For which values of the real parameter $a$ does the improper integral ${\int }_0^1{x}^a\ln xdx$ converge? When it converges, find its value.

Step-by-step solution

1. Examine convergence near $x=0$. For $x\to 0^{+}$, $x^a\ln x\sim x^a\ln x$; integrable iff $a>-1$ (since ${\int }_0{x}^{a}dx$ converges for $a>-1$). Near $x=1$ the integrand is bounded, so no issue.
2. Thus the integral converges for all real $a$ such that $a>-1$.
3. Compute the integral using integration by parts: let $u=\ln x$, $dv=x^{a}dx$; then $du=\frac{1}{x}dx$ and $v=\frac{x^{a+1}}{a+1}$ (valid for $a\ne -1$).
4. Apply the formula: $I=uv{|}_0^1-{\int }_0^{1}vdu={\frac{x^{a+1}}{a+1}\ln x|}_0^1-{\int }_0^1\frac{x^{a+1}}{a+1}\cdot \frac{1}{x}dx$.
5. The boundary term evaluates to $0$ (since $x^{a+1}\ln x\to 0$ as $x\to 0^{+}$ and equals $0$ at $x=1$). The remaining integral is $\frac{1}{a+1}{\int }_0^1{x}^{a}dx=\frac{1}{a+1}\cdot \frac{1}{a+1}=\frac{1}{(a+1{)}^2}$.
6. Hence $I=-\frac{1}{(a+1{)}^2}$.
7. So for $a>-1$ the integral equals $-\frac{1}{(a+1{)}^2}$.
8. If $a\le -1$, the integral diverges.

Concept tested: Improper integral with parameter

Published 14 September 2026 · How this solution was prepared and checked