Investigate the convergence of the integral and find its

Investigate the convergence of the integral and find its value: $$ \int_0^1 \frac{dx}{x^\alpha} $$

This is an improper integral at $x = 0$ when $\alpha > 0$, since $\frac{1}{x^\alpha} \to \infty$ as $x \to 0^+$.

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:

$$\int_0^1 \frac{dx}{x^\alpha} = \lim_{\varepsilon \to 0^+} \int_\varepsilon^1 x^{-\alpha} \, dx = \lim_{\varepsilon \to 0^+} \left[\frac{x^{1-\alpha}}{1-\alpha}\right]_\varepsilon^1$$

$$= \lim_{\varepsilon \to 0^+} \left(\frac{1^{1-\alpha}}{1-\alpha} - \frac{\varepsilon^{1-\alpha}}{1-\alpha}\right) = \frac{1}{1-\alpha} - \lim_{\varepsilon \to 0^+} \frac{\varepsilon^{1-\alpha}}{1-\alpha}$$

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

$$\lim_{\varepsilon \to 0^+} \varepsilon^{1-\alpha} = 0$$

Therefore:

$$\int_0^1 \frac{dx}{x^\alpha} = \frac{1}{1-\alpha} - 0 = \frac{1}{1-\alpha}$$

Conclusion for Case 1:

$$\boxed{\text{If } \alpha < 1: \quad \int_0^1 \frac{dx}{x^\alpha} = \frac{1}{1-\alpha} \quad \text{(converges)}}$$

Case 2: $\alpha = 1$

Analysis:

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

$$\int_0^1 \frac{dx}{x} = \lim_{\varepsilon \to 0^+} \int_\varepsilon^1 \frac{dx}{x} = \lim_{\varepsilon \to 0^+} [\ln x]_\varepsilon^1$$

$$= \lim_{\varepsilon \to 0^+} (\ln 1 - \ln \varepsilon) = 0 - \lim_{\varepsilon \to 0^+} \ln \varepsilon$$

Since $\lim_{\varepsilon \to 0^+} \ln \varepsilon = -\infty$:

$$\int_0^1 \frac{dx}{x} = 0 - (-\infty) = +\infty$$

Conclusion for Case 2:

$$\boxed{\text{If } \alpha = 1: \quad \int_0^1 \frac{dx}{x} = +\infty \quad \text{(diverges)}}$$

Case 3: $\alpha > 1$

Analysis:

For $\alpha > 1$:

$$\int_0^1 \frac{dx}{x^\alpha} = \lim_{\varepsilon \to 0^+} \int_\varepsilon^1 x^{-\alpha} \, dx = \lim_{\varepsilon \to 0^+} \left[\frac{x^{1-\alpha}}{1-\alpha}\right]_\varepsilon^1$$

$$= \lim_{\varepsilon \to 0^+} \left(\frac{1}{1-\alpha} - \frac{\varepsilon^{1-\alpha}}{1-\alpha}\right)$$

Since $\alpha > 1$, we have $1-\alpha < 0$, so as $\varepsilon \to 0^+$:

$$\varepsilon^{1-\alpha} = \frac{1}{\varepsilon^{\alpha-1}} \to +\infty$$

Therefore:

$$\frac{\varepsilon^{1-\alpha}}{1-\alpha} = \frac{1}{(1-\alpha)\varepsilon^{\alpha-1}} \to -\infty$$

(Note: $1-\alpha < 0$, so we're dividing by a negative number)

Thus:

$$\int_0^1 \frac{dx}{x^\alpha} = \frac{1}{1-\alpha} - (-\infty) = +\infty$$

Conclusion for Case 3:

$$\boxed{\text{If } \alpha > 1: \quad \int_0^1 \frac{dx}{x^\alpha} = +\infty \quad \text{(diverges)}}$$

Case 4: $\alpha \leq 0$

Analysis:

When $\alpha \leq 0$, we have $x^\alpha \geq 1$ for $x \in (0, 1]$, so $\frac{1}{x^\alpha} \leq 1$ is bounded.

The integrand has no singularity at $x = 0$:

$$\lim_{x \to 0^+} \frac{1}{x^\alpha} = \lim_{x \to 0^+} x^{-\alpha} = 0 \quad \text{if } \alpha < 0$$

$$\lim_{x \to 0^+} \frac{1}{x^0} = 1 \quad \text{if } \alpha = 0$$

The integral is proper (not improper):

$$\int_0^1 \frac{dx}{x^\alpha} = \left[\frac{x^{1-\alpha}}{1-\alpha}\right]_0^1 = \frac{1}{1-\alpha} - 0 = \frac{1}{1-\alpha}$$

Conclusion for Case 4:

$$\boxed{\text{If } \alpha \leq 0: \quad \int_0^1 \frac{dx}{x^\alpha} = \frac{1}{1-\alpha} \quad \text{(converges)}}$$

Summary of Results

Convergence of $\int_0^1 \frac{dx}{x^\alpha}$

$$\int_0^1 \frac{dx}{x^\alpha} = \begin{cases} \dfrac{1}{1-\alpha} & \text{if } \alpha < 1 \quad \text{(converges)}\\[10pt] +\infty & \text{if } \alpha \geq 1 \quad \text{(diverges)} \end{cases}$$

The critical value is $\alpha = 1$:

• For $\alpha < 1$: The integral converges
• For $\alpha \geq 1$: The integral diverges

Numerical examples

Let's compute some specific values:

Example 1: $\alpha = \frac{1}{2}$

$$\int_0^1 \frac{dx}{\sqrt{x}} = \frac{1}{1-\frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$$

Example 2: $\alpha = 0$

$$\int_0^1 \frac{dx}{x^0} = \int_0^1 1 \, dx = \frac{1}{1-0} = 1$$

Example 3: $\alpha = -1$

$$\int_0^1 \frac{dx}{x^{-1}} = \int_0^1 x \, dx = \frac{1}{1-(-1)} = \frac{1}{2}$$

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

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

Example 5: $\alpha = 2$ (divergent)

$$\int_0^1 \frac{dx}{x^2} = +\infty$$

Geometric interpretation

The integral $\int_0^1 \frac{dx}{x^\alpha}$ represents the area under the curve $y = \frac{1}{x^\alpha}$ from $x = 0$ to $x = 1$.

Behavior near $x = 0$:

• If $\alpha < 1$: The function grows to infinity as $x \to 0^+$, but not "too fast" — the area remains finite
• If $\alpha = 1$: The function $\frac{1}{x}$ is the critical case — the area is infinite
• If $\alpha > 1$: The function grows to infinity "too fast" — the area is infinite

The key is how quickly the width $dx$ shrinks compared to how fast the height $\frac{1}{x^\alpha}$ grows near $x = 0$.

Comparison test interpretation

We can understand the convergence using the comparison test:

Near $x = 0$, the "contribution" to the integral is approximately:

$$\text{Area} \approx \frac{1}{x^\alpha} \cdot dx = x^{-\alpha} \, dx$$

Integrating from $\varepsilon$ to 1:

$$\int_\varepsilon^1 x^{-\alpha} \, dx \sim \begin{cases} \varepsilon^{1-\alpha} & \text{if } \alpha < 1 \to 0 \text{ as } \varepsilon \to 0\\ \ln \varepsilon & \text{if } \alpha = 1 \to -\infty \text{ as } \varepsilon \to 0\\ \varepsilon^{1-\alpha} & \text{if } \alpha > 1 \to \infty \text{ as } \varepsilon \to 0 \end{cases}$$

Dual result for integrals to infinity

There is a dual result for integrals from 1 to infinity:

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

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

Notice the complementary behavior:

• At 0: Converges if $\alpha < 1$, diverges if $\alpha \geq 1$
• At $\infty$: Converges if $\alpha > 1$, diverges if $\alpha \leq 1$

The critical exponent $\alpha = 1$ is the boundary for both cases!

Physical interpretation

These integrals appear in physics:

Gravitational/Electric potential:

The potential at distance $r$ from a point source goes as $\frac{1}{r}$ (in 3D). Integrating over space involves powers of $r$.

Dimensional analysis:

In $n$ dimensions, the volume element is $r^{n-1} \, dr$. Combined with a potential $\frac{1}{r^\alpha}$, we get integrals of the form:

$$\int_0^R \frac{r^{n-1}}{r^\alpha} \, dr = \int_0^R r^{n-1-\alpha} \, dr$$

This converges at $r = 0$ if $n-1-\alpha > -1$, i.e., $\alpha < n$.

Summary

1. The integral $\int_0^1 \frac{dx}{x^\alpha}$ is improper when $\alpha > 0$
2. Convergence criterion: The integral converges if and only if $\alpha < 1$
3. Value when convergent: $\int_0^1 \frac{dx}{x^\alpha} = \dfrac{1}{1-\alpha}$ for $\alpha < 1$
4. The critical exponent is $\alpha = 1$ (the function $\frac{1}{x}$)
5. This result has a dual for integrals to infinity: converges if $\alpha > 1$

Key lessons:

• Power functions have critical exponents for convergence
• The exponent $\alpha = 1$ is special (logarithmic singularity)
• Convergence at 0 and at $\infty$ have opposite criteria
• The "speed" of growth/decay determines whether improper integrals converge
• This is a fundamental result used throughout analysis and physics