Investigate the convergence of the integral and find its

Investigate the convergence of the integral and find its value: $$ \int_a^b \frac{dx}{(b-x)^\alpha} \quad (a \neq b) $$

This integral has a potential singularity at $x = b$ when $\alpha > 0$, since $(b-x) \to 0$ as $x \to b^-$. We need to analyze the convergence for different values of $\alpha$.

We will consider the case $a < b$ (the case $a > b$ is similar).

Case 1: $\alpha < 1$

Analysis:

When $\alpha > 0$ and $\alpha \neq 1$, the integral is improper at $x = b$.

Using the substitution $u = b - x$, so $du = -dx$:

• When $x = a$: $u = b - a$
• When $x = b$: $u = 0$

$$\int_a^b \frac{dx}{(b-x)^\alpha} = \int_{b-a}^0 \frac{-du}{u^\alpha} = \int_0^{b-a} \frac{du}{u^\alpha}$$

This is the same type of integral we've studied before: $\int_0^{b-a} u^{-\alpha} \, du$

For $\alpha < 1$:

$$\int_0^{b-a} u^{-\alpha} \, du = \left[\frac{u^{1-\alpha}}{1-\alpha}\right]_0^{b-a} = \frac{(b-a)^{1-\alpha}}{1-\alpha} - 0 = \frac{(b-a)^{1-\alpha}}{1-\alpha}$$

Conclusion for Case 1:

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

Case 2: $\alpha = 1$

Analysis:

When $\alpha = 1$:

$$\int_a^b \frac{dx}{b-x} = \lim_{\varepsilon \to 0^+} \int_a^{b-\varepsilon} \frac{dx}{b-x}$$

Using the substitution $u = b - x$:

$$= \lim_{\varepsilon \to 0^+} \int_{b-a}^\varepsilon \frac{-du}{u} = \lim_{\varepsilon \to 0^+} \int_\varepsilon^{b-a} \frac{du}{u}$$

$$= \lim_{\varepsilon \to 0^+} [\ln u]_\varepsilon^{b-a} = \lim_{\varepsilon \to 0^+} (\ln(b-a) - \ln \varepsilon) = \ln(b-a) - (-\infty) = +\infty$$

Conclusion for Case 2:

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

Case 3: $\alpha > 1$

Analysis:

For $\alpha > 1$:

$$\int_0^{b-a} u^{-\alpha} \, du = \lim_{\varepsilon \to 0^+} \int_\varepsilon^{b-a} u^{-\alpha} \, du$$

$$= \lim_{\varepsilon \to 0^+} \left[\frac{u^{1-\alpha}}{1-\alpha}\right]_\varepsilon^{b-a} = \lim_{\varepsilon \to 0^+} \left(\frac{(b-a)^{1-\alpha}}{1-\alpha} - \frac{\varepsilon^{1-\alpha}}{1-\alpha}\right)$$

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

$$\varepsilon^{1-\alpha} = \frac{1}{\varepsilon^{\alpha-1}} \to +\infty \quad \text{as } \varepsilon \to 0^+$$

Therefore:

$$\frac{\varepsilon^{1-\alpha}}{1-\alpha} = \frac{1}{(1-\alpha)\varepsilon^{\alpha-1}} \to -\infty \quad \text{as } \varepsilon \to 0^+$$

(since $1-\alpha < 0$)

Thus, the integral diverges.

Conclusion for Case 3:

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

Case 4: $\alpha \leq 0$

Analysis:

When $\alpha \leq 0$, the function $(b-x)^{-\alpha} = (b-x)^{|\alpha|}$ is bounded on $[a, b]$:

$$0 < (b-x)^{|\alpha|} \leq (b-a)^{|\alpha|}$$

There is no singularity, and the integral is proper:

$$\int_a^b \frac{dx}{(b-x)^\alpha} = \int_a^b (b-x)^{-\alpha} \, dx$$

Using the substitution $u = b - x$:

$$= \int_0^{b-a} u^{-\alpha} \, du = \left[\frac{u^{1-\alpha}}{1-\alpha}\right]_0^{b-a} = \frac{(b-a)^{1-\alpha}}{1-\alpha}$$

Conclusion for Case 4:

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

Summary of Results

Convergence of $\int_a^b \frac{dx}{(b-x)^\alpha}$ for $a < b$

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

The critical value is $\alpha = 1$:

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

Case when $a > b$

If $a > b$, then $b - x < 0$ for $x \in [b, a]$, so we need to be careful with the absolute value.

For $x \in [b, a]$, we have $b - x = -(x - b)$, so:

$$(b-x)^\alpha = (-(x-b))^\alpha = (-1)^\alpha (x-b)^\alpha$$

For real-valued integrals, we typically assume $\alpha$ is such that this is well-defined.

Using the substitution $u = x - b$:

$$\int_a^b \frac{dx}{(b-x)^\alpha} = -\int_b^a \frac{dx}{(x-b)^\alpha} = \int_a^b \frac{dx}{(x-b)^\alpha}$$

This has a singularity at $x = b$ (the lower limit), and by similar analysis:

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

Numerical Examples

Assume $a = 0$ and $b = 1$ for simplicity:

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

$$\int_0^1 \frac{dx}{(1-x)^{1/2}} = \frac{(1-0)^{1-1/2}}{1-1/2} = \frac{1^{1/2}}{1/2} = 2$$

Example 2: $\alpha = 0$

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

Example 3: $\alpha = -1$

$$\int_0^1 \frac{dx}{(1-x)^{-1}} = \int_0^1 (1-x) \, dx = \frac{(1-0)^2}{1-(-1)} = \frac{1}{2}$$

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

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

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

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

Verification of Example 1

Let's verify $\int_0^1 \frac{dx}{\sqrt{1-x}} = 2$ directly:

$$\int_0^1 \frac{dx}{\sqrt{1-x}} = \lim_{\varepsilon \to 0^+} \int_0^{1-\varepsilon} (1-x)^{-1/2} \, dx$$

Let $u = 1 - x$, so $du = -dx$:

$$= \lim_{\varepsilon \to 0^+} \int_1^\varepsilon u^{-1/2} \cdot (-du) = \lim_{\varepsilon \to 0^+} \int_\varepsilon^1 u^{-1/2} \, du$$

$$= \lim_{\varepsilon \to 0^+} [2u^{1/2}]_\varepsilon^1 = \lim_{\varepsilon \to 0^+} (2 - 2\sqrt{\varepsilon}) = 2$$

This confirms our formula! ✓

Geometric Interpretation

The integral $\int_a^b \frac{dx}{(b-x)^\alpha}$ represents the area under the curve $y = \frac{1}{(b-x)^\alpha}$ from $x = a$ to $x = b$.

Behavior near $x = b$:

• As $x \to b^-$, we have $(b-x) \to 0^+$, so $\frac{1}{(b-x)^\alpha} \to +\infty$ when $\alpha > 0$
• If $\alpha < 1$: The function grows to infinity, but not "too fast" — the area remains finite
• If $\alpha = 1$: The function $\frac{1}{b-x}$ is the critical case — the area is infinite
• If $\alpha > 1$: The function grows to infinity "too fast" — the area is infinite

Connection to Time-to-Collision Problems

This type of integral appears in physics when analyzing motion with singularities.

Example: A particle approaches a barrier at $x = b$ with velocity proportional to distance:

$$v = \frac{dx}{dt} = k(b-x)^\beta$$

The time to reach $x = b$ from $x = a$ is:

$$T = \int_a^b \frac{dx}{v} = \int_a^b \frac{dx}{k(b-x)^\beta} = \frac{1}{k} \int_a^b \frac{dx}{(b-x)^\beta}$$

This time is:

• Finite if $\beta < 1$ (the particle reaches the barrier in finite time)
• Infinite if $\beta \geq 1$ (the particle approaches asymptotically, never reaching the barrier)

This is known as Zeno's paradox in physics when $\beta = 1$.

Relationship to Standard Forms

This integral is equivalent to the standard form we studied earlier:

$$\int_a^b \frac{dx}{(b-x)^\alpha} = \int_0^{b-a} \frac{du}{u^\alpha}$$

This shows that the convergence criterion is the same: converges if and only if $\alpha < 1$.

General Formula

For any interval $[a, b]$ with $a < b$:

$$\int_a^b \frac{dx}{(b-x)^\alpha} = (b-a)^{1-\alpha} \int_0^1 \frac{dt}{t^\alpha}$$

where the second integral is the "unit" version. The factor $(b-a)^{1-\alpha}$ scales the result.

Summary

1. The integral $\int_a^b \frac{dx}{(b-x)^\alpha}$ has a singularity at the upper endpoint $x = b$ when $\alpha > 0$
2. Convergence criterion: The integral converges if and only if $\alpha < 1$
3. Value when convergent: $\int_a^b \frac{dx}{(b-x)^\alpha} = \dfrac{(b-a)^{1-\alpha}}{1-\alpha}$ for $\alpha < 1$
4. The critical exponent is $\alpha = 1$
5. This is equivalent to $\int_0^{b-a} \frac{du}{u^\alpha}$ via substitution

Key lessons:

• Singularities at finite endpoints follow the same convergence rules as singularities at 0
• The critical exponent $\alpha = 1$ determines convergence
• Substitution transforms endpoint singularities to the standard form at 0
• These integrals model physical situations like time-to-collision problems
• The length of the interval $(b-a)$ scales the result but doesn't affect convergence
• Convergence depends only on the behavior near the singularity, not the interval length