Substitute $u=\ln t$, so $t=e^u$ and $dt=e^{u}du$:

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

Split using $1-\cos u=2{\sin }^2(u/2)$ and $\sin u=2\sin (u/2)\cos (u/2)$:

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

Let $g(u)=-\cot (u/2)$. Then $g^{\prime }(u)=\frac{1}{2}{\csc }^2(u/2)$, so the integrand is $[g(u)+g^{\prime }(u)]e^u$.

Using $\int [g(u)+g^{\prime }(u)]e^{u}du=g(u)e^u+C$:

$f=-e^u\cot (u/2)+C=-t\cot \left(\frac{\ln t}{2}\right)+C$.

Determine $C$: $f(e^{\pi /2})=-e^{\pi /2}\cot (\pi /4)+C=-e^{\pi /2}+C=-e^{\pi /2}\Rightarrow C=0$.

Evaluate at $t=e^{\pi /4}$: $f(e^{\pi /4})=-e^{\pi /4}\cot (\pi /8)$. Using $\cot (\pi /8)=\sqrt{2}+1$:

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