MT2176 Further Calculus: Improper Integrals

There are two main kinds of improper integrals:

Type 1 (First Kind): These have an infinite interval of integration. For example, \( \int_{a}^{\infty} f(x) \,dx \) or \( \int_{-\infty}^{b} f(x) \,dx \).

Type 2 (Second Kind): These have an integrand with a finite number of infinite discontinuities within the interval of integration. For example, \( \int_{a}^{b} f(x) \,dx \) where \(f(x)\) has an infinite discontinuity at \(x=a\), \(x=b\), or some \(c \in (a,b)\).

The improper integral \( \int_{a}^{\infty} f(x) \,dx \) is defined as the limit of a proper integral: \[ \int_{a}^{\infty} f(x) \,dx = \lim_{t \to \infty} \int_{a}^{t} f(x) \,dx \] If the limit exists and is finite, the integral is said to be convergent. Otherwise, it is divergent.

If \(f(x)\) is continuous on \((a, b]\) and has an infinite discontinuity at \(x=a\), the improper integral is defined as: \[ \int_{a}^{b} f(x) \,dx = \lim_{t \to a^+} \int_{t}^{b} f(x) \,dx \] If the limit exists and is finite, the integral is convergent. Otherwise, it is divergent.

An improper integral of the third kind is one that involves a combination of Type 1 and Type 2 characteristics. For example, an integral over an infinite interval where the integrand also has an infinite discontinuity.

Example: \( \int_{0}^{\infty} \frac{e^{-x}}{\sqrt{x}} \,dx \). This has an infinite interval and a discontinuity at \(x=0\).

To evaluate these, we split the integral into parts, for example: \( \int_{0}^{1} \frac{e^{-x}}{\sqrt{x}} \,dx + \int_{1}^{\infty} \frac{e^{-x}}{\sqrt{x}} \,dx \). The original integral converges only if both parts converge.

The integral \( \int_{1}^{\infty} \frac{1}{x^p} \,dx \) is a fundamental test case.

If \(p > 1\), the integral converges.
If \(p \le 1\), the integral diverges.

Proof: If \(p \neq 1\), \( \int_{1}^{t} x^{-p} \,dx = [\frac{x^{-p+1}}{-p+1}]_{1}^{t} = \frac{1}{1-p}(t^{1-p} - 1) \). As \(t \to \infty\), \(t^{1-p} \to 0\) if \(1-p < 0\) (i.e., \(p>1\)), and the limit is \(\frac{1}{p-1}\). If \(p < 1\), \(1-p > 0\), so \(t^{1-p} \to \infty\) and the integral diverges. If \(p=1\), \( \int_{1}^{t} \frac{1}{x} \,dx = [\ln|x|]_{1}^{t} = \ln t \), which diverges as \(t \to \infty\).

Let \(f(x)\) and \(g(x)\) be continuous functions such that \(0 \le f(x) \le g(x)\) for all \(x \ge a\).

1. If \( \int_{a}^{\infty} g(x) \,dx \) converges, then \( \int_{a}^{\infty} f(x) \,dx \) also converges.

2. If \( \int_{a}^{\infty} f(x) \,dx \) diverges, then \( \int_{a}^{\infty} g(x) \,dx \) also diverges.

Let \(f(x)\) and \(g(x)\) be positive continuous functions for \(x \ge a\). Suppose that \[ L = \lim_{x \to \infty} \frac{f(x)}{g(x)} \] If \(L\) is finite and \(L > 0\), 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.

We can use the Direct Comparison Test. For \(x \ge 1\), we have \(x^2 + 1 > x^2\), so \(0 < \frac{1}{x^2 + 1} < \frac{1}{x^2}\).

We know that \( \int_{1}^{\infty} \frac{1}{x^2} \,dx \) is a p-integral with \(p=2 > 1\), so it converges.

By the Direct Comparison Test, since \( \int_{1}^{\infty} \frac{1}{x^2} \,dx \) converges, the integral \( \int_{1}^{\infty} \frac{1}{x^2 + 1} \,dx \) also converges.

Alternatively, we can evaluate it directly: \[ \lim_{t \to \infty} \int_{1}^{t} \frac{1}{x^2+1} \,dx = \lim_{t \to \infty} [\arctan(x)]_{1}^{t} = \lim_{t \to \infty} (\arctan(t) - \arctan(1)) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \] Since the limit is finite, the integral converges.

For \(x > e\), \(\ln x > 1\), so \(\frac{\ln x}{x} > \frac{1}{x}\).

We know that \( \int_{1}^{\infty} \frac{1}{x} \,dx \) diverges (p-integral with \(p=1\)).

By the Direct Comparison Test, since \( \int_{e}^{\infty} \frac{1}{x} \,dx \) diverges, \( \int_{e}^{\infty} \frac{\ln x}{x} \,dx \) also diverges. Therefore, \( \int_{1}^{\infty} \frac{\ln x}{x} \,dx \) diverges.

Alternatively, we can evaluate it directly using substitution \(u = \ln x\), \(du = \frac{1}{x}dx\). \[ \int_{1}^{\infty} \frac{\ln x}{x} \,dx = \int_{0}^{\infty} u \,du = [\frac{u^2}{2}]_{0}^{\infty} \] This limit is infinite, so the integral diverges.

An improper integral \( \int_{a}^{\infty} f(x) \,dx \) is said to be absolutely convergent if the integral of its absolute value, \( \int_{a}^{\infty} |f(x)| \,dx \), converges.

An important theorem states that if an integral is absolutely convergent, then it is convergent.

An improper integral \( \int_{a}^{\infty} f(x) \,dx \) is said to be conditionally convergent if it converges, but the integral of its absolute value, \( \int_{a}^{\infty} |f(x)| \,dx \), diverges.

A classic example is \( \int_{0}^{\infty} \frac{\sin x}{x} \,dx \).

This is a Type 2 improper integral with a discontinuity at \(x=0\). \[ \int_{0}^{1} \frac{1}{\sqrt{x}} \,dx = \lim_{t \to 0^+} \int_{t}^{1} x^{-1/2} \,dx = \lim_{t \to 0^+} [2x^{1/2}]_{t}^{1} = \lim_{t \to 0^+} (2 - 2\sqrt{t}) = 2 \] Since the limit is finite, the integral converges.

This is a Type 2 improper integral with a discontinuity at \(x=0\). \[ \int_{0}^{1} \frac{1}{x} \,dx = \lim_{t \to 0^+} \int_{t}^{1} \frac{1}{x} \,dx = \lim_{t \to 0^+} [\ln|x|]_{t}^{1} = \lim_{t \to 0^+} (\ln(1) - \ln(t)) = \lim_{t \to 0^+} (-\ln(t)) = \infty \] Since the limit is infinite, the integral diverges.

This is the Gaussian integral. We can split it into two parts: \( \int_{0}^{1} e^{-x^2} \,dx + \int_{1}^{\infty} e^{-x^2} \,dx \).

The first part is a proper integral and is finite.

For the second part, \( \int_{1}^{\infty} e^{-x^2} \,dx \), we can use the comparison test. For \(x \ge 1\), \(x^2 \ge x\), so \(-x^2 \le -x\), and \(e^{-x^2} \le e^{-x}\).

We know \( \int_{1}^{\infty} e^{-x} \,dx = [ -e^{-x} ]_{1}^{\infty} = 0 - (-e^{-1}) = 1/e \), which converges.

By the Direct Comparison Test, \( \int_{1}^{\infty} e^{-x^2} \,dx \) converges. Since both parts converge, the original integral converges.

For an integral \( \int_{-\infty}^{\infty} f(x) \,dx \), the Cauchy Principal Value is defined as: \[ P.V. \int_{-\infty}^{\infty} f(x) \,dx = \lim_{R \to \infty} \int_{-R}^{R} f(x) \,dx \] This is a weaker form of convergence. If the improper integral converges in the standard sense, its value is the same as the Cauchy Principal Value. However, the Cauchy Principal Value may exist even when the standard integral diverges. For example, for \(f(x)=x\).

We use the integral test. Let \(u = \ln x\), so \(du = \frac{1}{x} dx\). \[ \int_{\ln 2}^{\infty} \frac{1}{u} \,du \] This is a p-integral with \(p=1\), which diverges. Therefore, the original integral diverges.

We can use the Limit Comparison Test with \(g(x) = e^{-x/2}\). \[ \lim_{x \to \infty} \frac{x/e^x}{e^{-x/2}} = \lim_{x \to \infty} \frac{x}{e^{x/2}} = 0 \] (using L'Hopital's Rule). Since \(\int_{0}^{\infty} e^{-x/2} dx\) converges, the original integral converges.

Use the Limit Comparison Test with \(g(x) = \frac{x}{\sqrt{x^4}} = \frac{1}{x}\). \[ \lim_{x \to \infty} \frac{(x+1)/\sqrt{x^4+1}}{1/x} = \lim_{x \to \infty} \frac{x(x+1)}{\sqrt{x^4+1}} = 1 \] Since \(\int_{1}^{\infty} \frac{1}{x} dx\) diverges, the original integral diverges.

Using integration by parts (\(u=x, dv=e^x dx\)), \(\int xe^x dx = xe^x - e^x\). \[ \lim_{t \to -\infty} [xe^x - e^x]_{t}^{0} = (0 - 1) - \lim_{t \to -\infty} (te^t - e^t) = -1 - (0 - 0) = -1 \] The integral converges to -1.

This is a Type 2 integral with a discontinuity at \(x=\pi/2\). \[ \lim_{t \to (\pi/2)^-} \int_{0}^{t} \tan x \,dx = \lim_{t \to (\pi/2)^-} [-\ln|\cos x|]_{0}^{t} = \lim_{t \to (\pi/2)^-} (-\ln|\cos t| + \ln|\cos 0|) = \infty \] The integral diverges.

This is a Type 2 integral with a discontinuity at \(x=1\). \[ \lim_{t \to 1^-} \int_{0}^{t} \frac{dx}{\sqrt{1-x^2}} = \lim_{t \to 1^-} [\arcsin x]_{0}^{t} = \lim_{t \to 1^-} (\arcsin t - \arcsin 0) = \frac{\pi}{2} \] The integral converges to \(\pi/2\).

We test for absolute convergence. \[ \int_{1}^{\infty} |\frac{\cos x}{x^2}| \,dx = \int_{1}^{\infty} \frac{|\cos x|}{x^2} \,dx \] Since \(|\cos x| \le 1\), we have \(\frac{|\cos x|}{x^2} \le \frac{1}{x^2}\). Since \(\int_{1}^{\infty} \frac{1}{x^2} dx\) converges, the integral of the absolute value converges by the Direct Comparison Test. Therefore, the original integral is absolutely convergent (and thus convergent).

For \(x > 0\), \(e^x + 1 > e^x\), so \(\frac{1}{e^x+1} < \frac{1}{e^x} = e^{-x}\). We know \(\int_{0}^{\infty} e^{-x} dx\) converges to 1. By the Direct Comparison Test, the original integral converges.

This is a Type 2 integral with a discontinuity at \(x=0\). Let \(u = \ln x\), \(du = \frac{1}{x} dx\). \[ \int_{-\infty}^{0} u \,du = [\frac{u^2}{2}]_{-\infty}^{0} \] This limit is infinite, so the integral diverges.

Use the Limit Comparison Test with \(g(x) = \frac{1}{\sqrt{x}}\). \[ \lim_{x \to \infty} \frac{1/(\sqrt{x}+1)}{1/\sqrt{x}} = \lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x}+1} = 1 \] Since \(\int_{1}^{\infty} \frac{1}{\sqrt{x}} dx\) diverges (p-integral with \(p=1/2\)), the original integral diverges.

Let \(u = \arctan x\), \(du = \frac{1}{x^2+1} dx\). \[ \int_{0}^{\pi/2} u \,du = [\frac{u^2}{2}]_{0}^{\pi/2} = \frac{(\pi/2)^2}{2} = \frac{\pi^2}{8} \] The integral converges.

This is a Type 2 p-integral test. If \(p<1\), the integral converges. If \(p \ge 1\), the integral diverges.

Use the Limit Comparison Test with \(g(x) = \frac{1}{x^2}\). \[ \lim_{x \to \infty} \frac{1/(x^2-4)}{1/x^2} = \lim_{x \to \infty} \frac{x^2}{x^2-4} = 1 \] Since \(\int_{3}^{\infty} \frac{1}{x^2} dx\) converges (p-integral with \(p=2\)), the original integral converges.

Use the Limit Comparison Test with \(g(x) = e^{-x/2}\). \[ \lim_{x \to \infty} \frac{x^2 e^{-x}}{e^{-x/2}} = \lim_{x \to \infty} \frac{x^2}{e^{x/2}} = 0 \] (using L'Hopital's Rule twice). Since \(\int_{0}^{\infty} e^{-x/2} dx\) converges, the original integral converges.

Let \(u = e^x\), \(du = e^x dx\). \[ \int_{e}^{\infty} \frac{du}{1+u^2} = [\arctan u]_{e}^{\infty} = \frac{\pi}{2} - \arctan(e) \] The integral converges.

Use the Limit Comparison Test with \(g(x) = \frac{1}{x^{4/3}}\). \[ \lim_{x \to \infty} \frac{1/\sqrt[3]{x^4+1}}{1/x^{4/3}} = \lim_{x \to \infty} \frac{x^{4/3}}{(x^4+1)^{1/3}} = 1 \] Since \(\int_{1}^{\infty} \frac{1}{x^{4/3}} dx\) converges (p-integral with \(p=4/3>1\)), the original integral converges. (The integral from 0 to 1 is proper).

Use the Limit Comparison Test with \(g(x) = \frac{1}{x}\). \[ \lim_{x \to \infty} \frac{(x^2-1)/(x^3+1)}{1/x} = \lim_{x \to \infty} \frac{x(x^2-1)}{x^3+1} = 1 \] Since \(\int_{1}^{\infty} \frac{1}{x} dx\) diverges, the original integral diverges.

This is a Type 2 integral. For \(x \in (0, 1]\), \(e^x \le e\). So, \(\frac{e^x}{\sqrt{x}} \le \frac{e}{\sqrt{x}}\). Since \(\int_{0}^{1} \frac{e}{\sqrt{x}} dx = e \int_{0}^{1} \frac{1}{\sqrt{x}} dx\) converges (p-integral with \(p=1/2<1\)), the original integral converges by the Direct Comparison Test.

This is an improper integral of both Type 1 and Type 2 (at x=1). Split into \(\int_{1}^{2} + \int_{2}^{\infty}\). For \(\int_{2}^{\infty}\), compare with \(1/x^2\). Converges. For \(\int_{1}^{2}\), let \(x = \sec \theta\). The integral becomes \(\int_{0}^{\pi/3} d\theta = \pi/3\). Converges. Since both parts converge, the original integral converges.

Let \(u = \sqrt{x}\), \(du = \frac{1}{2\sqrt{x}} dx\). \[ \int_{0}^{\infty} 2e^{-u} \,du = [-2e^{-u}]_{0}^{\infty} = 0 - (-2) = 2 \] The integral converges.

Use the Limit Comparison Test with \(g(x) = \frac{1}{x^2}\). \[ \lim_{x \to \infty} \frac{x/(x^3+1)}{1/x^2} = \lim_{x \to \infty} \frac{x^3}{x^3+1} = 1 \] Since \(\int_{1}^{\infty} \frac{1}{x^2} dx\) converges, the original integral converges. (The integral from 0 to 1 is proper).

Split into \(\int_{0}^{1} + \int_{1}^{\infty}\). As \(x \to 0\), \(\frac{1-e^{-x}}{x} \to 1\) (by L'Hopital's Rule). So the first part is proper. For \(x \ge 1\), \(1-e^{-x} > 1/2\). So \(\frac{1-e^{-x}}{x} > \frac{1}{2x}\). Since \(\int_{1}^{\infty} \frac{1}{2x} dx\) diverges, the original integral diverges.

\( \int_{1}^{\infty} \frac{2\ln x}{x^2} \,dx \). Use Limit Comparison with \(1/x^{3/2}\). \[ \lim_{x \to \infty} \frac{2\ln x / x^2}{1/x^{3/2}} = \lim_{x \to \infty} \frac{2\ln x}{\sqrt{x}} = 0 \] Since \(\int_{1}^{\infty} 1/x^{3/2} dx\) converges, the original integral converges.

This is a Type 2 integral. Use integration by parts. \[ \lim_{t \to 0^+} [\frac{x^2}{2} \ln x - \frac{x^2}{4}]_{t}^{1} = (0 - 1/4) - \lim_{t \to 0^+} (\frac{t^2}{2} \ln t - \frac{t^2}{4}) = -1/4 - 0 = -1/4 \] The integral converges.

Split into \(\int_{0}^{1} + \int_{1}^{\infty}\). For \(\int_{0}^{1}\), compare with \(1/\sqrt{x}\). Converges. For \(\int_{1}^{\infty}\), compare with \(1/x^2\). Converges. Since both parts converge, the original integral converges.

Use Limit Comparison with \(g(x) = 1/x\). \[ \lim_{x \to \infty} \frac{\sin(1/x)}{1/x} = 1 \] (Let \(u=1/x\)). Since \(\int_{1}^{\infty} 1/x dx\) diverges, the original integral diverges.

Let \(u = \sqrt{x}\), \(x=u^2\), \(dx = 2u du\). \[ \int_{0}^{\infty} \frac{2u du}{(u^2+1)u} = \int_{0}^{\infty} \frac{2 du}{u^2+1} = [2 \arctan u]_{0}^{\infty} = 2(\pi/2) = \pi \] The integral converges.

This is a Dirichlet-type integral. It is known to be conditionally convergent. The convergence can be shown using integration by parts.

The integrand behaves like \(x^{p-q}\) for large \(x\). The integral converges if \(q-p > 1\).

As \(x \to 0\), \(e^{\sqrt{x}} - 1 \approx \sqrt{x}\). Use Limit Comparison with \(g(x) = 1/\sqrt{x}\). \[ \lim_{x \to 0^+} \frac{1/(e^{\sqrt{x}}-1)}{1/\sqrt{x}} = 1 \] Since \(\int_{0}^{1} 1/\sqrt{x} dx\) converges, the original integral converges.

Let \(u = \ln x\), \(du = dx/x\). \[ \int_{0}^{\infty} \frac{du}{1+u^2} = [\arctan u]_{0}^{\infty} = \pi/2 \] The integral converges.

Test for absolute convergence. \(|e^{-x} \cos x| \le e^{-x}\). Since \(\int_{0}^{\infty} e^{-x} dx\) converges, the original integral is absolutely convergent.

Discontinuities at \(x=0\) and \(x=1\). Let \(x = \sin^2 \theta\), \(dx = 2 \sin \theta \cos \theta d\theta\). \[ \int_{0}^{\pi/2} \frac{2 \sin \theta \cos \theta d\theta}{\sqrt{\sin^2 \theta \cos^2 \theta}} = \int_{0}^{\pi/2} 2 d\theta = \pi \] The integral converges.

For \(x \ge 2\), \(x^x \ge x^2\). So \(1/x^x \le 1/x^2\). Since \(\int_{1}^{\infty} 1/x^2 dx\) converges, the original integral converges by Direct Comparison.

Split into \(\int_{0}^{1} + \int_{1}^{\infty}\). As \(x \to 0\), integrand is like \(x^2\). Proper. As \(x \to \infty\), compare with \(x^3 e^{-x}\). Converges. The integral converges.

Split into \(\int_{0}^{1} + \int_{1}^{\infty}\). For \(\int_{0}^{1}\), \(\ln(1+x) \approx x\), so integrand is like \(1/\sqrt{x}\). Converges. For \(\int_{1}^{\infty}\), compare with \(\ln x / x^{3/2}\). Converges. The integral converges.

Compare with \(1/x^4\) for \(x \ge 1\). Since \(\int_{1}^{\infty} 1/x^4 dx\) converges, the original integral converges.

Compare with \(1/x^{3/2}\) for \(x \ge 1\). Since \(\int_{1}^{\infty} 1/x^{3/2} dx\) converges, the original integral converges.

Compare with \(1/x\). Since \(\int_{1}^{\infty} 1/x dx\) diverges, the original integral diverges.

Compare with \(1/x^2\). Since \(\int_{1}^{\infty} 1/x^2 dx\) converges, the original integral converges.

Split into \(\int_{0}^{1} + \int_{1}^{\infty}\). For \(\int_{0}^{1}\), compare with \(1/x\). Diverges. The integral diverges.

Split into \(\int_{0}^{1} + \int_{1}^{\infty}\). For \(\int_{0}^{1}\), compare with \(1/x\). Diverges. The integral diverges.

Compare with \(1/x^4\). Since \(\int_{1}^{\infty} 1/x^4 dx\) converges, the original integral converges.

Compare with \(x/x^{5/2} = 1/x^{3/2}\). Since \(\int_{1}^{\infty} 1/x^{3/2} dx\) converges, the original integral converges.

Split into \(\int_{0}^{1} + \int_{1}^{\infty}\). For \(\int_{0}^{1}\), integrand is bounded. Proper. For \(\int_{1}^{\infty}\), compare with \(\ln x / x^3\). Converges. The integral converges.

Split into \(\int_{0}^{1} + \int_{1}^{\infty}\). As \(x \to 0\), integrand is like \(1\). Proper. For \(\int_{1}^{\infty}\), compare with \(1/x^2\). Converges. The integral converges.

Split into \(\int_{0}^{1} + \int_{1}^{\infty}\). For \(\int_{0}^{1}\), compare with \(1/x^{1/3}\). Converges. For \(\int_{1}^{\infty}\), compare with \(1/x^{5/6}\). Diverges. The integral diverges.

This integral diverges. The integrand has spikes that prevent convergence. This is a more advanced example.

Let \(u=x^2\). The integral becomes \(\frac{1}{2} \int_{0}^{\infty} \frac{\sin u}{u} du\). This is the Dirichlet integral, which is known to be conditionally convergent.

Split into \(\int_{0}^{1} + \int_{1}^{\infty}\). For \(\int_{0}^{1}\), compare with \(1/x^q\). Converges if \(q<1\). For \(\int_{1}^{\infty}\), compare with \(1/x^p\). Converges if \(p>1\). Converges if \(q<1\) and \(p>1\).

Compare with \(e^{-x}\). Since \(\int_{0}^{\infty} e^{-x} dx\) converges, the original integral converges.

Compare with \(e^{-x}\). Since \(\int_{0}^{\infty} e^{-x} dx\) converges, the original integral converges.

The denominator has no real roots. Compare with \(1/x^2\). Since \(\int_{1}^{\infty} 1/x^2 dx\) converges, the original integral converges.

\(\cosh x = (e^x+e^{-x})/2\). For large \(x\), \(\cosh x \approx e^x/2\). Compare with \(2e^{-x}\). Converges. The integral converges.

Compare with \(1/x^4\). Since \(\int_{1}^{\infty} 1/x^4 dx\) converges, the original integral converges.

Compare with \(x^2/x^4 = 1/x^2\). Since \(\int_{1}^{\infty} 1/x^2 dx\) converges, the original integral converges.

Compare with \(x^3/x^5 = 1/x^2\). Since \(\int_{1}^{\infty} 1/x^2 dx\) converges, the original integral converges.

Compare with \(1/x^2\). Since \(\int_{1}^{\infty} 1/x^2 dx\) converges, the original integral converges.

Split into \(\int_{0}^{1} + \int_{1}^{\infty}\). For \(\int_{0}^{1}\), compare with \(1/x\). Diverges. The integral diverges.

Converges if \(p>1\).

Converges if \(q-p > 1\).

Compare with \(e^{-x}\). Since \(\int_{0}^{\infty} e^{-x} dx\) converges, the original integral converges.

\( \int_{0}^{\infty} e^{-x/2} dx \). Converges. The integral converges.

This integral diverges. The integrand has spikes that prevent convergence.

Compare with \(1/x^3\). Since \(\int_{1}^{\infty} 1/x^3 dx\) converges, the original integral converges.

Compare with \(1/x^4\). Since \(\int_{1}^{\infty} 1/x^4 dx\) converges, the original integral converges.

Discontinuity at \(x=2\). The integral diverges.

Discontinuity at \(x=1\). The integral diverges.

\( [\arctan x]_{0}^{\infty} = \pi/2 \). The integral converges.

Compare with \(1/x^3\). Since \(\int_{1}^{\infty} 1/x^3 dx\) converges, the original integral converges.

Compare with \(1/x^4\). Since \(\int_{1}^{\infty} 1/x^4 dx\) converges, the original integral converges.

Compare with \(1/x^5\). Since \(\int_{1}^{\infty} 1/x^5 dx\) converges, the original integral converges.

Compare with \(1/x^6\). Since \(\int_{1}^{\infty} 1/x^6 dx\) converges, the original integral converges.

The integral diverges for all \(p>0\).

The integral diverges.

The integral diverges.

The integral diverges.

The integral diverges.

The integral diverges.

The integral diverges.

The integral diverges.

The integral diverges.

The integral diverges.

The integral diverges.

The integral diverges.

We split the integral: \( \int_{0}^{1} \frac{\sin^2 x}{x^2} \,dx + \int_{1}^{\infty} \frac{\sin^2 x}{x^2} \,dx \).

For the first part, as \(x \to 0\), \(\frac{\sin x}{x} \to 1\), so \(\frac{\sin^2 x}{x^2} \to 1\). The integrand is bounded, so \( \int_{0}^{1} \frac{\sin^2 x}{x^2} \,dx \) is a proper integral and converges.

For the second part, \( \int_{1}^{\infty} \frac{\sin^2 x}{x^2} \,dx \). We know \(0 \le \sin^2 x \le 1\), so \(0 \le \frac{\sin^2 x}{x^2} \le \frac{1}{x^2}\). Since \( \int_{1}^{\infty} \frac{1}{x^2} \,dx \) converges (p-integral with \(p=2\)), by the Direct Comparison Test, \( \int_{1}^{\infty} \frac{\sin^2 x}{x^2} \,dx \) converges.

Since both parts converge, the original integral converges.