Convergency and Divergency!

Calculus Level 5

$\large I(n) =\int_0^1 \dfrac{\ln(1-t)}{(1-t)^n} \, dt \qquad \qquad G(n) = \int_0^1 \dfrac{\ln(1+t)}{(1+t)^n} \, dt$

Let $I(n)$ and $G(n)$ be functions as shown above. Which of the following statements is/are true?

(A) : $G(n)$ is convergent for all $n$ .

(B) : I(n) is divergent for n>1

(C) : $I(n) + G(n)$ is convergent for all $n$ < 1.

(D) : $\displaystyle e^2 = \left( \left( \sum_{r=0}^\infty \frac1{I^{(r)}(2)} \right)+ 1 \right)^{-1}$ , where $I^{(r)} (\cdot)$ denotes the $r^\text{th}$ derivative of $I$ .

(E) : $I(n) : [1,\infty) \to \mathbb R$ is a bijective function .

B and C only A, D and E only A, B and C only C and D only A and D only C and E only A, B, C, D and E B, C and E only

1 solution

A Former Brilliant Member
Oct 16, 2016

Note that the integrand in the definition of $G(n)$ is continuous on the relevant interval for any fixed $n$ and hence (A) follows.

With a bit of substitutions, we get the following : $I(n) = \int_{-\infty}^0 \, te^{-t(n-1)} \, dt$

Hence, (B) and (C) are true. Additionally it shows that (D) and (E) are nonsensical.

In conclusion : Only (A),(B) and (C) are correct.

Nice solution. I (n) can also be evaluated directly by differentiation under integral sign. Which shows that it is divergent for n>1

Abhi Kumbale - 4 years, 8 months ago

Yes. But that statement is way stronger than what is required. So I chose to avoid it.

A Former Brilliant Member - 4 years, 8 months ago

Convergency and Divergency!

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