Tính giá trị tích phân bất thường

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.

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

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.

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.