Question

Use integration by parts to evaluate the following integrals. $$\int_{1}^{\infty} \frac{\ln x}{x^{2}} d x$$

Short answer

Answer: The value of the definite integral is -1.

Step-by-step solution

Identify u and dv

Choose: $$u = \ln x \quad \Rightarrow \quad d u=(1/x) d x$$ and $$d v=x^{-2} d x \quad \Rightarrow \quad v = -\frac{1}{x}$$

Apply integration by parts formula

Using integration by parts formula: $$\int u \, dv = uv - \int v \, du$$ Substitute u, du, and v from above: $$\int \frac{\ln x}{x^2} dx = -\frac{\ln x}{x} - \int \frac{1}{x^2} dx$$

Evaluate the remaining integral

Now, evaluate the integral \(\int \frac{1}{x^2} dx\) by finding its antiderivative. $$\int \frac{1}{x^2} dx = \int x^{-2} dx = -\frac{1}{x} + C$$ Substitute back into the previous equation: $$\int \frac{\ln x}{x^2} dx = -\frac{\ln x}{x} + \frac{1}{x} + C$$

Evaluate the definite integral

Now, find the definite integral: $$\int_{1}^{\infty} \frac{\ln x}{x^2} dx = \lim_{b\to\infty} \int_{1}^{b} \frac{\ln x}{x^2} dx$$ Evaluate the antiderivative at the limits of integration: $$= \lim_{b\to\infty} \left[ -\frac{\ln x}{x} + \frac{1}{x} \right]_1^b = \lim_{b\to\infty} \left[ -\frac{\ln b}{b} + \frac{1}{b} + \frac{\ln 1}{1} - 1 \right]$$ Since \(\ln 1 = 0\) and \(\lim_{b\to\infty} \frac{\ln b}{b} = 0\) as well as \(\lim_{b\to\infty} \frac{1}{b}=0\), we have: $$= 0 + 0 + 0 - 1 = -1$$ The definite integral is equal to -1: $$\int_{1}^{\infty} \frac{\ln x}{x^2} dx = -1$$