Evaluating Improper Integrals
Introduced| English | Meaning |
|---|---|
| improper integral/ɪmˈprɒpə ˈɪntɪɡrəl/ | An integral defined by a limit because an integration interval is unbounded or the integrand is unbounded at a boundary or interior point. |
| infinite limit/ˈɪnfɪnət ˈlɪmɪt/ | Limit behaviour in which a function grows without bound in the positive or negative direction as its input approaches a specified value. |
| unbounded integrand/ʌnˈbaʊndɪd ˈɪntɪɡrænd/ | unbounded integrand |
Integrating over the infinite (or the unbounded)
- A normal definite integral has finite limits and a finite integrand. An improper integral 反常积分 breaks one of those.
- Two ways it goes improper: an infinite limit 无穷极限 of integration (like $\int_1^\infty$), or an unbounded integrand 无界被积函数 (a vertical asymptote inside the interval).
- We can't just "plug in $\infty$" or evaluate at a blow-up, so we use a limit.
- The integral converges if that limit is a finite number, and diverges otherwise.
An integral is improper when...
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Infinite limit or unbounded integrand → improper.
You can evaluate an improper integral by directly substituting $\infty$ into the antiderivative.
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You must take a limit instead.
Infinite limits → a limit
- Replace the infinite limit with a variable, integrate, then let it run to infinity:
-
$$\int_a^\infty f(x)\,dx=\lim_{b\to\infty}\int_a^b f(x)\,dx$$
- Compute the ordinary integral up to $b$, then take the limit as $b\to\infty$.
- If the limit is finite, the improper integral converges to it.
The area under 1/x² out to infinity is finite; under 1/x it is not
The infinite tail under 1/x²
y = 1/x²
The area under $1/x^2$ from $1$ to $\infty$ is finite ($=1$) even though the region never ends — a convergent improper integral.
$\displaystyle\int_a^\infty f\,dx$ is defined as...
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It is the limit of the proper integral.
Unbounded integrand → a limit at the trouble point
- If $f$ blows up at an endpoint $c$, approach it with a limit instead of evaluating there:
- $\displaystyle\int_a^c f\,dx=\lim_{t\to c^-}\int_a^t f\,dx$ (and similarly from the other side).
- $\int_0^1 \tfrac{1}{\sqrt x}\,dx$ is improper at $0$ — take $\lim_{t\to0^+}\int_t^1$.
- Same principle: sneak up on the singularity with a limit.
Evaluate $\displaystyle\int_1^\infty \dfrac{1}{x^2}\,dx$.
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$\lim_{b\to\infty}(1-\tfrac1b)=1$.
Converge or diverge
- Converges: the limit is a finite number — that number is the value.
- Diverges: the limit is infinite or does not exist — the integral has no finite value.
- A famous pair: $\int_1^\infty\tfrac{1}{x}\,dx$ diverges, but $\int_1^\infty\tfrac{1}{x^2}\,dx$ converges (to $1$).
- The difference — same-looking integrands, opposite fates — is the whole point.
What happens to $\displaystyle\int_1^\infty \dfrac{1}{x}\,dx$?
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$\lim_{b\to\infty}\ln b=\infty$ → diverges.
An improper integral ____ if its defining limit is a finite number.
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Finite limit → converges.
You cannot treat an improper integral like a normal one — plugging in $\infty$ or a blow-up point is meaningless. Always rewrite it as a limit and evaluate that limit. And spot interior singularities: if the integrand blows up inside $[a,b]$, split the integral at that point and handle each piece separately.
Evaluate $\displaystyle\int_1^\infty \dfrac{1}{x^2}\,dx$.
- Rewrite: $\displaystyle\lim_{b\to\infty}\int_1^b x^{-2}\,dx=\lim_{b\to\infty}\Big[-\tfrac1x\Big]_1^b$.
- $=\displaystyle\lim_{b\to\infty}\Big(-\tfrac1b+1\Big)=0+1=1$.
- The limit is finite, so the integral converges to $1$.
An improper integral has an infinite limit or an unbounded integrand. Rewrite it as a limit of a proper integral: $\int_a^\infty f=\lim_{b\to\infty}\int_a^b f$ (or a limit approaching a singularity). It converges if the limit is finite, and diverges otherwise.