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∫ 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)g(x) is given by g(x)=x1+(log⁡ex)(log⁡ex)⋯∞g(x)=\dfrac{x}{1+(\log_{e}x)(\log_{e}x)\cdots\infty}, where x∈[1,∞)x\in[1,\infty). Then ∫12eg(x) dx\displaystyle\int_{1}^{2e}g(x)\,dx is equal to
A12 ⁣(e−1e)\dfrac12\!\left(e-\dfrac1e\right)
B12 ⁣(e+1e)\dfrac12\!\left(e+\dfrac1e\right)
C12(e2−1)\dfrac12\big(e^{2}-1\big)correct
D12(e2+1)\dfrac12\big(e^{2}+1\big)
Solution
Step 1: Interpret the denominator. The repeated product (log⁡ex)(log⁡ex)⋯∞(\log_{e}x)(\log_{e}x)\cdots\infty is lim⁡m→∞(log⁡ex)m\displaystyle\lim_{m\to\infty}(\log_{e}x)^{m}. Its value depends on whether log⁡ex\log_{e}x is below, equal to, or above 11. Step 2: Evaluate the limit on the three regions. For 1<x<e1<x<e: 0<log⁡ex<10<\log_{e}x<1, so (log⁡ex)m→0(\log_{e}x)^{m}\to 0. For x=ex=e: log⁡ex=1\log_{e}x=1, so the product =1=1. For x>ex>e: log⁡ex>1\log_{e}x>1, so (log⁡ex)m→∞(\log_{e}x)^{m}\to\infty. Step 3: Substitute into g(x)=x1+(log⁡ex)∞g(x)=\dfrac{x}{1+(\log_{e}x)^{\infty}}:
g(x)={x1+0=x,1<x<e,e1+1=e2,x=e,x1+∞=0,x>e.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=ex=e does not affect the integral. Step 4: Integrate over [1,2e][1,2e]:
∫12eg(x) dx=∫1ex dx+∫e2e0 dx=[x22]1e=e2−12.\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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