Problem 56 - Calculus Single Variable

Question

Pach of the improper integrals is improper for two reasons. Determine whether the integral converges or diverges. If it converges, then evaluate it. $$ \int_{0}^{\infty} \frac{\exp (-\sqrt{x})}{\sqrt{x}} d x $$

Step-by-Step Solution

Answer

The integral converges and evaluates to 1.

Step 1: Examine the form of the integral

The integral is \( \int_{0}^{\infty} \frac{\exp(-\sqrt{x})}{\sqrt{x}} \, dx \). It is improper because it has an infinite limit and because the integrand is undefined at \( x = 0 \).

Step 2: Split the integral at point of discontinuity

Since there is a discontinuity at \( x = 0 \) and infinity at the upper limit, we need to treat these separately by using limits. Consider the integral \( \int_{0}^{eta} \frac{\exp(-\sqrt{x})}{\sqrt{x}} \, dx \) where \( \beta \) tends to infinity.

Step 3: Change variables for easier integration

Let \( u = \sqrt{x} \), then \( du = \frac{1}{2\sqrt{x}} \, dx \) or \( dx = 2u \, du \). The integral becomes \( \int_{0}^{\sqrt{\beta}} \exp(-u) \, du \).

Step 4: Evaluate the integral with new limits

Since \( \exp(-u) \) is an exponential function, its integral is \( -\exp(-u) \). Evaluating this from 0 to \( \sqrt{\beta} \) yields \( [-\exp(-u)]_0^{\sqrt{\beta}} = -(\exp(-\sqrt{\beta}) - 1) \).

Step 5: Take the limit as \( \beta \) approaches infinity

As \( \beta \) approaches infinity, \( \sqrt{\beta} \) also approaches infinity, making \( \exp(-\sqrt{\beta}) \) tend to zero. Thus, the expression \( -(0 - 1) = 1 \).

Step 6: Address the improper nature at \( x = 0 \)

For small \( x \), \( \frac{\exp(-\sqrt{x})}{\sqrt{x}} \) behaves as \( \frac{1}{\sqrt{x}} \), which is integrable from \( 0 \) to any positive number. Thus, the integral converges at the lower limit as well. The integral from \( 0 \) to infinity converges.

Key Concepts

Convergence

Improper integrals often bring up the critical question of convergence. But what exactly does convergence mean here? In simple terms, convergence tells us that even though the integral seems infinite due to its bounds or its function behavior, it still approaches a finite number.
When we say an integral converges:

• The integral resolves to a specific number.
• This number persists regardless of the infinite nature of the upper bound or difficulty at lower bounds.
• Convergence ensures a well-defined result when integrating over discontinuities or extremities.

Integrals like \(\int_{0}^{\infty} \frac{\exp(-\sqrt{x})}{\sqrt{x}} \, dx \) converge because, despite their infinite limits, they approach a stable number, allowing us to compute them to achieve the answer.

Divergence

Opposite to convergence, we have the notion of divergence. It's crucial to differentiate between the two when dealing with improper integrals. Divergence means the integral doesn't settle on a single value. Instead, it keeps growing indefinitely.
Consider an integral that diverges:

• The sum becomes infinite or undefined.
• It doesn't produce a single, predictable result across its domain.
• Divergence hints that the integral's nature or definition might need re-evaluation or special techniques to approach.

In the solution provided, the integral's bounds and behavior at critical points were properly addressed, ensuring it converged rather than diverged. Recognizing divergence early can save you the effort of performing unnecessary calculations.

Limits

Limits play a pivotal role when determining whether an integral is improper and how we can handle it. They help in bridging the gap between intuitive expectations and mathematical precision.
Using limits involves:

• Approaching a boundary value — like zero or infinity — without actually reaching it.
• Ensuring calculations only include parts of a graph where behaviors are well-defined.
• Allowing solutions to inherently undefined problems by making them approachable and solvable.

In our specific problem, limits help deal with infinity as the upper bound and the undefined nature at zero. By breaking the integral at these points using limits, the solution is made feasible. Whether it's at the start, like \( x = 0 \), or the end, like infinity, limits clear the path for finding converging solutions.

Variable Substitution

Variable substitution is a handy trick to simplify complex integrals, converting them into easier forms. By changing variables, we align the integral with simpler integration techniques.
Here’s how it works:

• Identify a difficult part of the integral and set it equal to a new variable.
• Change variables to make the integral easier to deal with.
• Solve the new, simplified integral using known methods.

In our case, using the substitution \( u = \sqrt{x} \) transformed the messy expression into a straightforward exponential function, \( \exp(-u) \). This makes finding the integral more approachable, facilitating straightforward evaluation without the initial complexity at zero or infinity. Substitution aligns variables and ideas into a form we can work with, changing the intimidating to the manageable.