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.
 A: Consider the integral
$$I(\alpha) = \int\limits_{0}^{1} \frac{\sin{\left(\alpha \log{\left(x \right)} \right)}}{\log{\left(x \right)}}\, dx$$
Your integral becomes $I(1)$. Note that
$$I(0) = \int\limits_{0}^{1} \sin(0)\, dx = 0 \tag{$\dagger$} \label{eq}$$
Let's differentiate $I(\alpha)$ with respect to $\alpha$
$$I'(\alpha) = \int\limits_{0}^{1} \cos{\left(\alpha \log{\left(x \right)} \right)}\, dx\\
= \left.\frac{\alpha x \sin{\left(\alpha \log{\left(x \right)} \right)}}{\alpha^{2} + 1} + \frac{x \cos{\left(\alpha \log{\left(x \right)} \right)}}{\alpha^{2} + 1}\right|_{x=0}^{x=1} = \frac{1}{\alpha^{2} + 1}$$
So integrating back we get:
$$I(\alpha) = \arctan(\alpha) + C$$
But remember we know by $\ref{eq}$ that $I(0) = 0$ so
$$0 = \arctan(0) + C = 0 + C \implies C = 0$$
So
$$I(\alpha) = \arctan(\alpha)$$
and specifically, your integral becomes
$$I(1) = \arctan(1) = \frac{\pi}{4}$$
The lesson: When in doubt, remember Feynman.
A: 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}
