Definite IntegrationmediumFree

∫ g(x) dx with infinite power of ln x | JEE

JEE Maths question with a full step-by-step solution.

Question

A function

g(x)

is given by

g(x)=\dfrac{x}{1+(\log_{e}x)(\log_{e}x)\cdots\infty}

, where

x\in[1,\infty)

. Then

\displaystyle\int_{1}^{2e}g(x)\,dx

is equal to

\dfrac12\!\left(e-\dfrac1e\right)

\dfrac12\!\left(e+\dfrac1e\right)

\dfrac12\big(e^{2}-1\big)

correct

\dfrac12\big(e^{2}+1\big)

Solution

Step 1: Interpret the denominator. The repeated product

(\log_{e}x)(\log_{e}x)\cdots\infty

\displaystyle\lim_{m\to\infty}(\log_{e}x)^{m}

. Its value depends on whether

\log_{e}x

is below, equal to, or above

1

. Step 2: Evaluate the limit on the three regions. For

1<x<e

0<\log_{e}x<1

, so

(\log_{e}x)^{m}\to 0

. For

x=e

\log_{e}x=1

, so the product

=1

. For

x>e

\log_{e}x>1

, so

(\log_{e}x)^{m}\to\infty

. Step 3: Substitute into

g(x)=\dfrac{x}{1+(\log_{e}x)^{\infty}}

g(x)=\begin{cases}\dfrac{x}{1+0}=x,&1<x<e,\\[6pt]\dfrac{e}{1+1}=\dfrac e2,&x=e,\\[6pt]\dfrac{x}{1+\infty}=0,&x>e.\end{cases}

The single point

x=e

does not affect the integral. Step 4: Integrate over

[1,2e]

\int_{1}^{2e}g(x)\,dx=\int_{1}^{e}x\,dx+\int_{e}^{2e}0\,dx=\left[\dfrac{x^{2}}{2}\right]_{1}^{e}=\dfrac{e^{2}-1}{2}.

Correct answer: (3)

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