Calculate the improper integral: integral from a to ()

Calculate the improper integral: $$ \int_a^\infty \frac{dx}{x\ln^2 x} $$ (where $a > 1$)

This is an improper integral at infinity. We assume $a > 1$ so that $\ln a > 0$ and the integrand is well-defined throughout $[a, \infty)$.

Step 1: Express as a limit

$$\int_a^\infty \frac{dx}{x\ln^2 x} = \lim_{R \to \infty} \int_a^R \frac{dx}{x\ln^2 x}$$

Step 2: Find the antiderivative using substitution

Let $u = \ln x$, so $du = \frac{dx}{x}$.

When $x = a$: $u = \ln a$

When $x = R$: $u = \ln R$

The integral becomes:

$$\int_a^R \frac{dx}{x\ln^2 x} = \int_{\ln a}^{\ln R} \frac{du}{u^2}$$

This is a standard integral:

$$\int \frac{du}{u^2} = \int u^{-2} \, du = -\frac{1}{u} + C$$

Therefore, the antiderivative of $\frac{1}{x\ln^2 x}$ is:

$$-\frac{1}{\ln x} + C$$

Step 3: Evaluate the definite integral

$$\int_a^R \frac{dx}{x\ln^2 x} = \left[-\frac{1}{\ln x}\right]_a^R = -\frac{1}{\ln R} - \left(-\frac{1}{\ln a}\right)$$

$$= \frac{1}{\ln a} - \frac{1}{\ln R}$$

Step 4: Take the limit

$$\int_a^\infty \frac{dx}{x\ln^2 x} = \lim_{R \to \infty} \left(\frac{1}{\ln a} - \frac{1}{\ln R}\right)$$

As $R \to \infty$:

$$\ln R \to +\infty \implies \frac{1}{\ln R} \to 0$$

Therefore:

$$\int_a^\infty \frac{dx}{x\ln^2 x} = \frac{1}{\ln a} - 0 = \frac{1}{\ln a}$$

Final Answer

$$\boxed{\int_a^\infty \frac{dx}{x\ln^2 x} = \frac{1}{\ln a} \quad \text{(for } a > 1 \text{)}}$$

The integral converges to $\frac{1}{\ln a}$.

Special cases

Case 1: $a = e$

$$\int_e^\infty \frac{dx}{x\ln^2 x} = \frac{1}{\ln e} = \frac{1}{1} = 1$$

Case 2: $a = 2$

$$\int_2^\infty \frac{dx}{x\ln^2 x} = \frac{1}{\ln 2} \approx 1.443$$

Case 3: $a = 10$

$$\int_{10}^\infty \frac{dx}{x\ln^2 x} = \frac{1}{\ln 10} \approx 0.434$$

Why the integral converges

The substitution $u = \ln x$ transforms the integral to:

$$\int_{\ln a}^\infty \frac{du}{u^2}$$

This is of the form $\int_b^\infty \frac{du}{u^p}$ with $p = 2 > 1$.

We know that $\int_1^\infty \frac{dx}{x^p}$ converges if and only if $p > 1$.

Since $p = 2 > 1$, the integral converges! ✓

Comparison with related integrals

Case 1: $\int_a^\infty \frac{dx}{x\ln x}$

Using $u = \ln x$:

$$= \int_{\ln a}^\infty \frac{du}{u} = \lim_{R \to \infty} [\ln u]_{\ln a}^{\ln R} = \lim_{R \to \infty} (\ln(\ln R) - \ln(\ln a)) = +\infty$$

This integral diverges! ✗

Case 2: $\int_a^\infty \frac{dx}{x\ln^2 x}$

This is our integral, which converges to $\frac{1}{\ln a}$. ✓

Case 3: $\int_a^\infty \frac{dx}{x\ln^3 x}$

Using $u = \ln x$:

$$= \int_{\ln a}^\infty \frac{du}{u^3} = \left[-\frac{1}{2u^2}\right]_{\ln a}^\infty = 0 - \left(-\frac{1}{2(\ln a)^2}\right) = \frac{1}{2(\ln a)^2}$$

This also converges. ✓

General rule:

$$\int_a^\infty \frac{dx}{x(\ln x)^n} \text{ converges} \iff n > 1$$

This is analogous to the standard $p$-test!

Connection to the integral test

This result is directly related to the convergence of series via the integral test.

Consider the series:

$$\sum_{n=2}^\infty \frac{1}{n(\ln n)^2}$$

By the integral test, this series converges if and only if:

$$\int_2^\infty \frac{dx}{x(\ln x)^2}$$

converges, which we've shown it does (to $\frac{1}{\ln 2}$).

Therefore, the series $\sum_{n=2}^\infty \frac{1}{n(\ln n)^2}$ converges! ✓

In contrast, the series $\sum_{n=2}^\infty \frac{1}{n\ln n}$ diverges because $\int_2^\infty \frac{dx}{x\ln x}$ diverges.

Geometric interpretation

The function $f(x) = \frac{1}{x\ln^2 x}$ for $x \geq a > 1$ has these properties:

• Decreasing function
• Approaches 0 as $x \to \infty$
• Decays faster than $\frac{1}{x\ln x}$ but slower than $\frac{1}{x^{1+\varepsilon}}$ for any $\varepsilon > 0$
• The area under the curve from $a$ to $\infty$ is finite (equals $\frac{1}{\ln a}$)

Behavior as $a$ varies

As $a$ increases from $e$ to $\infty$:

$$\frac{1}{\ln a} \text{ decreases from } 1 \text{ to } 0$$

• As $a \to e^+$: $\ln a \to 1^+$, so $\frac{1}{\ln a} \to 1$
• As $a \to \infty$: $\ln a \to +\infty$, so $\frac{1}{\ln a} \to 0$

This makes sense: starting the integral further out (larger $a$) captures less area.

Relationship to logarithmic decay

Functions that decay like $\frac{1}{x(\ln x)^n}$ represent "logarithmic decay" - slower than polynomial decay but faster than pure logarithmic growth.

The hierarchy of decay rates near infinity:

1. Polynomial: $\frac{1}{x^p}$ (converges for $p > 1$)
2. Logarithmic-polynomial: $\frac{1}{x(\ln x)^n}$ (converges for $n > 1$)
3. Pure logarithmic: $\frac{1}{\ln x}$ (diverges)

Direct evaluation (formal)

We can write formally:

$$\int_a^\infty \frac{dx}{x\ln^2 x} = \left[-\frac{1}{\ln x}\right]_a^\infty$$

At $x = \infty$:

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

At $x = a$:

$$-\frac{1}{\ln a}$$

Therefore:

$$0 - \left(-\frac{1}{\ln a}\right) = \frac{1}{\ln a}$$

Summary

1. The integral has an improper limit at $x = \infty$
2. Using the substitution $u = \ln x$ transforms it to: $$ \int_{\ln a}^\infty \frac{du}{u^2} $$
3. The antiderivative is $-\frac{1}{\ln x}$
4. Evaluating from $a$ to $\infty$: $$ \left[-\frac{1}{\ln x}\right]_a^\infty = 0 - \left(-\frac{1}{\ln a}\right) = \frac{1}{\ln a} $$
5. The integral converges because it's equivalent to $\int \frac{du}{u^2}$ with exponent 2 > 1
6. Result: $\int_a^\infty \frac{dx}{x\ln^2 x} = \frac{1}{\ln a}$ (for $a > 1$)

Key lessons:

• The substitution $u = \ln x$ is essential for logarithmic integrands
• General rule: $\int_a^\infty \frac{dx}{x(\ln x)^n}$ converges if and only if $n > 1$
• This parallels the power rule: $\int_1^\infty \frac{dx}{x^p}$ converges if and only if $p > 1$
• The critical exponent is $n = 1$ (logarithmic boundary)
• Useful for the integral test in series convergence
• Logarithmic factors provide a "borderline" between convergence and divergence