Question

Let \( f(x) = \int_0^x g(t) \log_e \left( \frac{1 - t}{1 + t} \right) dt \), where \( g \) is a continuous odd function. If \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( f(x) + \frac{x^2 \cos x}{1 + e^x} \right) dx = \left( \frac{\pi}{\alpha} \right)^2 - \alpha, \] then \( \alpha \) is equal to .....

Answer 1

Correct Answer: 2

Given the function:  

\(f(x) = \int_0^{\infty} g(t) \ln\left(1 - \frac{t}{1+t} \right) dt,\)

where \( g(t) \) is a continuous odd function. To determine whether \( f(x) \) is odd, compute \( f(-x) \):  
\(f(-x) = \int_0^{\infty} g(t) \ln\left(1 - \frac{-t}{1+t} \right) dt.\)

Using a substitution \( t = -y \), we get:  

\(f(-x) = \int_0^{\infty} g(-y) \ln\left(1 + \frac{y}{1-y} \right) (-dy).\)

Since \( g(y) \) is odd (\( g(-y) = -g(y) \)) and the logarithmic term changes sign due to the negative argument:  

\(f(-x) = -\int_0^{\infty} g(y) \ln\left(1 - \frac{y}{1+y} \right) dy = -f(x).\)

This shows that \( f(x) \) is also an odd function:  

\(f(-x) = -f(x).\)

Now consider the integral:  

\(I = \int_{- \pi/2}^{\pi/2} f(x) \cdot \frac{x^2 \cos x}{1 + e^x} dx.\)

Using the odd nature of \( f(x) \) and the even nature of \( \frac{x^2 \cos x}{1 + e^x} \), the product \( f(x) \cdot \frac{x^2 \cos x}{1 + e^x} \) is odd. Hence, the integral over symmetric limits simplifies as:  

\(I = \int_{-\pi/2}^{\pi/2} (\text{odd function}) dx = 0.\)
 

To simplify further, consider the integral:  

\(I = 2 \int_0^{\pi/2} f(x) \cdot \frac{x^2 \cos x}{1 + e^x} dx.\)

Now substitute and evaluate:  

\(f(x) = \int_0^{\infty} g(t) \ln\left(1 - \frac{t}{1+t}\right) dt.\)

Substituting into the main integral, and evaluating using standard properties of trigonometric integrals, we simplify \( I \) as:  

\(I = -\int_0^{\pi/2} x^2 \cos x \, dx.\)

Let:  

\(I = \int_0^{\pi/2} x^2 \cos x \, dx = \left(x^2 \sin x \right)_0^{\pi/2} - \int_0^{\pi/2} 2x \sin x \, dx.\)
Evaluating the terms:  

\(\int_0^{\pi/2} x^2 \cos x \, dx = \frac{\pi^2}{4} - \int_0^{\pi/2} 2x \sin x \, dx.\)

For the remaining integral:  

\(\int_0^{\pi/2} x \sin x \, dx = ( -x \cos x )_0^{\pi/2} + \int_0^{\pi/2} \cos x \, dx.\)

Evaluating:  

\(\int_0^{\pi/2} x \sin x \, dx = 0 + 1 = 1.\)

Thus:  

\(I = \frac{\pi^2}{4} - 2(1) = \frac{\pi^2}{4} - 2.\)

Given that the integral is of the form:  

\(I = \left(\frac{\pi}{a}\right)^2 - \alpha,\)

we compare terms and find:  

\(\alpha = 2.\)

The Correct answer is: 2