Let $$f(t)=\int_{}^{}\left(\frac{1-\sin(\log_{e}{t})}{1-\cos(\log_{e}{t})}\right)dt,t > 1$$.
If $$f(e^{\pi/2})=-e^{\pi/2}\text{ and }f(e^{\pi/4})=\alpha e^{\pi/4}$$, then $$\alpha$$ equals

We need to find $$\alpha$$ where $$f(t) = \int \frac{1 - \sin(\ln t)}{1 - \cos(\ln t)} \,dt$$ with $$f(e^{\pi/2}) = -e^{\pi/2}$$ and $$f(e^{\pi/4}) = \alpha \cdot e^{\pi/4}$$.

Substituting $$u = \ln t$$ gives $$t = e^u$$ and $$dt = e^u\,du$$.

$$f = \int \frac{1 - \sin u}{1 - \cos u}\,e^u \,du$$

Using half-angle formulas $$1 - \cos u = 2\sin^2\!\tfrac{u}{2}$$ and $$1 - \sin u = 1 - 2\sin\!\tfrac{u}{2}\cos\!\tfrac{u}{2}$$, we have

$$\frac{1 - \sin u}{1 - \cos u} = \frac{1 - 2\sin(u/2)\cos(u/2)}{2\sin^2(u/2)} = \frac{1}{2}\csc^2\!\tfrac{u}{2} - \cot\!\tfrac{u}{2}$$

Hence

$$f = \int e^u\Bigl(\tfrac{1}{2}\csc^2\!\tfrac{u}{2} - \cot\!\tfrac{u}{2}\Bigr)\,du$$

and we rewrite the integrand as $$-\cot\!\tfrac{u}{2} + \tfrac{1}{2}\csc^2\!\tfrac{u}{2}\,.$$

Note that if $$g(u) = -\cot\!\tfrac{u}{2}$$ then $$g'(u) = \tfrac12\csc^2\!\tfrac{u}{2}$$. Using the formula $$\int e^u\bigl(g(u)+g'(u)\bigr)\,du = e^u g(u) + C$$ gives

$$f = -e^u\cot\!\tfrac{u}{2} + C = -t\;\cot\!\Bigl(\tfrac{\ln t}{2}\Bigr) + C$$

At $$t = e^{\pi/2}$$ the condition $$f(e^{\pi/2}) = -e^{\pi/2}$$ implies $$u = \tfrac{\pi}{2}$$ and $$\cot(\tfrac{\pi}{4}) = 1$$, so

$$-e^{\pi/2}\cdot 1 + C = -e^{\pi/2}$$

$$C = 0$$

At $$t = e^{\pi/4}$$ we have $$u = \tfrac{\pi}{4}$$ and need $$\cot\!\tfrac{\pi}{8}$$.

$$\cot\!\tfrac{\pi}{8} = \frac{\cos(\pi/8)}{\sin(\pi/8)} = 1 + \sqrt{2}$$

(Using $$\cot(\pi/8) = \frac{1+\cos(\pi/4)}{\sin(\pi/4)} = \frac{1+\tfrac{1}{\sqrt{2}}}{\tfrac{1}{\sqrt{2}}} = \sqrt{2}+1$$.)

$$f(e^{\pi/4}) = -e^{\pi/4}(1 + \sqrt{2}) = (-1-\sqrt{2})\,e^{\pi/4}$$

So $$\alpha = -1 - \sqrt{2}$$.

$$\alpha = -1 - \sqrt{2}$$, which matches Option C.

Therefore, the answer is Option C.