Question

Many improper double integrals may be handled using the techniques for improper integrals in one variable (Section 7.8 ). For example, under suitable conditions on \(f\), \(\int_{a}^{\infty} \int_{g(x)}^{h(x)} f(x, y) d y d x=\lim _{b \rightarrow \infty} \int_{a}^{b} \int_{g(x)}^{h(x)} f(x, y) d y d x\). Use or extend the one-variable methods for improper integrals to evaluate the following integrals. $$\int_{0}^{\infty} \int_{0}^{\infty} e^{-x-y} d y d x$$

Short answer

Question: Evaluate the improper double integral. $$\int_{0}^{\infty} \int_{0}^{\infty} e^{-x-y} d y d x$$ Answer: $$-1$$

Step-by-step solution

Change the double integral into iterated improper single integrals

To handle this improper double integral, we can rewrite it as an iterated integral by separating the limits: $$\int_{0}^{\infty} \int_{0}^{\infty} e^{-x-y} d y d x = \lim_{a \rightarrow \infty} \lim_{b \rightarrow \infty} \int_{0}^{a} \int_{0}^{b} e^{-x-y} d y d x$$

Evaluate the inner integral

Let's evaluate the inner integral first: $$\int_{0}^{b} e^{-x-y} d y = \left[-e^{-x-y}\right]_{0}^{b} = -e^{-x-b} + e^{-x}$$

Evaluate the outer integral

Now, we can evaluate the outer integral with respect to x: $$\int_{0}^{a} (-e^{-x-b} + e^{-x}) d x = \left[\frac{e^{-b} - e^{-x}}{e^{x}}\right]_{0}^{a} = (e^{-a} - e^{-(a+b)}) - (1 - e^{-b})$$

Evaluate the limits

Now, we find the limits as a and b approach infinity: $$\lim_{a \rightarrow \infty} \lim_{b \rightarrow \infty} (e^{-a} - e^{-(a+b)}) - (1 - e^{-b})$$ First, let's find the limit for b: $$\lim_{b \rightarrow \infty} (1 - e^{-b}) = 1$$ Now, let's find the limit for a: $$\lim_{a \rightarrow \infty} (e^{-a} - e^{-(a+b)}) = 0$$ Finally, calculate the overall limit: $$0 - 1 = -1$$ The answer for the improper double integral is: $$\int_{0}^{\infty} \int_{0}^{\infty} e^{-x-y} d y d x = -1$$

Key concepts

Double Integrals

Double integrals are essential when you want to calculate the volume under a surface in three-dimensional space. They involve two integrations performed over a two-variable function, typically denoted as \(f(x, y)\). Each variable corresponds to a different axis on the plane.

For instance, a double integral of a function \( f(x, y) \) over a rectangle [a, b] x [c, d] is expressed as:\[\int_{a}^{b} \int_{c}^{d} f(x, y) \, dy \, dx\]Here's what's happening:

• The inner integral takes the slice of the volume at a fixed \(x\) along the \(y\)-direction, computing the area under the curve for that value of \(x\).
• The outer integral then sums up all those areas by slicing in the \(x\)-direction, combining these results to form the total volume beneath the surface.

Double integrals allow us to examine real-world situations where we consider the cumulative effect over a surface area. The calculation can become complex, but the key is to focus on understanding the geometry involved in breaking down how the integral samples the function's value over its region.

Iterated Integrals

Iterated integrals simplify double integrals by iterating over one variable at a time. They are evaluated as one integration followed by another. This step-by-step processing makes dealing with complex integrals more manageable.

To transform a double integral into an iterated integral, you consider the integration limits for each variable separately. In our problem, the given double integral is expressed iteratively as:\[\int_{0}^{\infty} \left(\int_{0}^{\infty} e^{-x-y} \, dy\right) \, dx\]Here's how it unfolds:

• The inner integral \(\int_{0}^{\infty} e^{-x-y} \, dy\) is evaluated first, treating \(x\) as a constant. This finds the component of the volume slice in the \(y\)-direction.
• The outer integral \(\int_{0}^{\infty} \ldots \, dx\) follows by considering each \(x\) slice from the previous step to sum up all accumulated values.

Iterated integrals require careful management of boundaries and integration order, but they untangle multidimensional problems, making them accessible for practical evaluation.

Limit Evaluation

Limit evaluation is crucial when handling improper integrals, which are often characterized by infinite bounds. It allows for calculating values that otherwise approach infinity.

In our example, improper double integrals involved replacing infinity with a variable (say \(a\) or \(b\)), performing traditional integration, and then using limits to find the value as these variables approach infinity.

Here's how it's done:

• First, we take the inner integral over \(y\) and identify the limit as \(b\) tends towards infinity. This gives us the expression evaluated at \(b\) instead of infinity initially.
• Next, evaluate the outer integral over \(x\), replacing its upper bound with \(a\).
• Finally, assess the behavior of the resulting expression as \(a\) and \(b\) each head to infinity. The computed limits give the integral's converging value.

Understanding each step in the limit evaluation process is crucial as it converts potentially unmanageable expressions into finite results, providing a complete picture of the integral’s behavior across infinite bounds.