Interesting Integral including $\ln x$ I would like to evaluate this integral:  $$\int_0^1  \frac{\sin(\ln(x))}{\ln(x)}\,dx$$
I tried a lot, started by integral uv formula [integration-by-parts?] and it went quite lengthy and never ending.
Then, thought may be expansion of $\sin x$ but it didn't made any sense, it's a infinite series and there's no way here that it will end..
It is highly appreciated if you give some hints to solve it.
I am a high school student,so I expect it to be simple but tricky:-)

Observe that $\displaystyle \frac{\sin(\ln(x))  }{\ln{x}} = \int_0^1 \cos(t \ln{x}) \, \mathrm dt$. Then:
$$\begin{aligned} \int_0^1\frac{\sin(\ln(x))  }{\ln{x}}\, \mathrm dx &= \int_0^1 \int_0^1 \cos(t\ln{x})\;{\mathrm dt}\;{\mathrm dx} \\& = \ \int_0^1 \int_0^1 \cos(t\ln{x})\;{\mathrm dx}\;{\mathrm dt} \\&= \int_0^1 \frac{1}{t^2+1} \;{\mathrm dt} \\& =  \frac{\pi}{4}. \end{aligned}$$

Equivalently, consider the function
$$\displaystyle f(t) = \int_0^1\frac{\sin(t\ln(x))  }{\ln{x}}\, \mathrm dx.$$
Then
$$\displaystyle f'(t) = \int_0^1 \cos(t \ln{x})\, \mathrm{d}x = \frac{1}{1+t^2}.$$
Therefore $f(t) = \arctan(t)+C$. But $f(0) = 0$ so $C = 0$.
Hence $f(t) = \arctan(t)$. We seek $\displaystyle f(1) = \arctan(1) = \frac{\pi}{4}$.

Series solution:
$\displaystyle I = \int_0^1\frac{\sin(\ln(x))  }{\ln{x}}\, \mathrm dx = \int_0^1\sum_{k \ge 0} \frac{(-1)^k \ln^{2k}{x}}{(2k+1)!}\, \mathrm dx  = \sum_{k \ge 0} \frac{(-1)^k }{(2k+1)!} \int_0^1 \ln^{2k}{x}\, \mathrm dx $
Then we calculate $\displaystyle \int_0^1 \ln^nx \,\mathrm{d}x$ via integration by parts to find that it's equal to $(-1)^n n!$
Or consider $\displaystyle f(m) = \int_0^1 x^m \,{dx} = \frac{1}{1+m}.$
Then taking the $n$-th derivative of both sides:
$\displaystyle f^{(n)}(m) = \int_0^1 x^m \ln^{n}{x} \,{dx} = \frac{(-1)^n n! }{(1+m)^{n+1}}.$
In either case we get $\displaystyle \int_0^1 \ln^{2k}{x}\, \mathrm dx = (2k)!$.  Hence:
$\displaystyle I = \sum_{k \ge 0} \frac{(-1)^k (2k)!}{(2k+1)!}  = \sum_{k \ge 0} \frac{(-1)^k (2k)!}{(2k+1)(2k)!} = \sum_{k \ge 0} \frac{(-1)^k }{(2k+1)} = \frac{\pi}{4}.$
To prove the last equality, consider $\displaystyle \frac{1}{2k+1} = \int_0^1 x^{2k} \, \mathrm dx$
and the geometric series $\displaystyle \sum_{k \ge 0}(-1)^kx^{2k} = \frac{1}{1+x^2}$. Then
$\begin{aligned} \displaystyle \sum_{k \ge 0} \frac{(-1)^k}{2k+1} & = \sum_{k \ge 0}{(-1)^k}\int_0^1 x^{2k}\,{\mathrm dx} \\& = \int_0^1 \sum_{k \ge 0}(-1)^k x^{2k} \, \mathrm dx \\& = \int_0^1 \frac{1}{1+x^2}\,\mathrm dx \\& = \frac{\pi}{4}.\end{aligned}$

Regarding the integral $\displaystyle I = \int_0^1 \cos(t \ln{x})\, \mathrm{d}x $, we let $x = e^{-y}$. Then $\displaystyle I  = \int_0^\infty e^{-y}\cos(ty)\,\mathrm{d}y.$
We get the answer by applying integration by parts (twice). Or we can consider the real part, if we're familiar with complex numbers:
\begin{align}
I & = \int_0^\infty e^{-y}\cos(ty)\,\mathrm{d}y
=\Re\left(\int_0^\infty e^{-(1-it)y}\mathrm{d}y\right)\\
&=\Re\left(\int_0^\infty e^{-(1+t^2)y}\mathrm{d}(1+it)y\right)\\
&=\Re\left(\frac{1+it}{1+t^2}\int_0^\infty e^{-(1+t^2)y}\mathrm{d}(1+t^2)y\right)\\
&=\Re\left(\frac{1+it}{1+t^2}\right)\\& =\frac{1}{1+t^2}.
\end{align}
