Evaluation of Definite Integral with Exponential and Logarithmic Limits

To find the value of the given definite integral, we first determine the minimum value of $f(a)$, denoted as $\alpha$.

Step 1: Determine the function $f(a)$ and its minimum value $\alpha$

We are given that $\left(2^{1-a} + 2^{1+a}\right)$, $f(a)$, and $\left(3^a + 3^{-a}\right)$ are in Arithmetic Progression (A.P.).

By the definition of an A.P., the middle term is the arithmetic mean of the outer terms: $2 f(a) = \left(2^{1-a} + 2^{1+a}\right) + \left(3^a + 3^{-a}\right)$

Simplifying the term $2^{1-a} + 2^{1+a}$: $2^{1-a} + 2^{1+a} = 2 \cdot 2^{-a} + 2 \cdot 2^a = 2\left(2^a + 2^{-a}\right)$

Substituting this back gives: $2 f(a) = 2\left(2^a + 2^{-a}\right) + \left(3^a + 3^{-a}\right)$ $f(a) = \left(2^a + 2^{-a}\right) + \frac{1}{2}\left(3^a + 3^{-a}\right)$

Using the Arithmetic Mean - Geometric Mean (AM-GM) inequality for positive real numbers, for any $x > 0$: $x + \frac{1}{x} \ge 2$ with equality holding if and only if $x = 1$.

Applied to $2^a$ and $3^a$:

• $2^a + 2^{-a} \ge 2$, with equality when $2^a = 1 \implies a = 0$.
• $3^a + 3^{-a} \ge 2$, with equality when $3^a = 1 \implies a = 0$.

Since both components achieve their respective minimum values simultaneously at $a = 0$, the minimum value of $f(a)$ is: $\alpha = f(0) = \left(2^0 + 2^0\right) + \frac{1}{2}\left(3^0 + 3^0\right) = 2 + \frac{1}{2}(2) = 3$

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Step 2: Evaluate the Definite Integral

With $\alpha = 3$, the limits of the integral are:

• Lower limit: $\log_e(\alpha - 1) = \log_e(3 - 1) = \log_e 2$
• Upper limit: $\log_e(\alpha) = \log_e 3$

The integral to evaluate is: $I = \int_{\log_e 2}^{\log_e 3} \frac{dx}{e^{2x} - e^{-2x}}$

Multiplying the numerator and denominator by $e^{2x}$: $I = \int_{\log_e 2}^{\log_e 3} \frac{e^{2x}}{e^{4x} - 1} \, dx$

Let $u = e^{2x}$. Differentiating both sides gives: $du = 2 e^{2x} \, dx \implies e^{2x} \, dx = \frac{du}{2}$

Now, change the limits of integration in terms of $u$:

• When $x = \log_e 2$: $u = e^{2\log_e 2} = e^{\log_e 4} = 4$
• When $x = \log_e 3$: $u = e^{2\log_e 3} = e^{\log_e 9} = 9$

Substituting these into the integral: $I = \int_{4}^{9} \frac{1}{u^2 - 1} \frac{du}{2} = \frac{1}{2} \int_{4}^{9} \frac{du}{u^2 - 1}$

Using the standard integration formula $\int \frac{du}{u^2 - a^2} = \frac{1}{2a} \log_e \left| \frac{u-a}{u+a} \right|$: $I = \frac{1}{2} \left[ \frac{1}{2} \log_e \left( \frac{u - 1}{u + 1} \right) \right]_{4}^{9}$ $I = \frac{1}{4} \left[ \log_e \left( \frac{9 - 1}{9 + 1} \right) - \log_e \left( \frac{4 - 1}{4 + 1} \right) \right]$ $I = \frac{1}{4} \left[ \log_e \left( \frac{8}{10} \right) - \log_e \left( \frac{3}{5} \right) \right]$ $I = \frac{1}{4} \left[ \log_e \left( \frac{4}{5} \right) - \log_e \left( \frac{3}{5} \right) \right]$

Applying logarithmic properties $\log_e(A) - \log_e(B) = \log_e\left(\frac{A}{B}\right)$: $I = \frac{1}{4} \log_e \left( \frac{4/5}{3/5} \right) = \frac{1}{4} \log_e \left( \frac{4}{3} \right)$

Hence, the correct option is B.