Calculate the improper integral (or establish its

Calculate the improper integral (or establish its divergence): $$ \int_0^e x|\ln x| \, dx $$

This is an improper integral because $\ln x \to -\infty$ as $x \to 0^+$, making $|\ln x| \to +\infty$ near $x = 0$.

Step 1: Analyze the behavior near $x = 0$

As $x \to 0^+$:

• $\ln x \to -\infty$
• $|\ln x| = -\ln x \to +\infty$
• $x|\ln x| = -x\ln x \to 0$ (since $x$ goes to 0 faster than $\ln x$ goes to $-\infty$)

More precisely, using L'Hôpital's rule:

$$\lim_{x \to 0^+} x\ln x = \lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} \frac{-x^2}{x} = \lim_{x \to 0^+} (-x) = 0$$

So the singularity is removable, but we still need to evaluate as an improper integral.

Step 2: Handle the absolute value

For $x \in (0, e]$, we need to determine the sign of $\ln x$:

• When $0 < x < 1$: $\ln x < 0$, so $|\ln x| = -\ln x$
• When $x = 1$: $\ln 1 = 0$, so $|\ln 1| = 0$
• When $1 < x \leq e$: $\ln x > 0$, so $|\ln x| = \ln x$

Therefore, we split the integral at $x = 1$:

$$\int_0^e x|\ln x| \, dx = \int_0^1 x|\ln x| \, dx + \int_1^e x|\ln x| \, dx$$

$$= \int_0^1 x(-\ln x) \, dx + \int_1^e x\ln x \, dx$$

$$= -\int_0^1 x\ln x \, dx + \int_1^e x\ln x \, dx$$

Step 3: Evaluate $\int x\ln x \, dx$ using integration by parts

Let $u = \ln x$ and $dv = x \, dx$

Then: $du = \frac{1}{x} \, dx$ and $v = \frac{x^2}{2}$

Using integration by parts:

$$\int x\ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx$$

$$= \frac{x^2}{2} \ln x - \int \frac{x}{2} \, dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$$

$$= \frac{x^2}{4}(2\ln x - 1) + C$$

Step 4: Evaluate the first integral $-\int_0^1 x\ln x \, dx$

$$-\int_0^1 x\ln x \, dx = -\lim_{\varepsilon \to 0^+} \int_\varepsilon^1 x\ln x \, dx$$

$$= -\lim_{\varepsilon \to 0^+} \left[\frac{x^2}{4}(2\ln x - 1)\right]_\varepsilon^1$$

At $x = 1$:

$$\frac{1^2}{4}(2\ln 1 - 1) = \frac{1}{4}(0 - 1) = -\frac{1}{4}$$

At $x = \varepsilon$:

$$\frac{\varepsilon^2}{4}(2\ln \varepsilon - 1)$$

We need to find:

$$\lim_{\varepsilon \to 0^+} \frac{\varepsilon^2}{4}(2\ln \varepsilon - 1)$$

As $\varepsilon \to 0^+$:

$$\varepsilon^2 \to 0 \quad \text{and} \quad (2\ln \varepsilon - 1) \to -\infty$$

This is an indeterminate form $0 \cdot (-\infty)$. Rewrite:

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

We know $\lim_{\varepsilon \to 0^+} \varepsilon^2 = 0$, and using L'Hôpital's rule:

$$\lim_{\varepsilon \to 0^+} \varepsilon^2\ln \varepsilon = \lim_{\varepsilon \to 0^+} \frac{\ln \varepsilon}{1/\varepsilon^2} = \lim_{\varepsilon \to 0^+} \frac{1/\varepsilon}{-2/\varepsilon^3} = \lim_{\varepsilon \to 0^+} \frac{-\varepsilon^3}{2\varepsilon} = \lim_{\varepsilon \to 0^+} \frac{-\varepsilon^2}{2} = 0$$

Therefore:

$$\lim_{\varepsilon \to 0^+} \varepsilon^2(2\ln \varepsilon - 1) = 2 \cdot 0 - 0 = 0$$

So:

$$-\int_0^1 x\ln x \, dx = -\left(-\frac{1}{4} - 0\right) = \frac{1}{4}$$

Step 5: Evaluate the second integral $\int_1^e x\ln x \, dx$

$$\int_1^e x\ln x \, dx = \left[\frac{x^2}{4}(2\ln x - 1)\right]_1^e$$

At $x = e$:

$$\frac{e^2}{4}(2\ln e - 1) = \frac{e^2}{4}(2 \cdot 1 - 1) = \frac{e^2}{4}(1) = \frac{e^2}{4}$$

At $x = 1$:

$$\frac{1}{4}(2\ln 1 - 1) = \frac{1}{4}(0 - 1) = -\frac{1}{4}$$

Therefore:

$$\int_1^e x\ln x \, dx = \frac{e^2}{4} - \left(-\frac{1}{4}\right) = \frac{e^2}{4} + \frac{1}{4} = \frac{e^2 + 1}{4}$$

Step 6: Combine the results

$$\int_0^e x|\ln x| \, dx = -\int_0^1 x\ln x \, dx + \int_1^e x\ln x \, dx$$

$$= \frac{1}{4} + \frac{e^2 + 1}{4} = \frac{1 + e^2 + 1}{4} = \frac{e^2 + 2}{4}$$

Final Answer

$$\boxed{\int_0^e x|\ln x| \, dx = \frac{e^2 + 2}{4}}$$

Numerically: $\frac{e^2 + 2}{4} = \frac{7.389... + 2}{4} = \frac{9.389...}{4} \approx 2.347$

The integral converges.

Verification of the antiderivative

Let's verify that $\frac{d}{dx}\left[\frac{x^2}{4}(2\ln x - 1)\right] = x\ln x$:

$$\frac{d}{dx}\left[\frac{x^2}{4}(2\ln x - 1)\right] = \frac{1}{4}\frac{d}{dx}[x^2(2\ln x - 1)]$$

Using the product rule:

$$= \frac{1}{4}\left[2x(2\ln x - 1) + x^2 \cdot \frac{2}{x}\right]$$

$$= \frac{1}{4}\left[2x(2\ln x - 1) + 2x\right] = \frac{1}{4}\left[4x\ln x - 2x + 2x\right]$$

$$= \frac{1}{4} \cdot 4x\ln x = x\ln x$$

Correct! ✓

Geometric interpretation

The function $f(x) = x|\ln x|$ has these properties:

• At $x = 0^+$: $f(x) \to 0$ (removable singularity)
• At $x = 1$: $f(1) = 1 \cdot 0 = 0$ (minimum)
• At $x = e$: $f(e) = e \cdot 1 = e \approx 2.718$
• For $0 < x < 1$: $f(x) = -x\ln x$ (positive, since $\ln x < 0$)
• For $x > 1$: $f(x) = x\ln x$ (positive, since $\ln x > 0$)

The function is continuous and positive on $(0, e]$, with a minimum at $x = 1$.

Why the integral converges

Despite the apparent singularity at $x = 0$ (where $|\ln x| \to \infty$), the integral converges because:

1. The factor $x \to 0$ dominates the growth of $|\ln x|$

2. Near $x = 0$: $x|\ln x| = x(-\ln x) \sim x \cdot |\ln x|$, but $x$ decreases faster than $\ln x$ increases

3. The limit $\lim_{x \to 0^+} x\ln x = 0$ shows the singularity is removable

Alternative form of the answer

We can also write:

$$\int_0^e x|\ln x| \, dx = \frac{e^2 + 2}{4} = \frac{e^2}{4} + \frac{1}{2}$$

Summary

1. The integral has an apparent singularity at $x = 0$, but it's removable since $\lim_{x \to 0^+} x\ln x = 0$
2. We split the integral at $x = 1$ to handle $|\ln x|$: $$ \int_0^e x|\ln x| \, dx = -\int_0^1 x\ln x \, dx + \int_1^e x\ln x \, dx $$
3. Using integration by parts: $\int x\ln x \, dx = \frac{x^2}{4}(2\ln x - 1) + C$
4. Evaluating both parts and combining: $$ \int_0^e x|\ln x| \, dx = \frac{1}{4} + \frac{e^2 + 1}{4} = \frac{e^2 + 2}{4} $$
5. The integral converges to $\frac{e^2 + 2}{4} \approx 2.347$

Key lessons:

• Not all apparent singularities cause divergence
• Absolute values require splitting the integral based on where the argument changes sign
• Integration by parts is essential for integrals involving logarithms
• The limit $\lim_{x \to 0^+} x\ln x = 0$ is a fundamental result
• Careful analysis of limits at singularities determines convergence
• The factor $x$ near 0 dominates the logarithmic growth, ensuring convergence