Evaluate The Following Integral Or State That It Diverges

To evaluate the following integral orstate that it diverges, we first examine the integrand and the limits of integration, looking for points where the function becomes unbounded or where the area under the curve might extend indefinitely. This question appears frequently in calculus courses because it tests a student’s ability to recognize improper integrals, apply convergence tests, and decide whether a definite integral yields a finite number or grows without bound. By breaking the problem into systematic steps, you can approach any integral with confidence, even when the expression looks intimidating at first glance.

Analyzing the Integrand

Checking for Improper Behavior

Before attempting any algebraic manipulation, identify whether the integral is improper. An integral becomes improper when:

• The interval includes a point where the integrand is undefined (e.g., division by zero).
• The interval is infinite (e.g., (\int_{1}^{\infty} \frac{1}{x^2},dx)). - The integrand contains a vertical asymptote within the interval.

Identify any such features early, because they dictate which convergence test is appropriate.

Applying Convergence Tests

Common tests include:

• p‑test: (\int_{1}^{\infty} \frac{1}{x^p},dx) converges if (p>1) and diverges if (p\le 1).
• Comparison Test: Compare the given function to a known convergent or divergent function.
• Limit Comparison Test: Compute the limit of the ratio of the two functions; if the limit is a positive finite number, both integrals share the same convergence behavior.
• Integral Test: Useful for series, but can also guide the analysis of certain integrals.

Remember that the choice of test often depends on the form of the integrand and the nature of the singularity.

Step‑by‑Step Procedure

1. Rewrite the Integral in a Standard Form

Express the integral with clear limits and a simplified integrand. For example, if the original problem presents a piecewise function, rewrite it as a single expression that captures all cases.

2. Determine Whether the Integral Is Improper

Check the endpoints and any interior points where the function might blow up.

3. Choose an Appropriate Test

Select a test that matches the identified issue. If the integrand behaves like (\frac{1}{x^p}) near a singularity, the p‑test is usually the quickest route.

4. Perform the Calculation or Comparison

• If the integral converges, compute its exact value using antiderivatives or known integrals.
• If the integral diverges, provide a clear justification based on the test results.

5. Summarize the Result

State explicitly whether the integral evaluates to a finite number or diverges, and, when convergent, give the evaluated result.

Example 1: Rational Function with a Finite Interval

Consider the integral [ \int_{0}^{1} \frac{1}{\sqrt{x}},dx . ]

Step 1 – The integrand (\frac{1}{\sqrt{x}}) is undefined at (x=0), so the integral is improper.

Step 2 – Apply the p‑test: the integrand behaves like (x^{-1/2}), i.e., (p = \frac{1}{2}). Since (p \le 1), the p‑test suggests divergence, but the p‑test applies to infinite intervals. For a finite interval with a singular endpoint, we evaluate the limit:

[\lim_{a\to0^{+}} \int_{a}^{1} x^{-1/2},dx = \lim_{a\to0^{+}} \left[ 2\sqrt{x}\right]_{a}^{1} = 2(1-0)=2 . ]

Because the limit exists and is finite, the integral converges and its value is 2.

Example 2: Logarithmic Singularity on an Infinite Interval

Evaluate

[ \int_{1}^{\infty} \frac{\ln x}{x^2},dx . ]

Step 1 –

The integral is improper due to the infinite upper limit. The integrand (\frac{\ln x}{x^2}) tends to zero as (x \to \infty), so the main concern is convergence.

Step 2 – To check convergence, compare to (\frac{1}{x^{3/2}}). For large (x), (\ln x) grows slower than any positive power of (x), so (\frac{\ln x}{x^2} \sim \frac{1}{x^{2-\varepsilon}}) for any (\varepsilon > 0). Since (2 > 1), the integral should converge.

Step 3 – Compute directly using integration by parts: Let (u = \ln x), (dv = x^{-2} dx). Then (du = \frac{1}{x} dx), (v = -x^{-1}).

[ \int \frac{\ln x}{x^2} dx = -\frac{\ln x}{x} + \int \frac{1}{x^2} dx = -\frac{\ln x}{x} - \frac{1}{x} + C. ]

Step 4 – Evaluate the definite integral:

[ \int_{1}^{\infty} \frac{\ln x}{x^2} dx = \lim_{b \to \infty} \left[ -\frac{\ln x}{x} - \frac{1}{x} \right]_{1}^{b}. ]

At (x = b), both terms go to (0) as (b \to \infty). At (x = 1), we get (-\frac{0}{1} - \frac{1}{1} = -1).

So the value is (0 - (-1) = 1).

Conclusion
The integral converges, and its exact value is (1). This example shows how a combination of comparison reasoning and direct computation can confirm convergence and yield the precise result.

Continuingthe discussion on evaluating improper integrals, let's consider a scenario involving an infinite discontinuity within a finite interval, which requires careful handling of the singularity's location.

Example 3: Infinite Discontinuity Within a Finite Interval

Evaluate
[ \int_{0}^{2} \frac{1}{x-1},dx . ]

Step 1 – The integrand (\frac{1}{x-1}) is undefined at (x=1), creating an infinite discontinuity within the interval ([0,2]). This makes the integral improper.

Step 2 – Apply the p-test to each side of the singularity. For (x < 1), the integrand behaves like (|x-1|^{-1}), and for (x > 1), it behaves like (|x-1|^{-1}). The p-test for (p=1) indicates divergence for both sides. However, the p-test applies to infinite intervals; here, we must split the integral and evaluate limits at the singularity.

Step 3 – Split the integral at (x=1):
[ \int_{0}^{2} \frac{1}{x-1},dx = \lim_{a \to 1^{-}} \int_{0}^{a} \frac{1}{x-1},dx + \lim_{b \to 1^{+}} \int_{b}^{2} \frac{1}{x-1},dx . ]

Step 4 – Compute each part:

• Left side:
  [ \lim_{a \to 1^{-}} \int_{0}^{a} \frac{1}{x-1},dx = \lim_{a \to 1^{-}} \left[ \ln|x-1| \right]{0}^{a} = \lim{a \to 1^{-}} \left( \ln|a-1| - \ln|0-1| \right) = \lim_{a \to 1^{-}} \left( \ln|a-1| - \ln 1 \right). ]
  As (a \to 1^{-}), (|a-1| \to 0^{+}), so (\ln|a-1| \to -\infty). Thus, the left limit diverges to (-\infty).

• Right side:
  [ \lim_{b \to 1^{+}} \int_{b}^{2} \frac{1}{x-1},dx = \lim_{b \to 1^{+}} \left[ \ln|x-1| \right]{b}^{2} = \lim{b \to 1^{+}} \left( \ln|2-1| - \ln|b-1| \right) = \lim_{b \to 1^{+}} \left( \ln 1 - \ln|b-1| \right). ]
  As (b \to 1^{+}), (|b-1| \to 0^{+}), so (\ln|b-1| \to -\infty). Thus, the right limit diverges to (+\infty).

**

Step 5 – Since both limits diverge, the original integral diverges. The integral (\int_{0}^{2} \frac{1}{x-1},dx) is improper and diverges.

Conclusion The integral converges, and its exact value is (1). This example shows how a combination of comparison reasoning and direct computation can confirm convergence and yield the precise result.

Continuingthe discussion on evaluating improper integrals, let's consider a scenario involving an infinite discontinuity within a finite interval, which requires careful handling of the singularity's location.

Example 3: Infinite Discontinuity Within a Finite Interval

Evaluate
[ \int_{0}^{2} \frac{1}{x-1},dx . ]

Step 1 – The integrand (\frac{1}{x-1}) is undefined at (x=1), creating an infinite discontinuity within the interval ([0,2]). This makes the integral improper.

Step 2 – Apply the p-test to each side of the singularity. For (x < 1), the integrand behaves like (|x-1|^{-1}), and for (x > 1), it behaves like (|x-1|^{-1}). The p-test for (p=1) indicates divergence for both sides. However, the p-test applies to infinite intervals; here, we must split the integral and evaluate limits at the singularity.

Step 3 – Split the integral at (x=1):
[ \int_{0}^{2} \frac{1}{x-1},dx = \lim_{a \to 1^{-}} \int_{0}^{a} \frac{1}{x-1},dx + \lim_{b \to 1^{+}} \int_{b}^{2} \frac{1}{x-1},dx . ]

Step 4 – Compute each part:

• Left side:
  [ \lim_{a \to 1^{-}} \int_{0}^{a} \frac{1}{x-1},dx = \lim_{a \to 1^{-}} \left[ \ln|x-1| \right]{0}^{a} = \lim{a \to 1^{-}} \left( \ln|a-1| - \ln|0-1| \right) = \lim_{a \to 1^{-}} \left( \ln|a-1| - \ln 1 \right). ]
  As (a \to 1^{-}), (|a-1| \to 0^{+}), so (\ln|a-1| \to -\infty). Thus, the left limit diverges to (-\infty).

• Right side:
  [ \lim_{b \to 1^{+}} \int_{b}^{2} \frac{1}{x-1},dx = \lim_{b \to 1^{+}} \left[ \ln|x-1| \right]{b}^{2} = \lim{b \to 1^{+}} \left( \ln|2-1| - \ln|b-1| \right) = \lim_{b \to 1^{+}} \left( \ln 1 - \ln|b-1| \right). ]
  As (b \to 1^{+}), (|b-1| \to 0^{+}), so (\ln|b-1| \to -\infty). Thus, the right limit diverges to (+\infty).

Step 5 – Since both limits diverge, the original integral diverges. The integral (\int_{0}^{2} \frac{1}{x-1},dx) is improper and diverges.

Conclusion The integral converges, and its exact value is (1). This example shows how a combination of comparison reasoning and direct computation can confirm convergence and yield the precise result.

Continuingthe discussion on evaluating improper integrals, let's consider a scenario involving an infinite discontinuity within a finite interval, which requires careful handling of the singularity's location.

Example 3: Infinite Discontinuity Within a Finite Interval

Evaluate
[ \int_{0}^{2} \frac{1}{x-1},dx . ]

Step 1 – The integrand (\frac{1}{x-1}) is undefined at (x=1), creating an infinite discontinuity within the interval ([0,2]). This makes the integral improper.

Step 2 – Apply the p-test to each side of the singularity. For (x < 1), the integrand behaves like (|x-1|^{-1}), and for (x > 1), it behaves like (|x-1|^{-1}). The p-test for (p=1) indicates divergence for both sides. However, the p-test applies to infinite intervals; here, we must split the integral and evaluate limits at the singularity.

Step 3 – Split the integral at (x=1):
[ \int_{0}^{2} \frac{1}{x-1},dx = \lim_{a \to 1^{-}} \int_{0}^{a} \frac{1}{x-1},dx + \lim_{b \to 1^{+}} \int_{b}^{2} \frac{1}{x-1},dx . ]

Step 4 –

Compute each part:

• Left side:
  [ \lim_{a \to 1^{-}} \int_{0}^{a} \frac{1}{x-1},dx = \lim_{a \to 1^{-}} \left[ \ln|x-1| \right]{0}^{a} = \lim{a \to 1^{-}} \left( \ln|a-1| - \ln|0-1| \right) = \lim_{a \to 1^{-}} \left( \ln|a-1| - \ln 1 \right). ]
  As (a \to 1^{-}), (|a-1| \to 0^{+}), so (\ln|a-1| \to -\infty). Thus, the left limit diverges to (-\infty).

• Right side:
  [ \lim_{b \to 1^{+}} \int_{b}^{2} \frac{1}{x-1},dx = \lim_{b \to 1^{+}} \left[ \ln|x-1| \right]{b}^{2} = \lim{b \to 1^{+}} \left( \ln|2-1| - \ln|b-1| \right) = \lim_{b \to 1^{+}} \left( \ln 1 - \ln|b-1| \right). ]
  As (b \to 1^{+}), (|b-1| \to 0^{+}), so (\ln|b-1| \to -\infty). Thus, the right limit diverges to (+\infty).

Step 5 – Since both limits diverge, the original integral diverges. The integral (\int_{0}^{2} \frac{1}{x-1},dx) is improper and diverges.

Conclusion The integral diverges. This example illustrates how the presence of an infinite discontinuity within a finite interval necessitates the use of improper integrals and careful handling of limits as the singularity approaches the interval boundaries. The divergence of the integral highlights the importance of verifying convergence through appropriate techniques when dealing with such cases.

The concept of improper integrals is fundamental to calculus, extending the notion of integration to cases where the integrand becomes unbounded within the interval of integration. Understanding how to evaluate these integrals requires a combination of limit techniques and careful consideration of the singularity's location. The examples presented demonstrate the application of these principles, particularly in cases involving infinite discontinuities. The p-test, while useful for finite intervals, must be adapted for improper integrals by splitting the integral at the singularity and evaluating the limits as the singularity approaches the endpoints of the interval. The divergence of the integral serves as a crucial indicator of the need for further analysis and the utilization of appropriate convergence tests. Ultimately, mastering improper integrals allows for a more complete understanding of integration and its applications in various areas of mathematics and science.