Determine all real values of the parameter $a$ for which the improper integral ${\int }_0^{\infty }\frac{\ln x}{(1+x^2{)}^a}dx$ converges. For those values, evaluate the integral when $a=2$.

Step-by-step solution

1. Split the integral at $x=1$: ${\int }_0^{\infty }={\int }_0^1+{\int }_1^{\infty }$.
2. Examine convergence near $x=0$. As $x\to 0^{+},\ln x\sim -\infty$ but $(1+x^2{)}^a\approx 1$. The integrand behaves like $\ln x$. The integral ${\int }_0^1\ln xdx$ converges (its value is $-1$). Hence no restriction from the lower limit.
3. Examine convergence as $x\to \infty$. For large $x$, $\ln x$ grows slowly, while $(1+x^2{)}^a\sim x^{2a}$. The integrand behaves like $\frac{\ln x}{x^{2a}}$.
4. Compare with the p‑test: ${\int }_1^{\infty }\frac{\ln x}{x^p}dx$ converges if $p>1$. Here $p=2a$. Therefore we need $2a>1$ i.e. $a>\frac{1}{2}$.
5. Thus the integral converges exactly for $a>\frac{1}{2}$.
6. Now evaluate the integral for $a=2$. Write $I={\int }_0^{\infty }\frac{\ln x}{(1+x^2{)}^2}dx$.
7. Make the substitution $x=\frac{1}{t}$; then $dx=-\frac{1}{t^2}dt$ and $\ln x=-\ln t$. Also $1+x^2=1+\frac{1}{t^2}=\frac{t^2+1}{t^2}$. Hence $(1+x^2{)}^2=\frac{(t^2+1{)}^2}{t^4}$.
8. Transform the integral: $I={\int }_{\infty }^0\frac{-\ln t}{\frac{(t^2+1{)}^2}{t^4}}\cdot \left(-\frac{1}{t^2}\right)dt={\int }_0^{\infty }\frac{\ln t}{(1+t^2{)}^2}dt$.
9. Add the original integral and the transformed one: $2I={\int }_0^{\infty }\frac{\ln x+\ln x}{(1+x^2{)}^2}dx=0$ because the integrand is an odd function of $\ln x$ under the substitution $x\to 1/x$. More formally, the two expressions cancel each other, giving $I=0$.
10. Hence for $a=2$ the value of the integral is $0$.

Concept tested: Convergence of improper integrals with parameters

Published 13 September 2026 · How this solution was prepared and checked