Question

Consider the improper integrals \(I_\lambda = \displaystyle\int_0^1 \dfrac{dx}{(1-x)^\lambda}\) and \(J_\lambda = \displaystyle\int_1^\infty \dfrac{dx}{x^\lambda}\), \(\lambda \in \mathbb{R}\), and the statements:
(I) For \(\lambda = 1\), \(I_\lambda\) converges, but \(J_\lambda\) diverges.
(II) For \(\lambda = 2\), \(I_\lambda\) diverges, but \(J_\lambda\) converges.
Choose the correct option.

• A. Only (I) is true
• B. Only (II) is true
• C. Both (I) and (II) are true
• D. Neither (I) nor (II) is true

Hint:

Substitute u = 1-x in I_lambda to see both integrals are p-type integrals with the same convergence rule lambda < 1 versus lambda > 1.

Answer 1

The Correct Option is B

Step 1: Convert I_lambda to a standard p-integral. In \(I_\lambda = \displaystyle\int_0^1 \dfrac{dx}{(1-x)^\lambda}\), substitute \(u = 1-x\), so \(du = -dx\), and as \(x\) runs from \(0\) to \(1\), \(u\) runs from \(1\) to \(0\). This gives \[I_\lambda = \int_0^1 \frac{du}{u^\lambda}.\]

Step 2: State the convergence rule. The integral \(\displaystyle\int_0^1 u^{-\lambda}\,du\) converges if and only if \(\lambda < 1\) (it is improper at \(u=0\)). Likewise \(J_\lambda = \displaystyle\int_1^\infty x^{-\lambda}\,dx\) converges if and only if \(\lambda > 1\) (improper at infinity).

Step 3: Test statement (I) at lambda = 1. \(I_1 = \displaystyle\int_0^1 \frac{du}{u} = \big[\ln u\big]\), which diverges to \(-\infty\) as \(u \to 0^+\) (logarithmic divergence). \(J_1 = \displaystyle\int_1^\infty \frac{dx}{x} = \big[\ln x\big]\), which diverges to \(\infty\) as \(x\to\infty\). So both \(I_1\) and \(J_1\) diverge; statement (I), which claims \(I_1\) converges, is false.

Step 4: Test statement (II) at lambda = 2. Since \(\lambda = 2 > 1\), by Step 2, \(I_2\) (which needs \(\lambda < 1\) to converge) diverges. Concretely, \(I_2 = \displaystyle\int_0^1 u^{-2}\,du = \left[-\dfrac{1}{u}\right]_0^1\), and as \(u\to0^+\), \(-1/u \to -\infty\), confirming divergence.

Step 5: Compute J_2. \(J_2 = \displaystyle\int_1^\infty x^{-2}\,dx = \left[-\dfrac1x\right]_1^\infty = 0-(-1) = 1\), a finite value, so \(J_2\) converges. This matches statement (II) exactly: \(I_2\) diverges and \(J_2\) converges.

Step 6: Conclude. Statement (I) is false and statement (II) is true. \[\boxed{\text{Only (II) is true}}\]