Problem 9 - 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. $$ \int_{-\infty}^{\infty} \frac{x}{\left(1+x^{2}\right)^{2}} d x $$

Step-by-Step Solution

Answer

The integral is improper due to infinite limits and evaluates to 0, as the function is odd and symmetric about the y-axis.

Step 1: Identify Why Integral is Improper

The given integral \( \int_{-\infty}^{\infty} \frac{x}{(1+x^2)^2} \, dx \) is improper because the limits of integration are infinite. This requires special techniques to evaluate as a proper limit cannot be directly calculated.

Step 2: Use Limits to Define the Integral from Negative to Positive Infinity

To evaluate the integral, express it as a limit: \[ \lim_{a \to -\infty} \lim_{b \to \infty} \int_{a}^{b} \frac{x}{(1+x^2)^2} \, dx. \] Evaluate this integral by considering the nature of the function \( \frac{x}{(1+x^2)^2} \).

Step 3: Observe the Symmetry of the Function

The function \( \frac{x}{(1+x^2)^2} \) is an odd function because \( f(-x) = -f(x) \). This means the area under the curve from \(-a\) to \(a\) symmetrically cancels out on both sides of the origin, making the integral of any odd function over symmetric limits around zero equal to zero.

Step 4: Conclude the Integral's Value from the Symmetry Consideration

From the previous observation, since the function is symmetric and odd and the integral is taken from \(-\infty\) to \(\infty\), the value of the integral is zero because the positive half and the negative half cancel each other out.

Key Concepts

Limits of Integration

When we deal with integrals, the 'limits of integration' are the points from which we start and stop. These limits often define the interval over which we calculate the area under a curve. In the exercise above, the limits of integration are

• \( -\infty \) and \( \infty \),
• representing two extreme ends all the way to negative and positive infinity, respectively.

These infinite boundaries are a key feature that classifies our integrals as 'improper integrals'. Having limits from minus infinity to infinity challenges traditional calculation techniques because an actual numerical end-point is never reached.
Consequently, we incorporate calculus techniques, like limits, to evaluate them, essentially reforming the calculation to be about continuously approaching these points rather than directly reaching them. This substitution allows us to mathematically 'handle' these giant, indefinite numbers.

Infinite Limits

Infinite limits play a crucial role in evaluating improper integrals. In essence, they account for integrals with endpoints extending to infinity. To analyze these,

• we use limits as a tool, replacing the infinite limits with finite variables that approach infinity.
• This turns our integral into a limit problem, such that the expression \( \lim_{a \to -\infty} \lim_{b \to \infty} \int_{a}^{b} \ldots \) is mathematically formed.

Using limits allows us to carefully evaluate the behavior of functions as they stretch to never-ending spaces. When the actual integration is done, its main task is to look at how the function behaves as it softly and smoothly reaches out past the observable. The resulting value reveals the actual 'area' under the curve from theoretical negative infinity to positive infinity. This approach helps us tackle problems where the standard definition of definite integrals just wouldn't hold up, offering insight into these boundless measures.

Symmetry in Functions

Symmetry often simplifies integrals by using the properties of functions. A key type of symmetry is that of odd functions, where

• \( f(-x) = -f(x) \) for every point \( x \).
• This symmetry means when you graph the function, it mirrors evenly across the origin (think of looking in a mirror where left is right and right is left, but flipped).

In the exercise, the function is odd, rendered as \( \frac{x}{(1+x^2)^2} \). With limits from \(-\infty\) to \(\infty\), this symmetry tells us that whatever area the function creates on the positive half of the axis, it is counterbalanced by an identical, negative area on the other side. Thus, the integral of any odd function over symmetric limits equals zero since the areas cancel.
Using symmetry allows us to leap directly to solutions without undergoing extensive calculations, as the natural balance within the function neatly wraps its net area to zero, offering a beautiful real-world application of mathematical elegance.