Calculate the improper integral: integral from 0 to (1/2)

Calculate the improper integral: $$ \int_0^{1/2} \frac{dx}{x\ln^2 x} $$

This is an improper integral because the integrand has a singularity at $x = 0$.

Step 1: Analyze the singularity

As $x \to 0^+$:

• $x \to 0$
• $\ln x \to -\infty$
• $\ln^2 x = (\ln x)^2 \to +\infty$
• $x \ln^2 x \to 0$ (but the product goes to 0)

The behavior near $x = 0$ is complex. Let's analyze more carefully:

$$\frac{1}{x\ln^2 x} = \frac{1}{x} \cdot \frac{1}{\ln^2 x}$$

Both factors go to infinity as $x \to 0^+$, so we need to determine which dominates.

Step 2: Express as a limit

$$\int_0^{1/2} \frac{dx}{x\ln^2 x} = \lim_{\varepsilon \to 0^+} \int_\varepsilon^{1/2} \frac{dx}{x\ln^2 x}$$

Step 3: Find the antiderivative using substitution

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

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

When $x = \frac{1}{2}$: $u = \ln \frac{1}{2} = -\ln 2$

The integral becomes:

$$\int_\varepsilon^{1/2} \frac{dx}{x\ln^2 x} = \int_{\ln \varepsilon}^{-\ln 2} \frac{du}{u^2}$$

This is a standard integral:

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

Step 4: Evaluate the definite integral

$$\int_{\ln \varepsilon}^{-\ln 2} \frac{du}{u^2} = \left[-\frac{1}{u}\right]_{\ln \varepsilon}^{-\ln 2}$$

$$= -\frac{1}{-\ln 2} - \left(-\frac{1}{\ln \varepsilon}\right) = \frac{1}{\ln 2} + \frac{1}{\ln \varepsilon}$$

Step 5: Take the limit

$$\int_0^{1/2} \frac{dx}{x\ln^2 x} = \lim_{\varepsilon \to 0^+} \left(\frac{1}{\ln 2} + \frac{1}{\ln \varepsilon}\right)$$

As $\varepsilon \to 0^+$:

$$\ln \varepsilon \to -\infty \implies \frac{1}{\ln \varepsilon} \to 0^-$$

More precisely: $\frac{1}{\ln \varepsilon} \to 0$ from the negative side.

Wait, let me reconsider. Since $\ln \varepsilon < 0$ for $\varepsilon < 1$, and $\ln \varepsilon \to -\infty$:

$$\frac{1}{\ln \varepsilon} \to \frac{1}{-\infty} = 0^-$$

Therefore:

$$\lim_{\varepsilon \to 0^+} \left(\frac{1}{\ln 2} + \frac{1}{\ln \varepsilon}\right) = \frac{1}{\ln 2} + 0 = \frac{1}{\ln 2}$$

Final Answer

$$\boxed{\int_0^{1/2} \frac{dx}{x\ln^2 x} = \frac{1}{\ln 2}}$$

The integral converges to $\frac{1}{\ln 2} \approx 1.443$.

Verification

Let's verify our antiderivative. If $F(x) = -\frac{1}{\ln x}$, then:

$$F'(x) = -\frac{d}{dx}\left[\frac{1}{\ln x}\right] = -\frac{d}{dx}[(\ln x)^{-1}]$$

Using the chain rule:

$$= -(-1)(\ln x)^{-2} \cdot \frac{1}{x} = \frac{1}{x(\ln x)^2} = \frac{1}{x\ln^2 x}$$

Correct! ✓

Why the integral converges

Despite both $\frac{1}{x}$ and $\frac{1}{\ln^2 x}$ going to infinity as $x \to 0^+$, their product behaves in a way that allows convergence.

The key is the substitution $u = \ln x$ transforms the problem to:

$$\int_{\ln \varepsilon}^{-\ln 2} \frac{du}{u^2}$$

This is of the form $\int \frac{du}{u^2}$ with exponent 2, which converges when integrated from a large negative number (like $-\infty$) to a finite value.

More precisely, as $\varepsilon \to 0^+$:

$$\ln \varepsilon \to -\infty$$

And we're evaluating:

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

Near $u = -\infty$, the integrand $\frac{1}{u^2}$ decays like $\frac{1}{u^2}$, which is integrable.

Comparison with related integrals

Case 1: $\int_0^{1/2} \frac{dx}{x\ln x}$

Using $u = \ln x$:

$$= \int_{\ln \varepsilon}^{-\ln 2} \frac{du}{u} = [\ln|u|]_{\ln \varepsilon}^{-\ln 2} = \ln|-\ln 2| - \ln|\ln \varepsilon|$$

As $\varepsilon \to 0^+$: $\ln|\ln \varepsilon| = \ln(-\ln \varepsilon) \to +\infty$

This integral diverges! ✗

Case 2: $\int_0^{1/2} \frac{dx}{x\ln^2 x}$

This is our integral, which converges to $\frac{1}{\ln 2}$. ✓

Case 3: $\int_0^{1/2} \frac{dx}{x\ln^3 x}$

Using $u = \ln x$:

$$= \int_{\ln \varepsilon}^{-\ln 2} \frac{du}{u^3} = \left[-\frac{1}{2u^2}\right]_{\ln \varepsilon}^{-\ln 2}$$

This also converges. ✓

General pattern:

$$\int_0^{a} \frac{dx}{x(\ln x)^n} \text{ converges} \iff n > 1$$

This is analogous to the $p$-test for integrals!

Geometric interpretation

The function $f(x) = \frac{1}{x\ln^2 x}$ on $(0, \frac{1}{2}]$ has these properties:

• As $x \to 0^+$: $f(x) \to +\infty$ (vertical asymptote)
• At $x = \frac{1}{2}$: $f(\frac{1}{2}) = \frac{1}{\frac{1}{2} \cdot (\ln \frac{1}{2})^2} = \frac{2}{(\ln 2)^2} \approx 4.16$
• The function is decreasing on $(0, \frac{1}{2}]$
• The singularity is "weak enough" that the area converges

Alternative form of the answer

We can also write:

$$\frac{1}{\ln 2} = \frac{1}{\ln 2} \cdot \frac{\ln e}{\ln e} = \frac{\log_2 e}{\ln e} = \log_2 e$$

So:

$$\int_0^{1/2} \frac{dx}{x\ln^2 x} = \log_2 e = \frac{1}{\ln 2} \approx 1.443$$

Direct evaluation without limits (formal)

We can also write formally:

$$\int_0^{1/2} \frac{dx}{x\ln^2 x} = \left[-\frac{1}{\ln x}\right]_0^{1/2}$$

At $x = \frac{1}{2}$:

$$-\frac{1}{\ln \frac{1}{2}} = -\frac{1}{-\ln 2} = \frac{1}{\ln 2}$$

At $x = 0^+$:

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

Therefore:

$$\frac{1}{\ln 2} - 0 = \frac{1}{\ln 2}$$

Summary

1. The integral has a singularity at $x = 0$
2. Using the substitution $u = \ln x$ transforms it to: $$ \int_{\ln \varepsilon}^{-\ln 2} \frac{du}{u^2} $$
3. The antiderivative is $-\frac{1}{\ln x}$
4. Evaluating from 0 to $\frac{1}{2}$: $$ \left[-\frac{1}{\ln x}\right]_0^{1/2} = \frac{1}{\ln 2} - 0 = \frac{1}{\ln 2} $$
5. The integral converges because it's equivalent to $\int \frac{du}{u^2}$ with exponent 2 > 1
6. Result: $\int_0^{1/2} \frac{dx}{x\ln^2 x} = \frac{1}{\ln 2} \approx 1.443$

Key lessons:

• The substitution $u = \ln x$ is powerful for integrands involving logarithms
• The general rule: $\int_0^a \frac{dx}{x(\ln x)^n}$ converges if and only if $n > 1$
• This parallels the $p$-test: $\int \frac{dx}{x^p}$ behavior
• Logarithmic singularities can be more subtle than power singularities
• The antiderivative $-\frac{1}{\ln x}$ should be memorized
• Even complex-looking singularities may lead to convergent integrals