Determine for which values of the parameter a the improper integral ∫_0^1x^-aln x dx converges, and give its value when it converges.

Converges for a<1 and its value is -1/(1-a)^2; diverges for a≥1

Determine for which values of the parameter $a$ the improper integral $\int_{0}^{1} x^{- a} ln x d x$ converges, and give its value when it converges.

hard 4 marks Class 12 · Mathematics · Integrals

Answer

Converges for

a < 1

and its value is

- \frac{1}{( 1 - a ) ^{2}}

; diverges for

a \geq 1

The correct answer is Converges for $a < 1$ and its value is $- \frac{1}{( 1 - a ) ^{2}}$ ; diverges for $a \geq 1$ — If $a \geq 1$ , the condition $b > 0$ fails, so the integral diverges.

Step-by-step solution

Rewrite $x^{- a}$ as $x^{(- a)} = x^{b - 1}$ with $b = 1 - a$ .
Recognize the standard integral $\int_{0}^{1} x^{b - 1} ln x d x = - \frac{1}{b ^{2}}$ for $b > 0$ .
Determine the condition for convergence: $b > 0 ⟹ 1 - a > 0 ⟹ a < 1$ .
For $a < 1$ , substitute $b = 1 - a$ to get the value $- \frac{1}{( 1 - a ) ^{2}}$ .
If $a \geq 1$ , the condition $b > 0$ fails, so the integral diverges.

Common mistake: Using

\int_{0}^{1} x^{b - 1} ln x d x = - 1/ b

instead of

- 1/ b^{2}

or overlooking the condition

b > 0

Concept tested: Convergence of logarithmic improper integrals

Published 7 September 2026 · How this solution was prepared and checked

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Determine for which values of the parameter a the improper integral ∫01​x−alnxdx converges, and give its value when it converges.

Step-by-step solution

Determine for which values of the parameter $a$ the improper integral $\int_{0}^{1} x^{- a} ln x d x$ converges, and give its value when it converges.