a value of \(-\frac{1}{2}\). This means that despite extending to an infinite upper bound, the integral's area is finite.

Evaluation of Integrals

Evaluating integrals involves finding the antiderivative or the indefinite integral, and then applying the limits of integration to get the definite integral result. In our improper integral \( \int_{a}^{0} x \exp(-x^2) \, dx \), the first step is to find an antiderivative.

• Use the substitution method, setting \( u = -x^2 \) and \( du = -2x \, dx \), so the substitution would rewrite \( x \, dx \) as \(-\frac{1}{2} du\).
• This changes the integral to \( -\frac{1}{2} \int e^{u} \, du \), which evaluates to \(-\frac{1}{2} e^{u}\).

Returning to the original variable, the antiderivative becomes \(-\frac{1}{2} e^{-x^2}\). Applying the limits from \(a\) to \(0\), and taking the limit as \(a\) approaches negative infinity allows us to calculate that the integral equals \(-\frac{1}{2}\). This value represents the accumulated area under the curve from \(-\infty\) to \(0\).

Odd Functions

Odd functions are a fascinating aspect of mathematics where the function is symmetric about the origin. This means for any \( x \), \( f(-x) = -f(x) \). You can visualize odd functions as having a mirrored shape across both the y-axis and x-axis.
The integrand in our integral, \( x \exp(-x^2) \), is an example of an odd function. When you integrate an odd function over symmetric limits about zero, such as from \(-a\) to \(0\), the positive and negative areas usually cancel each other out, which suggests the integral may converge to zero.
However, the integral being improper requires checking convergence to ensure the smaller positive rest near zero outbalances the more spread-out negative rest towards \(-\infty\). Despite this, because \(x \exp(-x^2)\) is integrable, it turns out these "mirror" parts don't exactly cancel due to the area distribution as addressed through substitution, and it indeed converges to \(-\frac{1}{2}\).

Substitution Method

The substitution method is a powerful technique used to simplify integration, especially when dealing with more complex functions. By changing variables, you can transform an integral into an easier form to evaluate.
The substitution method was utilized in solving the given integral \( \int x \exp(-x^2) \, dx \).
Here's a brief rundown of how substitution helps in this context:

• Choose a substitution based on the complexity of the exponent: \( u = -x^2 \).
• Calculate the differential: \( du = -2x \, dx \).
• Replace \( x \, dx \) in the integral with \( -\frac{1}{2} du \) to yield \( -\frac{1}{2} \int e^{u} \, du \), which is straightforward to integrate.

This simplification reduces the problem to something more manageable while leading you straight to an antiderivative of the original function in the bounds of interest. Substituting back after integration and applying limits completes the process, showcasing the elegance and efficiency of this method for solving complex integrals.