Step 1: Use the property of definite integrals
The integral \( \int_{-a}^a f(x) \, dx = 0 \) if the function \( f(x) \) is odd. A function \( f(x) \) is odd if: \[ f(-x) = -f(x). \]
Step 2: Simplify the integral for odd functions
For odd functions: \[ \int_{-a}^a f(x) \, dx = \int_{-a}^0 f(x) \, dx + \int_{0}^a f(x) \, dx. \]
Since \( f(-x) = -f(x) \), the integral over \( [-a, 0] \) cancels with the integral over \( [0, a] \), resulting in: \[ \int_{-a}^a f(x) \, dx = 0. \]
Step 3: {Conclude the result}
The given integral equals 0 when \( f(-x) = -f(x) \).
Let \[ f(t)=\int \left(\frac{1-\sin(\log_e t)}{1-\cos(\log_e t)}\right)dt,\; t>1. \] If $f(e^{\pi/2})=-e^{\pi/2}$ and $f(e^{\pi/4})=\alpha e^{\pi/4}$, then $\alpha$ equals