Evaluating this definite integral Given:
$$I=\int_0^1 \frac{\sin(\ln(x))}{\ln(x)}dx$$
Now I have tried every strategy, I could think of, substitution, by parts (it will return the same $I$ after a few times), etc.
In the solution provided, they did introduce fancy complex numbers, which I have no idea of. Note that I am a normal high school student, so I cannot understand high-level calculus. Please help here.

Using
$$ \sin x=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^{2n-1}}{(2n-1)!}, \int_0^1(\ln x)^ndx=(-1)^nn! $$
one has
\begin{eqnarray}
I&=&\int_0^1 \frac{\sin(\ln(x))}{\ln(x)}dx\\
&=&\int_0^1\sum_{n=1}^\infty\frac{(-1)^{n-1}(\ln x)^{2n-2}}{(2n-1)!}dx\\
&=&\sum_{n=1}^\infty\frac{(-1)^{n-1}(2n-2)!}{(2n-1)!}\\
&=&\sum_{n=1}^\infty\frac{(-1)^{n-1}}{2n-1}\\\\
&=&\frac\pi4.
\end{eqnarray}
