Investigate the convergence of the integral and find its

Investigate the convergence of the integral and find its value: $$ \int_1^\infty \frac{dx}{x^\alpha} $$

This is an improper integral at $x = \infty$. We will analyze the convergence for different values of $\alpha$ and compute the value when the integral converges.

Case 1: $\alpha > 1$

Analysis:

For $\alpha \neq 1$, we can use the power rule:

$$\int x^{-\alpha} \, dx = \frac{x^{-\alpha+1}}{-\alpha+1} + C = \frac{x^{1-\alpha}}{1-\alpha} + C$$

Evaluation:

$$\int_1^\infty \frac{dx}{x^\alpha} = \lim_{R \to \infty} \int_1^R x^{-\alpha} \, dx = \lim_{R \to \infty} \left[\frac{x^{1-\alpha}}{1-\alpha}\right]_1^R$$

$$= \lim_{R \to \infty} \left(\frac{R^{1-\alpha}}{1-\alpha} - \frac{1^{1-\alpha}}{1-\alpha}\right) = \lim_{R \to \infty} \frac{R^{1-\alpha}}{1-\alpha} - \frac{1}{1-\alpha}$$

Since $\alpha > 1$, we have $1-\alpha < 0$, so:

$$\lim_{R \to \infty} R^{1-\alpha} = \lim_{R \to \infty} \frac{1}{R^{\alpha-1}} = 0$$

Therefore:

$$\int_1^\infty \frac{dx}{x^\alpha} = \frac{0}{1-\alpha} - \frac{1}{1-\alpha} = -\frac{1}{1-\alpha} = \frac{1}{\alpha-1}$$

Conclusion for Case 1:

$$\boxed{\text{If } \alpha > 1: \quad \int_1^\infty \frac{dx}{x^\alpha} = \frac{1}{\alpha-1} \quad \text{(converges)}}$$

Case 2: $\alpha = 1$

Analysis:

When $\alpha = 1$, the integrand is $\frac{1}{x}$:

$$\int_1^\infty \frac{dx}{x} = \lim_{R \to \infty} \int_1^R \frac{dx}{x} = \lim_{R \to \infty} [\ln x]_1^R$$

$$= \lim_{R \to \infty} (\ln R - \ln 1) = \lim_{R \to \infty} \ln R = +\infty$$

Conclusion for Case 2:

$$\boxed{\text{If } \alpha = 1: \quad \int_1^\infty \frac{dx}{x} = +\infty \quad \text{(diverges)}}$$

Case 3: $\alpha < 1$

Analysis:

For $\alpha < 1$:

$$\int_1^\infty \frac{dx}{x^\alpha} = \lim_{R \to \infty} \int_1^R x^{-\alpha} \, dx = \lim_{R \to \infty} \left[\frac{x^{1-\alpha}}{1-\alpha}\right]_1^R$$

$$= \lim_{R \to \infty} \left(\frac{R^{1-\alpha}}{1-\alpha} - \frac{1}{1-\alpha}\right)$$

Since $\alpha < 1$, we have $1-\alpha > 0$, so:

$$\lim_{R \to \infty} R^{1-\alpha} = +\infty$$

Therefore:

$$\int_1^\infty \frac{dx}{x^\alpha} = \frac{+\infty}{1-\alpha} - \frac{1}{1-\alpha} = +\infty$$

Conclusion for Case 3:

$$\boxed{\text{If } \alpha < 1: \quad \int_1^\infty \frac{dx}{x^\alpha} = +\infty \quad \text{(diverges)}}$$

Summary of Results

Convergence of $\int_1^\infty \frac{dx}{x^\alpha}$

$$\int_1^\infty \frac{dx}{x^\alpha} = \begin{cases} \dfrac{1}{\alpha-1} & \text{if } \alpha > 1 \quad \text{(converges)}\\[10pt] +\infty & \text{if } \alpha \leq 1 \quad \text{(diverges)} \end{cases}$$

The critical value is $\alpha = 1$:

• For $\alpha > 1$: The integral converges
• For $\alpha \leq 1$: The integral diverges

Numerical examples

Let's compute some specific values:

Example 1: $\alpha = 2$

$$\int_1^\infty \frac{dx}{x^2} = \frac{1}{2-1} = 1$$

Example 2: $\alpha = 3$

$$\int_1^\infty \frac{dx}{x^3} = \frac{1}{3-1} = \frac{1}{2}$$

Example 3: $\alpha = \frac{3}{2}$

$$\int_1^\infty \frac{dx}{x^{3/2}} = \frac{1}{\frac{3}{2}-1} = \frac{1}{\frac{1}{2}} = 2$$

Example 4: $\alpha = 1$ (divergent)

$$\int_1^\infty \frac{dx}{x} = +\infty$$

Example 5: $\alpha = \frac{1}{2}$ (divergent)

$$\int_1^\infty \frac{dx}{\sqrt{x}} = +\infty$$

Example 6: $\alpha = 0$ (divergent)

$$\int_1^\infty \frac{dx}{x^0} = \int_1^\infty 1 \, dx = +\infty$$

Verification of Example 1

Let's verify $\int_1^\infty \frac{dx}{x^2} = 1$ directly:

$$\int_1^\infty \frac{dx}{x^2} = \lim_{R \to \infty} \left[-\frac{1}{x}\right]_1^R = \lim_{R \to \infty} \left(-\frac{1}{R} - \left(-\frac{1}{1}\right)\right)$$

$$= \lim_{R \to \infty} \left(-\frac{1}{R} + 1\right) = 0 + 1 = 1$$

This confirms our formula! ✓

Geometric interpretation

The integral $\int_1^\infty \frac{dx}{x^\alpha}$ represents the area under the curve $y = \frac{1}{x^\alpha}$ from $x = 1$ to $x = \infty$.

Behavior as $x \to \infty$:

• If $\alpha > 1$: The function decays fast enough (like $\frac{1}{x^2}$ or faster) — the area is finite
• If $\alpha = 1$: The function $\frac{1}{x}$ is the critical case — the area is infinite
• If $\alpha < 1$: The function decays too slowly (slower than $\frac{1}{x}$) — the area is infinite

The key is how quickly the height $\frac{1}{x^\alpha}$ decreases compared to how fast we move along the $x$-axis toward infinity.

Comparison with the integral from 0 to 1

Complementary Behavior

The integrals at 0 and at $\infty$ have opposite convergence criteria:

• $\int_0^1 \frac{dx}{x^\alpha}$ converges if and only if $\alpha < 1$
• $\int_1^\infty \frac{dx}{x^\alpha}$ converges if and only if $\alpha > 1$

Summary table:

For the entire line:

$$\int_0^\infty \frac{dx}{x^\alpha} = \int_0^1 \frac{dx}{x^\alpha} + \int_1^\infty \frac{dx}{x^\alpha}$$

This integral converges only if both parts converge, which never happens! For any $\alpha$, at least one part diverges.

Physical applications

These integrals appear throughout physics:

Gravitational potential:

The gravitational potential energy of a particle at distance $r$ from a mass $M$ is:

$$U(r) = -\frac{GMm}{r}$$

The work to move from $r = R$ to $r = \infty$ is:

$$W = \int_R^\infty \frac{GMm}{r^2} \, dr = GMm \int_R^\infty \frac{dr}{r^2} = GMm \cdot \frac{1}{2-1} \cdot \frac{1}{R} = \frac{GMm}{R}$$

This integral converges because $\alpha = 2 > 1$.

Electric field:

Similar analysis applies to the electric field of a point charge.

Series convergence:

The integral test for series convergence uses this result:

$$\sum_{n=1}^\infty \frac{1}{n^\alpha} \text{ converges} \iff \int_1^\infty \frac{dx}{x^\alpha} \text{ converges} \iff \alpha > 1$$

This is the famous $p$-series test.

The $p$-series connection

The convergence of $\int_1^\infty \frac{dx}{x^p}$ is directly related to the convergence of the $p$-series:

$$\sum_{n=1}^\infty \frac{1}{n^p}$$

Harmonic series ($p = 1$):

$$\sum_{n=1}^\infty \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots = \infty$$

Basel problem ($p = 2$):

$$\sum_{n=1}^\infty \frac{1}{n^2} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots = \frac{\pi^2}{6}$$

The integral gives $\int_1^\infty \frac{dx}{x^2} = 1$, which is close to but not equal to $\frac{\pi^2}{6} \approx 1.645$.

Rate of decay and convergence

The integral $\int_1^\infty \frac{dx}{x^\alpha}$ converges when the function decays faster than $\frac{1}{x}$ at infinity.

General principle:

If $f(x) \sim \frac{1}{x^\alpha}$ as $x \to \infty$, then:

$$\int_1^\infty f(x) \, dx \text{ converges} \iff \alpha > 1$$

This is a fundamental comparison test used throughout analysis.

Alternative proof using substitution

We can also analyze this using the substitution $u = \frac{1}{x}$:

When $x = 1$: $u = 1$

When $x \to \infty$: $u \to 0^+$

Also, $x = \frac{1}{u}$, so $dx = -\frac{du}{u^2}$:

$$\int_1^\infty \frac{dx}{x^\alpha} = \int_1^0 \frac{1}{(1/u)^\alpha} \cdot \left(-\frac{du}{u^2}\right) = \int_0^1 \frac{u^\alpha}{u^2} \, du = \int_0^1 u^{\alpha-2} \, du$$

This integral converges at 0 if and only if $\alpha - 2 > -1$, i.e., $\alpha > 1$.

This gives the same result and shows the duality between integrals at 0 and at $\infty$!

Summary

1. The integral $\int_1^\infty \frac{dx}{x^\alpha}$ is improper at $x = \infty$
2. Convergence criterion: The integral converges if and only if $\alpha > 1$
3. Value when convergent: $\int_1^\infty \frac{dx}{x^\alpha} = \dfrac{1}{\alpha-1}$ for $\alpha > 1$
4. The critical exponent is $\alpha = 1$ (the function $\frac{1}{x}$)
5. This result is dual to $\int_0^1 \frac{dx}{x^\alpha}$: converges if $\alpha < 1$
6. The convergence criterion is fundamental for the $p$-series test

Key lessons:

• Functions must decay faster than $\frac{1}{x}$ at infinity for convergence
• The exponent $\alpha = 1$ is the critical boundary
• Convergence at 0 and at $\infty$ have opposite requirements
• This result connects integral convergence to series convergence
• The substitution $u = \frac{1}{x}$ transforms one problem into the other
• These are among the most important test cases in analysis