Improper Integrals: How to Tell If They Converge or Diverge

An integral is improper when either:

• a limit of integration is infinite, like \( \int_1^{\infty}\frac{dx}{x^2} \), or
• the integrand blows up somewhere on the interval, like \( \int_0^1\frac{dx}{\sqrt x} \).

We handle both with a limit. If the limit is a finite number the integral converges; otherwise it diverges.

The intuition: infinite regions with finite area

It sounds impossible that a region stretching off to infinity can have a finite area, but you already know a discrete version of this: \( \frac12 + \frac14 + \frac18 + \cdots = 1 \) (a geometric series). Infinitely many pieces can add up to a finite total, as long as the pieces shrink quickly enough.

The same thing happens with area. The graph of \( \frac{1}{x^2} \) drops so fast that the area from 1 to 10 is 0.9, from 1 to 100 is 0.99, from 1 to 1000 is 0.999. Each extra stretch adds less and less, and the totals settle toward 1. The graph of \( \frac1x \) drops too, but more slowly: the area from 1 to \( t \) is \( \ln t \), which keeps growing without bound. Deciding which of these two behaviors you have is the whole subject.

That is why the definition uses a limit. You cannot integrate “up to infinity” directly, so you integrate up to a finite \( t \), where the Fundamental Theorem of Calculus applies, and then ask what happens as \( t \) grows.

Type 1: infinite limits

$$\int_a^{\infty}f(x)\,dx = \lim_{t\to\infty}\int_a^t f(x)\,dx$$

Example 1. \( \int_1^{\infty}\frac{dx}{x^2} = \lim_{t\to\infty}\left[-\frac1x\right]_1^t = \lim_{t\to\infty}\left(1 - \frac1t\right) = 1 \). Converges.

Example 2. \( \int_1^{\infty}\frac{dx}{x} = \lim_{t\to\infty}\ln t = \infty \). Diverges (see integral of 1/x).

Example 3. \( \int_0^{\infty}e^{-x}\,dx = \lim_{t\to\infty}\left(1 - e^{-t}\right) = 1 \).

Example 4 (both ends infinite). Split at any point:

$$\int_{-\infty}^{\infty}\frac{dx}{1+x^2} = \lim_{s\to-\infty}\int_s^0 + \lim_{t\to\infty}\int_0^t = \frac\pi2 + \frac\pi2 = \pi$$

Both halves must converge separately.

Example 5 (parts plus a limit). \( \int_0^{\infty}xe^{-x}\,dx \). Integration by parts gives the antiderivative \( -xe^{-x} - e^{-x} \), so

$$\int_0^t xe^{-x}\,dx = 1 - te^{-t} - e^{-t}$$

As \( t \to \infty \), \( e^{-t} \to 0 \), and \( te^{-t} = \frac{t}{e^t} \to 0 \) as well, because exponentials outgrow polynomials (you can confirm it with L’Hôpital’s rule). The integral converges to 1.

Example 6 (substitution plus a limit). \( \int_e^{\infty}\frac{dx}{x(\ln x)^2} \). With \( u = \ln x \), the antiderivative is \( -\frac{1}{\ln x} \), so

$$\int_e^t\frac{dx}{x(\ln x)^2} = 1 - \frac{1}{\ln t} \;\longrightarrow\; 1$$

This one is instructive: the integrand shrinks only slightly faster than \( \frac1x \), yet that is enough for convergence.

Type 2: infinite integrand

If \( f \) blows up at \( x = a \):

$$\int_a^b f(x)\,dx = \lim_{s\to a^{+}}\int_s^b f(x)\,dx$$

Example 7. \( \int_0^1\frac{dx}{\sqrt x} = \lim_{s\to0^+}\left[2\sqrt x\right]_s^1 = 2 \). Converges, even though the function is unbounded.

Example 8. \( \int_{-1}^{1}\frac{dx}{x^2} \). The blow-up is at \( x = 0 \), inside the interval. Split there: \( \int_0^1\frac{dx}{x^2} = \infty \), so the whole integral diverges. (Naively plugging in the antiderivative gives \( -2 \), which is wrong: the area is positive.)

Example 9. \( \int_0^1\ln x\,dx \). The log heads to \( -\infty \) at 0. Using the antiderivative \( x\ln x - x \),

$$\int_s^1\ln x\,dx = -1 - s\ln s + s$$

As \( s \to 0^+ \), \( s\ln s \to 0 \), so the integral converges to \( -1 \). The answer is negative because the graph lies below the axis on \( (0, 1) \).

The p-integral test

This one fact settles a huge number of problems:

$$\int_1^{\infty}\frac{dx}{x^p}\ \text{converges} \iff p > 1, \qquad \int_0^{1}\frac{dx}{x^p}\ \text{converges} \iff p < 1$$

For example, \( \int_1^{\infty}\frac{dx}{x^3} = \frac12 \), while \( \int_1^{\infty}\frac{dx}{\sqrt x} \) diverges.

Why it is true. For \( p \ne 1 \), the power rule gives

$$\int_1^t x^{-p}\,dx = \frac{t^{1-p} - 1}{1 - p}$$

If \( p > 1 \), the exponent \( 1 - p \) is negative, so \( t^{1-p} \to 0 \) and the integral converges to \( \frac{1}{p-1} \). If \( p < 1 \), the exponent is positive, \( t^{1-p} \to \infty \), and the integral diverges. The borderline \( p = 1 \) gives \( \ln t \), which also diverges. The interval \( [0, 1] \) works the same way with the roles reversed, since there you care about small \( x \) instead of large \( x \).

Comparison test

If \( 0 \le f(x) \le g(x) \) and \( \int g \) converges, so does \( \int f \). If \( \int f \) diverges, so does \( \int g \).

Example: \( \int_1^{\infty}\frac{dx}{x^2 + 1} \) converges because \( \frac{1}{x^2+1} < \frac{1}{x^2} \). (Its exact value happens to be \( \frac\pi4 \).)

A divergent comparison. \( \int_1^{\infty}\frac{2 + \sin x}{x}\,dx \) diverges, because the numerator is always at least 1, so the integrand is at least \( \frac1x \), whose integral diverges.

The comparison test is how you decide convergence when no antiderivative is available. Think of it as asking “what does this function behave like for large \( x \)?” and then comparing with a p-integral or an exponential. The same reasoning drives limits at infinity.

Limit comparison. Sometimes the inequality is awkward to set up. If \( f \) and \( g \) are positive and \( \frac{f(x)}{g(x)} \) approaches a finite, nonzero number as \( x \to \infty \), then the two integrals share the same fate. For \( \int_1^\infty\frac{x + 1}{x^3 + 5}\,dx \), the ratio with \( \frac{1}{x^2} \) tends to 1, so the integral converges just like the p-integral with \( p = 2 \).

A checklist for any improper integral

1. Find every problem point: infinite limits, and any \( x \) in the interval where the integrand is undefined or unbounded.
2. Split the integral so each piece has exactly one problem point, at one end.
3. For each piece, either compute it with a limit or decide convergence by comparison.
4. Combine: the original integral converges only if every piece converges. If even one piece diverges, stop; the whole integral diverges.

Common mistakes

• Missing a blow-up inside the interval. Always check the integrand for vertical asymptotes between the limits, as in Example 8.
• Treating \( \infty \) as a number. Write the limit, evaluate the finite integral, then take the limit. Never plug \( \infty \) into an antiderivative.
• Cancelling infinities. \( \int_{-\infty}^{\infty}x\,dx \) is not 0. Each half diverges on its own, so the whole integral diverges, even though the graph is symmetric.
• Assuming “goes to zero” means convergence. The integrand must go to zero fast enough; \( \frac1x \) does not.
• Reversing the comparison. Being smaller than a divergent integral tells you nothing, and neither does being larger than a convergent one.

Famous improper integrals

Integral	Value
\( \int_0^{\infty}e^{-x}dx \)	\( 1 \)
\( \int_0^{\infty}xe^{-x}dx \)	\( 1 \)
\( \int_{-\infty}^{\infty}e^{-x^2}dx \)	\( \sqrt\pi \) (Gaussian integral)
\( \int_{-\infty}^{\infty}\frac{dx}{1+x^2} \)	\( \pi \)
\( \int_0^1\ln x\,dx \)	\( -1 \)

Where they are used

Improper integrals are everywhere in probability, where a density must integrate to 1 over the whole real line, and expected values are integrals like Example 5. In physics, the work needed to move a rocket “to infinity” against gravity is an improper integral of \( \frac{1}{r^2} \), which converges, and that is where escape velocity comes from. The Laplace transform and the gamma function are both defined by improper integrals.

Evaluate improper integrals online

Type inf or -inf as a bound. The solver maps the infinite interval onto a finite one and integrates numerically, and tells you if the integral diverges.

The same engine powers the definite integral calculator.

Practice problems

Try these before checking the answers.

1. \( \int_1^{\infty}\frac{dx}{x^4} \)
2. \( \int_0^{\infty}e^{-2x}\,dx \)
3. \( \int_0^{8}\frac{dx}{\sqrt[3]{x}} \)
4. \( \int_1^{\infty}x^{-3/2}\,dx \)
5. \( \int_0^{\infty}\frac{dx}{x^2 + 4} \)
6. \( \int_0^1\frac{dx}{x} \)

Answers: (1) \( \frac13 \); (2) \( \frac12 \); (3) \( 6 \); (4) \( 2 \); (5) \( \frac\pi4 \); (6) diverges.

For (5), the antiderivative is \( \frac12\arctan\frac x2 \), and \( \arctan \) approaches \( \frac\pi2 \) as its input grows. For (6), this is a p-integral on \( [0, 1] \) with \( p = 1 \), which is not less than 1.

FAQ

If the function goes to zero, does the integral converge?

Not necessarily. \( \frac1x \to 0 \) but \( \int_1^\infty\frac{dx}{x} \) diverges. The function must shrink fast enough.

How are improper integrals related to series?

The integral test says \( \sum f(n) \) and \( \int_1^\infty f(x)\,dx \) converge or diverge together (for positive decreasing \( f \)). See the ratio test for another series tool.

Is the integral of x from minus infinity to infinity zero?

No. By definition you must split it and both halves must converge, but \( \int_0^\infty x\,dx \) diverges. The symmetric limit \( \lim_{t\to\infty}\int_{-t}^{t}x\,dx = 0 \) is a different object, called the Cauchy principal value.

What does “diverges” mean exactly?

It means the defining limit does not exist as a finite number. It may go to \( \infty \) or \( -\infty \), or it may oscillate, as \( \int_0^t\sin x\,dx = 1 - \cos t \) does, never settling on a value.

Can a calculator decide convergence for me?

A numerical tool is a good check, but it cannot replace the reasoning. Slowly divergent integrals such as \( \int_1^\infty\frac{dx}{x} \) grow so gently that a finite computation can look like it is settling down. Use the p-test or a comparison to decide convergence, then use a calculator to confirm the value.

Further reading

• Improper Integrals (Paul’s Online Math Notes) — both types of improper integral with many worked limits.
• Comparison Test for Improper Integrals (Paul’s Online Math Notes) — eight examples of choosing a good comparison function.