The general form of a sine function is: \[ y = \sin(kx) \] where the period of the sine function is given by: \[ {Period} = \frac{2\pi}{|k|} \] Here, \( k \) is the coefficient of \( x \) in the argument of the sine function. In our case, the function is \( \sin\left( \frac{\pi x}{4} \right) \), so \( k = \frac{\pi}{4} \).
Using the formula for the period, we get: \[ {Period} = \frac{2\pi}{\left|\frac{\pi}{4}\right|} = \frac{2\pi}{\frac{\pi}{4}} = 8 \] Thus, the period of the function \( \sin\left( \frac{\pi x}{4} \right) \) is \( 8 \). Thus, the correct answer is \( \boxed{8} \), corresponding to option (D).
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