Improper Integrals & Rigorous Convergence Tests

1. Rigorous Definition of Type I Improper Integrals

An integral is called a Type I improper integral if the interval of integration is unbounded. It is defined as the analytical limit of proper Riemann integrals over truncated compact intervals:

• Upper Infinite Limit: For $f$ continuous on $[a, \infty)$: $$\int_a^\infty f(x) \, dx = \lim_{M \to \infty} \int_a^M f(x) \, dx$$
• Lower Infinite Limit: For $f$ continuous on $(-\infty, b]$: $$\int_{-\infty}^b f(x) \, dx = \lim_{N \to -\infty} \int_N^b f(x) \, dx$$
• Doubly Infinite Domain: For $f$ continuous on $(-\infty, \infty)$ and any partitioning point $c \in \mathbb{R}$: $$\int_{-\infty}^\infty f(x) \, dx = \int_{-\infty}^c f(x) \, dx + \int_c^\infty f(x) \, dx = \lim_{N \to -\infty} \int_N^c f(x) \, dx + \lim_{M \to \infty} \int_c^M f(x) \, dx$$ The doubly infinite integral converges if and only if both one-sided limits converge independently.

2. The Foundational $p$-Integral Convergence Test

Proof: For $p \ne 1$:

As $M \to \infty$:

• If $p > 1$, then $1 - p < 0 \implies \lim_{M \to \infty} M^{1-p} = 0$, so $\int_1^\infty x^{-p} \, dx = \frac{1}{p - 1}$ (Convergent).
• If $p < 1$, then $1 - p > 0 \implies \lim_{M \to \infty} M^{1-p} = \infty$ (Divergent).
• If $p = 1$, $\int_1^M \frac{1}{x} \, dx = \ln(M) \to \infty$ as $M \to \infty$ (Divergent). $\blacksquare$

1. Unbounded Integrands on Bounded Intervals

An integral $\int_a^b f(x) \, dx$ is called a Type II improper integral if the integrand $f(x)$ has an infinite singularity ($\lim |f(x)| = \infty$) at one or more points in $[a, b]$:

• Singularity at Upper Endpoint $b$: $$\int_a^b f(x) \, dx = \lim_{\varepsilon \to 0^+} \int_a^{b - \varepsilon} f(x) \, dx$$
• Singularity at Lower Endpoint $a$: $$\int_a^b f(x) \, dx = \lim_{\varepsilon \to 0^+} \int_{a + \varepsilon}^b f(x) \, dx$$
• Interior Singularity at $c \in (a, b)$: $$\int_a^b f(x) \, dx = \lim_{\varepsilon_1 \to 0^+} \int_a^{c - \varepsilon_1} f(x) \, dx + \lim_{\varepsilon_2 \to 0^+} \int_{c + \varepsilon_2}^b f(x) \, dx$$ The integral converges if and only if both limits exist independently.

2. The Bounded Interval $p$-Singularity Test

Notice the critical duality: for unbounded domains $[1, \infty)$, convergence requires $p > 1$; for singular origins $[0, 1]$, convergence requires $p < 1$. The reciprocal function $\int_0^\infty \frac{1}{x^p} dx$ diverges for all $p \in \mathbb{R}$.

1. Definition of the Cauchy Principal Value

When an improper integral diverges in the standard independent-limit sense, symmetric cancellation can produce a finite Cauchy Principal Value (P.V.):

• For Type I Integrals: $$\operatorname{P.V.} \int_{-\infty}^\infty f(x) \, dx = \lim_{R \to \infty} \int_{-R}^R f(x) \, dx$$
• For Type II Integrals with Interior Singularity at $c$: $$\operatorname{P.V.} \int_a^b f(x) \, dx = \lim_{\varepsilon \to 0^+} \left( \int_a^{c - \varepsilon} f(x) \, dx + \int_{c + \varepsilon}^b f(x) \, dx \right)$$

2. Strict Distinction from Standard Convergence

Consider $f(x) = x$ over $(-\infty, \infty)$:

However, the integral $\int_{-\infty}^\infty x \, dx$ diverges under standard definition because $\lim_{M \to \infty} \int_0^M x dx = \infty$. If an integral converges normally, its Cauchy Principal Value exists and is identical; the converse is strictly false.

1. The Direct Comparison Test

Let $f, g$ be continuous functions on $[a, \infty)$ satisfying $0 \le f(x) \le g(x)$ for all $x \ge a$:

1. Convergence Inheritance: If $\int_a^\infty g(x) \, dx$ converges, then $\int_a^\infty f(x) \, dx$ converges, and $\int_a^\infty f(x)dx \le \int_a^\infty g(x)dx$.
2. Divergence Inheritance: If $\int_a^\infty f(x) \, dx$ diverges to $\infty$, then $\int_a^\infty g(x) \, dx$ diverges to $\infty$.

2. The Limit Comparison Test

Let $f(x) \ge 0$ and $g(x) > 0$ for all $x \ge a$. Suppose the asymptotic limit exists:

• If $0 < L < \infty$, then $\int_a^\infty f(x) \, dx$ and $\int_a^\infty g(x) \, dx$ either both converge or both diverge.
• If $L = 0$ and $\int_a^\infty g(x) \, dx$ converges, then $\int_a^\infty f(x) \, dx$ converges.
• If $L = \infty$ and $\int_a^\infty g(x) \, dx$ diverges, then $\int_a^\infty f(x) \, dx$ diverges.

3. Absolute vs Conditional Convergence

If $\int_a^\infty f(x) \, dx$ converges but $\int_a^\infty |f(x)| \, dx = \infty$, the integral is conditionally convergent.

1. Dirichlet's Test for Improper Integrals

1. The antiderivative of $f$ is uniformly bounded: $\exists M > 0$ such that $\left| \int_a^x f(t) \, dt \right| \le M$ for all $x \ge a$.
2. $g(x)$ is monotonic and decays to zero: $g'(x) \le 0$ (or $\ge 0$) and $\lim_{x \to \infty} g(x) = 0$.

Proof: Integrating by parts over $[a, B]$: $\int_a^B f(x)g(x) dx = F(B)g(B) - F(a)g(a) - \int_a^B F(x)g'(x)dx$. Since $|F(B)g(B)| \le M|g(B)| \to 0$ and $\int_a^\infty |F(x)g'(x)| dx \le M \int_a^\infty |g'(x)|dx = M g(a) < \infty$, the limit exists. $\blacksquare$

2. The Dirichlet Sinc Integral: $\int_0^\infty \frac{\sin x}{x} \, dx = \frac{\pi}{2}$

By Dirichlet's test with $f(x) = \sin(x)$ (bounded antiderivative $-\cos x \in [-1, 1]$) and $g(x) = \frac{1}{x} \to 0$ monotonically, $\int_1^\infty \frac{\sin x}{x} dx$ converges. Near $x = 0$, $\lim_{x \to 0} \frac{\sin x}{x} = 1$, so the integral is finite on $[0, 1]$. Hence $\int_0^\infty \frac{\sin x}{x} dx$ converges (conditionally, as $\int_0^\infty \frac{|\sin x|}{x} dx = \infty$).