Calculate the improper integral (or establish its

Calculate the improper integral (or establish its divergence): $$ \int_0^5 \frac{dx}{x^7} $$

This is an improper integral because the integrand $\frac{1}{x^7}$ has a vertical asymptote at $x = 0$.

Step 1: Identify the singularity

The function $f(x) = \frac{1}{x^7} = x^{-7}$ is undefined at $x = 0$ and:

$$\lim_{x \to 0^+} \frac{1}{x^7} = +\infty$$

Therefore, we must evaluate this as an improper integral with a limit.

Step 2: Express as a limit

$$\int_0^5 \frac{dx}{x^7} = \lim_{\varepsilon \to 0^+} \int_\varepsilon^5 \frac{dx}{x^7}$$

Step 3: Check the convergence criterion

We know from the general theory that:

$$\int_0^a \frac{dx}{x^\alpha} \text{ converges} \iff \alpha < 1$$

In our case, $\alpha = 7$, so:

$$7 > 1 \implies \text{The integral diverges}$$

Step 4: Verify by direct calculation

Let's verify this by computing the limit explicitly.

The antiderivative of $x^{-7}$ is:

$$\int x^{-7} \, dx = \frac{x^{-7+1}}{-7+1} = \frac{x^{-6}}{-6} = -\frac{1}{6x^6}$$

Verify:

$$\frac{d}{dx}\left[-\frac{1}{6x^6}\right] = -\frac{1}{6} \cdot \frac{d}{dx}[x^{-6}] = -\frac{1}{6} \cdot (-6)x^{-7} = x^{-7} = \frac{1}{x^7}$$

Correct! ✓

Step 5: Evaluate the definite integral

$$\int_\varepsilon^5 \frac{dx}{x^7} = \left[-\frac{1}{6x^6}\right]_\varepsilon^5 = -\frac{1}{6 \cdot 5^6} - \left(-\frac{1}{6\varepsilon^6}\right)$$

$$= -\frac{1}{6 \cdot 15625} + \frac{1}{6\varepsilon^6} = -\frac{1}{93750} + \frac{1}{6\varepsilon^6}$$

Step 6: Take the limit

$$\int_0^5 \frac{dx}{x^7} = \lim_{\varepsilon \to 0^+} \left(-\frac{1}{93750} + \frac{1}{6\varepsilon^6}\right)$$

As $\varepsilon \to 0^+$:

$$\frac{1}{6\varepsilon^6} = \frac{1}{6} \cdot \frac{1}{\varepsilon^6} \to +\infty$$

Therefore:

$$\int_0^5 \frac{dx}{x^7} = -\frac{1}{93750} + \infty = +\infty$$

Final Answer

$$\boxed{\int_0^5 \frac{dx}{x^7} = +\infty \quad \text{(diverges)}}$$

The integral diverges.

Explanation

The integral diverges because the exponent $\alpha = 7$ satisfies $\alpha > 1$.

Why does it diverge?

Near $x = 0$, the function $\frac{1}{x^7}$ grows to infinity so rapidly that the area under the curve, even over a small interval near 0, is infinite.

More precisely, for small $x$:

$$\frac{1}{x^7} \text{ is extremely large}$$

The "contribution" to the integral near $x = 0$ is approximately:

$$\int_0^\varepsilon \frac{dx}{x^7} \sim \frac{1}{\varepsilon^6}$$

As $\varepsilon \to 0$, this grows without bound.

Comparison with convergent cases

Let's compare with similar integrals:

Case 1: $\int_0^5 \frac{dx}{x^{1/2}}$ (converges since $\frac{1}{2} < 1$)

$$\int_0^5 \frac{dx}{\sqrt{x}} = [2\sqrt{x}]_0^5 = 2\sqrt{5} \approx 4.472 \quad \text{(finite)}$$

Case 2: $\int_0^5 \frac{dx}{x}$ (diverges since $\alpha = 1$)

$$\int_0^5 \frac{dx}{x} = [\ln x]_0^5 = \ln 5 - \ln 0 = \ln 5 - (-\infty) = +\infty$$

Case 3: $\int_0^5 \frac{dx}{x^2}$ (diverges since $2 > 1$)

$$\int_0^5 \frac{dx}{x^2} = \left[-\frac{1}{x}\right]_0^5 = -\frac{1}{5} - (-\infty) = +\infty$$

Case 4: $\int_0^5 \frac{dx}{x^7}$ (diverges since $7 > 1$)

$$\int_0^5 \frac{dx}{x^7} = +\infty$$

All cases with $\alpha \geq 1$ diverge.

Visual understanding

The function $y = \frac{1}{x^7}$ has these properties:

• At $x = 0$: vertical asymptote (function goes to $+\infty$)
• At $x = 1$: $y = 1$
• At $x = 5$: $y = \frac{1}{5^7} = \frac{1}{78125} \approx 0.0000128$ (very small)
• The function decreases very rapidly for $x > 0$

Despite the function being very small over most of the interval $[0, 5]$, the singularity at $x = 0$ is so strong that it causes the integral to diverge.

Rate of divergence

Near $x = 0$, the integral behaves like:

$$\int_\varepsilon^5 \frac{dx}{x^7} \approx \frac{1}{6\varepsilon^6}$$

This shows the integral diverges like $\frac{1}{\varepsilon^6}$ as $\varepsilon \to 0^+$.

For comparison:

• $\int_0^5 \frac{dx}{x}$ diverges like $\ln\frac{1}{\varepsilon}$ (logarithmically)
• $\int_0^5 \frac{dx}{x^2}$ diverges like $\frac{1}{\varepsilon}$ (linearly)
• $\int_0^5 \frac{dx}{x^7}$ diverges like $\frac{1}{\varepsilon^6}$ (much faster!)

The larger the exponent $\alpha > 1$, the faster the divergence.

Contrast with integral to infinity

It's interesting to contrast this with the dual integral:

$$\int_1^\infty \frac{dx}{x^7}$$

This integral converges because $7 > 1$:

$$\int_1^\infty \frac{dx}{x^7} = \left[-\frac{1}{6x^6}\right]_1^\infty = 0 - \left(-\frac{1}{6}\right) = \frac{1}{6}$$

Key difference:

• At 0: $\int_0^a \frac{dx}{x^\alpha}$ converges if $\alpha < 1$, diverges if $\alpha \geq 1$
• At $\infty$: $\int_1^\infty \frac{dx}{x^\alpha}$ converges if $\alpha > 1$, diverges if $\alpha \leq 1$

The criteria are opposite!

General principle

For $\int_0^a \frac{dx}{x^\alpha}$:

• Converges if the singularity is "weak" ($\alpha < 1$)
• Diverges if the singularity is "strong" ($\alpha \geq 1$)

The exponent $\alpha = 7$ creates a very strong singularity at 0, causing divergence.

Summary

1. The integral $\int_0^5 \frac{dx}{x^7}$ has a singularity at $x = 0$
2. The exponent is $\alpha = 7 > 1$, so by the convergence criterion, the integral diverges
3. Direct calculation confirms: $$ \int_0^5 \frac{dx}{x^7} = \lim_{\varepsilon \to 0^+} \left(-\frac{1}{93750} + \frac{1}{6\varepsilon^6}\right) = +\infty $$
4. The singularity at $x = 0$ is too strong for the integral to converge
5. The integral diverges like $\frac{1}{\varepsilon^6}$ as $\varepsilon \to 0^+$

Key lessons:

• Not all improper integrals converge
• High exponents ($\alpha \geq 1$) at 0 cause divergence
• The critical exponent is $\alpha = 1$
• Direct calculation confirms the theoretical criterion
• The rate of divergence increases with the exponent
• Understanding convergence criteria saves computation time