Calculate the improper integral: integral from 1 to ()

Calculate the improper integral: $$ \int_1^\infty \frac{x\ln x}{(1+x^2)^2} \, dx $$

This is an improper integral at infinity with a logarithmic numerator and polynomial-type denominator.

Step 1: Express as a limit

$$\int_1^\infty \frac{x\ln x}{(1+x^2)^2} \, dx = \lim_{R \to \infty} \int_1^R \frac{x\ln x}{(1+x^2)^2} \, dx$$

Step 2: Use substitution to find the antiderivative

Let $u = 1 + x^2$, so $du = 2x \, dx$, which means $x \, dx = \frac{1}{2} du$.

Also, from $u = 1 + x^2$, we get $x^2 = u - 1$, so $x = \sqrt{u-1}$ (for $x > 0$).

Therefore: $\ln x = \frac{1}{2}\ln(u-1)$

The integral becomes:

$$\int \frac{x\ln x}{(1+x^2)^2} \, dx = \int \frac{\ln x}{u^2} \cdot \frac{1}{2} \, du = \frac{1}{2}\int \frac{\frac{1}{2}\ln(u-1)}{u^2} \, du$$

$$= \frac{1}{4}\int \frac{\ln(u-1)}{u^2} \, du$$

This still requires integration by parts. Let's go back to the original integral.

Step 2 (Alternative): Integration by parts

Let's use integration by parts directly on:

$$I = \int \frac{x\ln x}{(1+x^2)^2} \, dx$$

Let $u = \ln x$ and $dv = \frac{x}{(1+x^2)^2} \, dx$

Then: $du = \frac{1}{x} \, dx$

For $v$, we need to find $\int \frac{x}{(1+x^2)^2} \, dx$.

Using substitution $w = 1 + x^2$, $dw = 2x \, dx$:

$$\int \frac{x}{(1+x^2)^2} \, dx = \frac{1}{2}\int \frac{dw}{w^2} = \frac{1}{2} \cdot \left(-\frac{1}{w}\right) = -\frac{1}{2w} = -\frac{1}{2(1+x^2)}$$

So $v = -\frac{1}{2(1+x^2)}$

By integration by parts:

$$I = uv - \int v \, du = \ln x \cdot \left(-\frac{1}{2(1+x^2)}\right) - \int \left(-\frac{1}{2(1+x^2)}\right) \cdot \frac{1}{x} \, dx$$

$$= -\frac{\ln x}{2(1+x^2)} + \frac{1}{2}\int \frac{1}{x(1+x^2)} \, dx$$

Step 3: Evaluate $\int \frac{1}{x(1+x^2)} \, dx$ using partial fractions

$$\frac{1}{x(1+x^2)} = \frac{A}{x} + \frac{Bx + C}{1+x^2}$$

Multiplying by $x(1+x^2)$:

$$1 = A(1+x^2) + (Bx+C)x = A + Ax^2 + Bx^2 + Cx$$

Comparing coefficients:

• Constant: $1 = A$
• $x$: $0 = C$
• $x^2$: $0 = A + B \implies B = -A = -1$

So: $\frac{1}{x(1+x^2)} = \frac{1}{x} - \frac{x}{1+x^2}$

Therefore:

$$\int \frac{1}{x(1+x^2)} \, dx = \int \frac{1}{x} \, dx - \int \frac{x}{1+x^2} \, dx$$

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

$$= \ln\frac{x}{\sqrt{1+x^2}} + C$$

Step 4: Complete the antiderivative

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

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

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

Step 5: Evaluate the definite integral

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

At $x = R$:

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

At $x = 1$:

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

Therefore:

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

Step 6: Take the limit as $R \to \infty$

$$\lim_{R \to \infty} \left[-\frac{\ln R}{2(1+R^2)} + \frac{\ln R}{2} - \frac{\ln(1+R^2)}{4} + \frac{\ln 2}{4}\right]$$

Let's analyze each term:

Term 1: $-\frac{\ln R}{2(1+R^2)}$

As $R \to \infty$: $\frac{\ln R}{1+R^2} \approx \frac{\ln R}{R^2} \to 0$ (logarithmic growth vs quadratic growth)

So: $-\frac{\ln R}{2(1+R^2)} \to 0$

Term 2: $\frac{\ln R}{2}$

As $R \to \infty$: $\frac{\ln R}{2} \to +\infty$

Term 3: $-\frac{\ln(1+R^2)}{4}$

As $R \to \infty$: $\ln(1+R^2) \approx \ln(R^2) = 2\ln R$

So: $-\frac{\ln(1+R^2)}{4} \approx -\frac{2\ln R}{4} = -\frac{\ln R}{2}$

Combining Terms 2 and 3:

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

More precisely:

$$\ln(1+R^2) = \ln(R^2(1/R^2 + 1)) = \ln R^2 + \ln(1 + 1/R^2) = 2\ln R + \ln(1 + 1/R^2)$$

As $R \to \infty$: $\ln(1 + 1/R^2) \to \ln 1 = 0$

Therefore:

$$\frac{\ln R}{2} - \frac{2\ln R + \ln(1+1/R^2)}{4} = \frac{\ln R}{2} - \frac{\ln R}{2} - \frac{\ln(1+1/R^2)}{4}$$

$$= -\frac{\ln(1+1/R^2)}{4} \to -\frac{\ln 1}{4} = 0$$

Step 7: Final result

$$\lim_{R \to \infty} \left[-\frac{\ln R}{2(1+R^2)} + \frac{\ln R}{2} - \frac{\ln(1+R^2)}{4} + \frac{\ln 2}{4}\right]$$

$$= 0 + 0 + \frac{\ln 2}{4} = \frac{\ln 2}{4}$$

Final Answer

$$\boxed{\int_1^\infty \frac{x\ln x}{(1+x^2)^2} \, dx = \frac{\ln 2}{4}}$$

The integral converges to $\frac{\ln 2}{4} \approx 0.173$.

Why the integral converges

For large $x$, the integrand behaves like:

$$\frac{x\ln x}{(1+x^2)^2} \approx \frac{x\ln x}{x^4} = \frac{\ln x}{x^3}$$

We know that $\int_1^\infty \frac{\ln x}{x^p} \, dx$ converges for $p > 1$.

Since $p = 3 > 1$, the integral converges. ✓

Verification of the limit calculation

Let's verify more carefully:

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

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

As $R \to \infty$:

$$\frac{R^2}{1+R^2} = \frac{1}{1/R^2 + 1} \to \frac{1}{0+1} = 1$$

Therefore:

$$\frac{\ln\frac{R^2}{1+R^2}}{4} \to \frac{\ln 1}{4} = 0$$

This confirms our calculation! ✓

Summary

1. The integral is improper at $x = \infty$
2. Using integration by parts with $u = \ln x$ and $dv = \frac{x}{(1+x^2)^2} \, dx$
3. Partial fractions for $\int \frac{1}{x(1+x^2)} \, dx$
4. Evaluating limits carefully, noting cancellations
5. The integral converges to $\frac{\ln 2}{4}$
6. Convergence assured by the $\frac{\ln x}{x^3}$ behavior at infinity

Key lessons:

• Integration by parts is essential for logarithmic integrands
• Partial fractions decompose rational functions
• Careful limit analysis reveals subtle cancellations
• Terms that individually diverge can cancel to give a finite result
• The $\frac{\ln x}{x^p}$ decay with $p > 1$ ensures convergence
• Beautiful result: the answer involves $\ln 2$, connecting to the lower limit