Improper Integral Practice Problems with Answers

These improper integral practice problems cover infinite limits, vertical asymptotes at an endpoint and inside the interval, and the comparison test. For each one, decide whether it converges, and if it does, find its value. The answers come first, then full solutions.

The definition for an infinite limit:

For an asymptote at x = a, use \lim_{t\to a^+}\int_t^{b} f(x)\,dx instead. For more examples, read the improper integrals guide.

Problems

Infinite limits

1. \displaystyle\int_1^{\infty}\frac{1}{x^2}\,dx
2. \displaystyle\int_1^{\infty}\frac{1}{x}\,dx
3. \displaystyle\int_1^{\infty}x^{-3/2}\,dx
4. \displaystyle\int_0^{\infty}e^{-2x}\,dx
5. \displaystyle\int_0^{\infty}x\,e^{-x}\,dx
6. \displaystyle\int_{-\infty}^{\infty}\frac{1}{1 + x^2}\,dx

Vertical asymptotes

7. \displaystyle\int_0^{1}\frac{1}{\sqrt{x}}\,dx
8. \displaystyle\int_0^{1}\frac{1}{x}\,dx
9. \displaystyle\int_0^{1}\ln x\,dx
10. \displaystyle\int_{-1}^{1}\frac{1}{x^2}\,dx
11. \displaystyle\int_0^{3}\frac{1}{(x - 1)^{2/3}}\,dx

Comparison

12. Does \displaystyle\int_1^{\infty}\frac{2 + \cos x}{x^2}\,dx converge?

Answers

Step-by-step solutions

1. \displaystyle\lim_{b\to\infty}\left[-\frac{1}{x}\right]_1^{b} = \lim_{b\to\infty}\left(1 - \frac{1}{b}\right) = 1

2. \displaystyle\lim_{b\to\infty}\big[\ln x\big]_1^{b} = \lim_{b\to\infty}\ln b = \infty, so it diverges.

3. An antiderivative of x^{-3/2} is -2x^{-1/2}:

This is a p-integral with p = \frac{3}{2} > 1, so convergence is expected.

4. \displaystyle\lim_{b\to\infty}\left[-\frac{1}{2}e^{-2x}\right]_0^{b} = \lim_{b\to\infty}\left(\frac{1}{2} - \frac{1}{2}e^{-2b}\right) = \frac{1}{2}

5. By parts, \int x\,e^{-x}\,dx = -(x + 1)e^{-x}. Then:

The term (b + 1)e^{-b} \to 0 because the exponential grows much faster than b + 1 (use L'Hôpital's rule if you want to check).

6. Split at 0. Each half is \lim\arctan b = \frac{\pi}{2}, so the total is \pi. Both halves must converge for the whole integral to converge.

7. The integrand blows up at x = 0:

8. \displaystyle\lim_{t\to 0^+}\big(\ln 1 - \ln t\big) = \lim_{t\to 0^+}(-\ln t) = \infty, so it diverges.

9. An antiderivative is x\ln x - x. At x = 1 it is -1. As t \to 0^+, t\ln t \to 0, so:

10. The asymptote at x = 0 is inside the interval, so split there. The right half is \lim_{t\to 0^+}\left(\frac{1}{t} - 1\right) = \infty, so the whole integral diverges. The tempting answer \left[-\frac{1}{x}\right]_{-1}^{1} = -2 is impossible, since the integrand is positive.

11. The asymptote is at x = 1, inside [0, 3]. An antiderivative is 3(x - 1)^{1/3}, which is continuous at x = 1:

Both halves converge, so the total is 3 + 3\sqrt[3]{2} \approx 6.78.

12. Since -1 \le \cos x \le 1, the integrand is between \frac{1}{x^2} and \frac{3}{x^2}. And \int_1^{\infty}\frac{3}{x^2}\,dx = 3 converges. A positive function below a convergent one also converges.

Common mistakes on these problems

• Missing an asymptote inside the interval. Problems 10 and 11 look like ordinary definite integrals. Always check where the integrand is undefined.
• Writing \infty as a number. Write the limit. Expressions like \frac{1}{\infty} hide mistakes.
• Comparing in the wrong direction. Being smaller than a divergent integral proves nothing. Being larger than a convergent one proves nothing.
• Splitting only one side. In problem 6, both halves must converge separately.

Keep practicing

For unlimited problems with instant feedback, try the infinite limits lesson, the infinite discontinuities lesson and the comparison test lesson. The same p rule comes back for series in the convergence test problems.