Determine all real values of the parameter $p$ for which the improper integral ${\int }_1^{\infty }\frac{\ln x}{x^p}dx$ converges. Then evaluate the integral when $p=3$.

Step-by-step solution

1. Examine the behaviour for large $x$: $\ln x$ grows slower than any power of $x$, so the integrand behaves like $1/x^p$.
2. The integral ${\int }_1^{\infty }{x}^{-p}dx$ converges only when $p>1$; therefore convergence requires $p>1$.
3. Near the lower limit $x=1$, $\ln x\to 0$, so no additional restriction arises.
4. Thus the given integral converges exactly for $p>1$.
5. To evaluate at $p=3$, set $I={\int }_1^{\infty }\frac{\ln x}{x^3}dx$.
6. Apply integration by parts: let $u=\ln x$, $dv=x^{-3}dx$.
7. Then $du=\frac{1}{x}dx$ and $v=\int x^{-3}dx=-\frac{1}{2x^2}$.
8. Write $I=uv{|}_1^{\infty }-{\int }_1^{\infty }vdu$.
9. Compute the boundary term: ${\lim }_{x\to \infty }(\ln x\cdot (-\frac{1}{2x^2}))=0$, and at $x=1$, $\ln 1=0$, so the term is $0$.
10. The remaining integral: $-{\int }_1^{\infty }(-\frac{1}{2x^2})\cdot \frac{1}{x}dx=\frac{1}{2}{\int }_1^{\infty }{x}^{-3}dx$.
11. Integrate: ${\int }_1^{\infty }{x}^{-3}dx=[\frac{x^{-2}}{-2}{]}_1^{\infty }=0-(-\frac{1}{2})=\frac{1}{2}$.
12. Multiply by $\frac{1}{2}$: $I=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$.

Concept tested: Convergence of improper integrals with parameters and integration by parts

Published 17 September 2026 · How this solution was prepared and checked