If you’re learning calculus, you’ve probably come across integrals with limits at infinity or negative infinity or that have a vertical asymptote in the interval of integration. These are called improper integrals. In this guide, we’ll walk through examples of improper integrals with solutions so you can see how to approach them step by step.
What Is an Improper Integral?
An improper integral is a definite integral where either the limits of integration are infinite, or the integrand has an infinite discontinuity somewhere in the interval of integration. We need to use limits to evaluate these integrals.
There are two main types of improper integrals:
Integrals with Infinite Limits of Integration
These occur when the interval of integration extends to positive or negative infinity. For example
$$\int_1^{\infty} \frac{1}{x^2} \, dx.$$
This isn’t a normal definite integral because you can’t plug \( \infty \) into the function. Instead, you rewrite it using a limit:
$$\int_1^{\infty} \frac{1}{x^2} \, dx = \lim_{b \to \infty} \int_1^b \frac{1}{x^2} \, dx.$$
We then evaluate the integral like usual and then take the limit.
Integrals with Discontinuous Integrands
Sometimes, the function you’re integrating is undefined at a certain point in the interval. For instance
$$\int_0^1 \frac{1}{\sqrt{x}} \, dx.$$
The function \( \frac{1}{\sqrt{x}} \) is undefined at \( x = 0 \), so we can’t evaluate this directly. Instead, we write
$$\int_0^1 \frac{1}{\sqrt{x}} \, dx = \lim_{a \to 0^+} \int_a^1 \frac{1}{\sqrt{x}} \, dx.$$
Again, the idea is to use a limit to handle the integral’s problematic part(s).
Steps to Evaluate Improper Integrals
Step 1: Split up the integral, if necessary
If the interval of integration has more than one point, that makes the integral improper, or if it’s an interior point, you must split the integral into multiple integrals. This can be done using one of the properties of definite integrals mentioned in the article Basic Integration Problems for Beginners.
Step 2: Rewrite as a limit
Rewrite the integral as a limit at the point which makes the integral improper.
Step 3: Evaluate the definite integral
Compute the definite integral. The Integral Calculus archive has resources to help you evaluate integrals using many techniques.
Step 4: Take the limit
Lastly, take the limit. For assistance, please refer to the article How to Calculate Limits in Calculus: Everything You Need to Know.
Improper Integrals with Infinite Limits of Integration
Improper integrals with infinite limits occur when the magnitude of either limit is infinite. Since infinity isn’t a number, you can’t plug it in directly. Instead, we rewrite these integrals using limits at infinity.
Example 1: Evaluate \( \int_{-\infty}^0 e^x \, dx \).
Solution: Rewriting as a limit gives
$$\lim_{a \to -\infty} \int_a^0 e^x dx.$$
Next, we evaluate the definite integral to get
$$\lim_{a \to -\infty} e^x|_a^0 = \lim_{a \to -\infty} e^0 – e^a = \lim_{a \to -\infty} 1 – e^a.$$
Taking the limit we obtain
$$1 – 0 = 1.$$
Example 2: Evaluate \( \int_1^\infty \frac{1}{x^2} \, dx \).
Solution: Rewriting as a limit gives
$$\lim_{b \to \infty} \int_1^b \frac{1}{x^2} \, dx.$$
Next, we evaluate the definite integral to get
$$\lim_{b \to \infty} -\frac{1}{x} |_1^b = \lim_{b \to \infty} -\frac{1}{b} + 1.$$
Taking the limit we obtain
$$0 + 1 = 1.$$
Example 3: Evaluate \( \int_{-\infty}^{\infty} \frac{1}{1 + x^2} \, dx \).
Solution: We start by splitting the integral into two as follows:
$$\int_{-\infty}^0 \frac{1}{1 + x^2} \, dx + \int_0^{\infty} \frac{1}{1 + x^2} \, dx.$$
Rewriting the integrals as limits gives
$$\lim_{a \to -\infty} \int_a^0 \frac{1}{1 + x^2} \, dx + \lim_{b \to \infty} \int_0^b \frac{1}{1 + x^2} \, dx.$$
Next, we evaluate the definite integrals to get
$$\lim_{a \to -\infty} \arctan{x}|_a^0 + \lim_{b \to \infty} \arctan{x}|_0^b = \lim_{a \to -\infty} \arctan{0} – \arctan{a} + \lim_{b \to \infty} \arctan{b} – \arctan{0} = – \lim_{a \to -\infty} \arctan{a} + \lim_{b \to \infty} \arctan{b}.$$
Taking the limits we obtain
$$-( -\frac{\pi}{2}) + \frac{\pi}{2} = \frac{\pi}{2} + \frac{\pi}{2} = {\pi}.$$
Improper Integrals with Discontinuous Integrands
This type of improper integral arises when the function is undefined at one or more points within the interval of integration. In this case, we rewrite the integral using one-sided limits.
Example 4: Evaluate \( \int_0^1 \frac{1}{\sqrt{x}} \, dx \).
Solution: The integrand is undefined at \( x = 0 \), so the integral is improper at the left endpoint. Rewriting as a limit gives
$$\lim_{a \to 0^+} \int_a^1 \frac{1}{\sqrt{x}} \, dx.$$
Next, we evaluate the definite integral to get
$$\lim_{a \to 0^+} \int_a^1 x^{-1/2} \, dx = \lim_{a \to 0^+} \left[ 2x^{1/2} \right]_a^1 = 2 – 2\sqrt{a}.$$
Taking the limit, we obtain
$$2 – 0 = 2.$$
Example 5: Evaluate \( \int_{-1}^1 \frac{1}{x} \, dx \).
Solution: The integrand is undefined at \( x = 0 \), which lies inside the interval of integration. This means the integral is improper at an interior point. We split the integral into two as follows:
$$\int_{-1}^0 \frac{1}{x} \, dx + \int_0^1 \frac{1}{x} \, dx.$$
We evaluate each integral individually. The first integral is
$$\int_{-1}^0 \frac{1}{x} \, dx = \lim_{a \to 0^-} \int_{-1}^a \frac{1}{x} \, dx = \lim_{a \to 0^-} \left[ \ln|x| \right]_{-1}^a = \lim_{a \to 0^-} (\ln|a| – \ln(1)) = -\infty.$$
There is no need to calculate the second integral. Since the first integral diverges, the entire integral diverges.
Conclusion
In this guide, we explored multiple examples of improper integrals with solutions, covering both types: those with unbounded intervals and those with discontinuities. Some integrals converged to a finite value, while others diverged.
