An integral is improper when you’re trying to take a definite integral in a situation where the usual Riemann integral definition breaks, so you must define it using limits instead.

Quick Scoop: What makes an integral improper?

There are two main ways this happens:

  1. Infinite limits of integration (unbounded interval)

This is when at least one of the bounds is +∞+\infty +∞ or −∞-\infty −∞.
Examples:

 * ∫1∞1x2 dx\displaystyle \int_{1}^{\infty}\frac{1}{x^2}\,dx∫1∞​x21​dx (upper limit is infinite)

 * ∫−∞∞e−x2 dx\displaystyle \int_{-\infty}^{\infty}e^{-x^2}\,dx∫−∞∞​e−x2dx (both limits infinite)

Here, you can’t just “plug infinity in” to the Fundamental Theorem of Calculus, so you redefine the integral as a limit, like

∫1∞1x2 dx=lim⁡b→∞∫1b1x2 dx.\int_{1}^{\infty}\frac{1}{x^2},dx=\lim_{b\to\infty}\int_{1}^{b}\frac{1}{x^2},dx.∫1∞​x21​dx=b→∞lim​∫1b​x21​dx.

If that limit exists and is finite, the improper integral converges ; if not, it diverges.

  1. Unbounded integrand or discontinuity on the interval

This is when the function blows up (goes to ±∞\pm\infty ±∞) or isn’t defined somewhere in the interval, including at the endpoints.
Examples:

 * ∫011x dx\displaystyle \int_{0}^{1}\frac{1}{\sqrt{x}}\,dx∫01​x​1​dx: integrand 1x\frac{1}{\sqrt{x}}x​1​ goes to ∞\infty ∞ as x→0+x\to 0^+x→0+.

 * ∫−111x dx\displaystyle \int_{-1}^{1}\frac{1}{x}\,dx∫−11​x1​dx: integrand has a vertical asymptote at x=0x=0x=0 inside the interval.

Again, you convert to a limit around the problematic point. For instance,

∫011x dx=lim⁡a→0+∫a11x dx.\int_{0}^{1}\frac{1}{\sqrt{x}},dx=\lim_{a\to 0^+}\int_{a}^{1}\frac{1}{\sqrt{x}},dx.∫01​x​1​dx=a→0+lim​∫a1​x​1​dx.

Mathematicians sometimes call:

  • “Infinite interval” cases Type I improper integrals.
  • “Blowing up / discontinuous integrand” cases Type II improper integrals.

One-sentence memory trick

An integral is improper if the interval or the integrand misbehaves : the interval is infinite, or the function becomes infinite or undefined somewhere in the interval, so you must interpret the integral via limits.

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