Improper Integral Calculator is loading…
Use this improper integral calculator to find out if an integral converges (has a finite value) or diverges (grows without end). It works with infinite limits like \(\int_1^\infty\) and with functions that blow up, like \(\frac{1}{\sqrt{x}}\) at 0. You get the value, the steps, and a graph.
How to Use It
- Type the function, for example
1/x^2. - Enter the limits. Type
inffor \(\infty\) and-inffor \(-\infty\). - Press Solve. A badge tells you Converges or Diverges.
What Makes an Integral “Improper”?
A definite integral is improper when one of these is true:
- Type 1: an infinite limit. Example: \(\int_1^\infty \frac{1}{x^2}\,dx\).
- Type 2: the function blows up at an end point or inside the interval. Example: \(\int_0^1 \frac{1}{\sqrt{x}}\,dx\), because \(\frac{1}{\sqrt{x}}\) goes to infinity at \(x = 0\).
We can’t plug infinity into a formula. So we replace the problem point with a letter, like \(t\), and take a limit:
\[\int_1^\infty f(x)\,dx = \lim_{t\to\infty} \int_1^t f(x)\,dx.\]
If that limit is a real number, the integral converges. If not, it diverges. To practice limits on their own, try the limit calculator.
Worked Examples
Example 1: \(\int_1^\infty \frac{1}{x^2}\,dx\) (converges)
\(\int_1^t x^{-2}\,dx = \left[-\frac{1}{x}\right]_1^t = 1 – \frac{1}{t}\).
As \(t \to \infty\), \(\frac{1}{t} \to 0\). So the limit is 1. The integral converges to 1.
Example 2: \(\int_1^\infty \frac{1}{x}\,dx\) (diverges)
\(\int_1^t \frac{1}{x}\,dx = \ln t – \ln 1 = \ln t\).
As \(t \to \infty\), \(\ln t\) keeps growing. So the integral diverges. Even though \(\frac1x\) gets small, it does not get small fast enough. Learn why in the antiderivative of 1/x.
Example 3: \(\int_0^1 \frac{1}{\sqrt{x}}\,dx\) (converges)
\(\int_t^1 x^{-1/2}\,dx = \left[2\sqrt{x}\right]_t^1 = 2 – 2\sqrt{t}\).
As \(t \to 0^+\), this goes to 2. The area is finite even though the graph shoots up.
Example 4: \(\int_{-\infty}^{\infty} e^{-x^2}\,dx\)
This is the famous Gaussian integral. It has no simple antiderivative, but its value is known: \(\sqrt{\pi} \approx 1.7725\). The calculator finds it with numerical methods and recognizes the exact value.
The p-Test: A Quick Shortcut
For integrals of \(\frac{1}{x^p}\), you can tell the answer without any work:
| Integral | Converges when | Diverges when |
|---|---|---|
| \(\int_1^\infty \frac{1}{x^p}\,dx\) | \(p > 1\) | \(p \le 1\) |
| \(\int_0^1 \frac{1}{x^p}\,dx\) | \(p < 1\) | \(p \ge 1\) |
For example, \(\int_1^\infty \frac{1}{x^3}\,dx\) converges because \(3 > 1\). It equals \(\frac{1}{p-1} = \frac12\).
Common Mistakes
- Missing a hidden problem point. \(\int_{-1}^{1}\frac{1}{x^2}\,dx\) looks normal, but \(\frac{1}{x^2}\) blows up at 0. You must split it at 0. Both halves diverge.
- Plugging in infinity. Always use a limit with a variable like \(t\).
- Thinking “the function goes to 0, so it converges.” \(\frac1x\) goes to 0 but its integral still diverges.
- Splitting \(\int_{-\infty}^{\infty}\) wrong. Split at any point, like 0. Both parts must converge on their own.
Learn the Math Behind the Improper Integral Calculator
Improper integrals have an infinite limit or a break in the function, so they are found with a limit. Some give a finite area; others diverge.
Frequently Asked Questions
How does the calculator decide if an integral converges?
It computes the integral up to closer and closer points (or larger and larger bounds) and watches the values. If they settle on a number, it converges. If they keep growing, it diverges. When an antiderivative exists, it uses it for an exact limit.
Can an improper integral be negative?
Yes. A convergent improper integral can be any real number, including negative values.
What does “diverges to infinity” mean?
The area under the curve is not finite. The partial integrals keep growing past any number you pick.
Where are improper integrals used?
They show up in probability (total probability equals 1), physics (escape velocity), and in tests for infinite series.
Related Guides
- Definite vs indefinite integrals
- Antiderivatives of exponential functions
- Why is integration so hard?