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Improper Integrals



What Makes an Integral Improper?
Infinite Limits of Integration
Discontinuous Integrands
Convergence vs Divergence
The p-Test
Comparison Test
Limit Comparison Test



Integrating Beyond Bounds


Standard definite integrals require finite intervals and bounded integrands. Improper integrals remove these restrictions, extending integration to infinite intervals and functions with vertical asymptotes.

The key idea: replace the problematic bound or point with a variable, compute the resulting proper integral, then take a limit. If the limit exists and is finite, the improper integral converges to that value. If not, it diverges.

Some infinite regions have finite area. The region under 1/x21/x^2 from 11 to ∞\infty has area exactly 11. Other infinite regions have infinite area—the region under 1/x1/x from 11 to ∞\infty diverges. The distinction matters throughout mathematics and physics.



What Makes an Integral Improper?


An integral is improper if it involves:

Infinite limits of integration: The interval extends to ∞\infty or −∞-\infty

∫1∞1x2 dx\int_1^{\infty} \frac{1}{x^2}\, dx


Unbounded integrand: The function has a vertical asymptote within or at the boundary of the interval

∫011x dx\int_0^1 \frac{1}{\sqrt{x}}\, dx


Both conditions can occur simultaneously. The integral

∫0∞1x(1+x) dx\int_0^{\infty} \frac{1}{\sqrt{x}(1 + x)}\, dx


has an infinite upper limit and an unbounded integrand at x=0x = 0.

Infinite Limits of Integration


Replace the infinite limit with a finite variable and take a limit.

Type 1: Upper limit infinite

∫a∞f(x) dx=lim⁡b→∞∫abf(x) dx\int_a^{\infty} f(x)\, dx = \lim_{b \to \infty} \int_a^b f(x)\, dx


Type 2: Lower limit infinite

∫−∞bf(x) dx=lim⁡a→−∞∫abf(x) dx\int_{-\infty}^b f(x)\, dx = \lim_{a \to -\infty} \int_a^b f(x)\, dx


Type 3: Both limits infinite

∫−∞∞f(x) dx=∫−∞cf(x) dx+∫c∞f(x) dx\int_{-\infty}^{\infty} f(x)\, dx = \int_{-\infty}^c f(x)\, dx + \int_c^{\infty} f(x)\, dx


Both integrals must converge independently. The choice of cc is arbitrary—any finite value works.

Discontinuous Integrands


When ff has a vertical asymptote at cc within [a,b][a, b], split the integral and use limits.

Asymptote at left endpoint:

∫abf(x) dx=lim⁡t→a+∫tbf(x) dx\int_a^b f(x)\, dx = \lim_{t \to a^+} \int_t^b f(x)\, dx


Asymptote at right endpoint:

∫abf(x) dx=lim⁡t→b−∫atf(x) dx\int_a^b f(x)\, dx = \lim_{t \to b^-} \int_a^t f(x)\, dx


Asymptote at interior point $c$:

∫abf(x) dx=lim⁡t→c−∫atf(x) dx+lim⁡s→c+∫sbf(x) dx\int_a^b f(x)\, dx = \lim_{t \to c^-} \int_a^t f(x)\, dx + \lim_{s \to c^+} \int_s^b f(x)\, dx


Both limits must exist independently.

Convergence vs Divergence


An improper integral converges if the defining limit exists and is finite. It diverges if the limit is infinite or fails to exist.

Convergent example:

∫1∞1x2 dx=lim⁡b→∞[−1x]1b=lim⁡b→∞(−1b+1)=1\int_1^{\infty} \frac{1}{x^2}\, dx = \lim_{b \to \infty} \left[-\frac{1}{x}\right]_1^b = \lim_{b \to \infty} \left(-\frac{1}{b} + 1\right) = 1


Divergent example:

∫1∞1x dx=lim⁡b→∞[ln⁡x]1b=lim⁡b→∞ln⁡b=∞\int_1^{\infty} \frac{1}{x}\, dx = \lim_{b \to \infty} [\ln x]_1^b = \lim_{b \to \infty} \ln b = \infty


Convergence depends on how fast the integrand decays. The 1/x21/x^2 decays fast enough; 1/x1/x does not.

The p-Test


The integrals of 1/xp1/x^p serve as benchmarks.

At infinity:

∫1∞1xp dx{convergesp>1divergesp≤1\int_1^{\infty} \frac{1}{x^p}\, dx \quad \begin{cases} \text{converges} & p > 1 \\ \text{diverges} & p \leq 1 \end{cases}


At zero:

∫011xp dx{convergesp<1divergesp≥1\int_0^1 \frac{1}{x^p}\, dx \quad \begin{cases} \text{converges} & p < 1 \\ \text{diverges} & p \geq 1 \end{cases}


The boundary case p=1p = 1 always diverges—∫1/x dx=ln⁡∣x∣\int 1/x\, dx = \ln|x|, which is unbounded both as x→∞x \to \infty and as x→0+x \to 0^+.

Comparison Test


Compare an unknown integral to one with known behavior.

Direct comparison: For f(x)≥0f(x) \geq 0 and g(x)≥0g(x) \geq 0:

If f(x)≤g(x)f(x) \leq g(x) and ∫g\int g converges, then ∫f\int f converges.

If f(x)≥g(x)f(x) \geq g(x) and ∫g\int g diverges, then ∫f\int f diverges.

Example: Does ∫1∞1x2+1 dx\int_1^{\infty} \frac{1}{x^2 + 1}\, dx converge?

Since 1x2+1<1x2\dfrac{1}{x^2 + 1} < \dfrac{1}{x^2} and ∫1∞1x2 dx\int_1^{\infty} \dfrac{1}{x^2}\, dx converges, the given integral converges by comparison.

Limit Comparison Test


When direct comparison is awkward, use limits.

For f(x)>0f(x) > 0 and g(x)>0g(x) > 0, if:

lim⁡x→∞f(x)g(x)=Lwhere 0<L<∞\lim_{x \to \infty} \frac{f(x)}{g(x)} = L \quad \text{where } 0 < L < \infty


then ∫f\int f and ∫g\int g both converge or both diverge.

Example: Does ∫1∞xx3+5 dx\int_1^{\infty} \frac{x}{x^3 + 5}\, dx converge?

Compare to g(x)=1/x2g(x) = 1/x^2:

lim⁡x→∞x/(x3+5)1/x2=lim⁡x→∞x3x3+5=1\lim_{x \to \infty} \frac{x/(x^3 + 5)}{1/x^2} = \lim_{x \to \infty} \frac{x^3}{x^3 + 5} = 1


Since ∫1∞1/x2 dx\int_1^{\infty} 1/x^2\, dx converges, so does the given integral.