Various methods used to evaluate $\int_0^1 \frac{\sin(\ln x)}{\ln x}\,dx$ The integral on focus here is:
$$\int_0^1 \frac{\sin(\ln x)}{\ln x}\,dx$$
I know a solution to this by considering a function 
$$f(a) = \int_0^1 \frac{\sin(\ln(ax))}{\ln x}\,dx$$ and then differentiating it w.r.t $a$ solving integral and then integrating again w.r.t. $a$.

Although I was wondering if you could perform this integration by any other method?

Mention the method in detail, please. :-)
 A: Substitute $\ln x = -t$,
\begin{align}
\int_0^1 \frac{\sin(\ln x)}{\ln x}\,dx
=& \int_0^\infty \frac{e^{-t}}t \sin t \ dt
=\int_0^\infty \int_1^\infty \sin t \ e^{-xt}dx \ dt\\
= &\int_1^\infty \frac1{1+x^2}dx =\frac\pi4\\
\end{align}
A: As in the post by @quanto, we beign by enforcing the substitution $x\mapsto e^{-x}$ to reveal
$$\int_0^1 \frac{\sin(\log(x))}{\log(x)}\,dx=\int_0^\infty \frac{e^{-x}\sin(x)}{x}\,dx\tag1$$

Next, we will use the identity from the Laplace Transform (See This)
$$\int_0^\infty f(x)g(x)\,dx=\int_0^\infty \mathscr{L}\{f\}(s)\,\,\mathscr{L}^{-1}\{g\}(s)\,ds\tag2$$

Let $f(x)=e^{-x}\sin(x)$ and $g(x)=\frac1x$.  Then, we see that
$$\mathscr{L}\{f\}(s)=\frac{1}{(s+1)^2+1}\tag3$$
and
$$\mathscr{L}^{-1}\{g\}(s)=1\tag4$$

Hence, using $(3)$ and $(4)$ in $(2)$ and applying to $(1)$ yields 
$$\begin{align}
\underbrace{\int_0^\infty \frac{e^{-x}\sin(x)}{x}\,dx}_{\int_0^\infty f(x)g(x)\,dx}&=\underbrace{\int_0^\infty \frac{1}{(s+1)^2+1}\,ds}_{\int_0^\infty \mathscr{L}\{f\}(s)\,\,\mathscr{L}^{-1}\{g\}(s)\,ds}\\\\
&=\frac{\pi}{4}
\end{align}$$

A: \begin{align}J&=\int_0^1 \frac{\sin(\ln x)}{\ln x}\,dx\\
&=\int_0^1 \left(\sum_{n=0}^\infty\frac{(-1)^n\ln^{2n} x}{(2n+1)!}\right)\,dx\\
&=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\left(\int_0^1 \ln^{2n} x\,dx\right)\\
&=\sum_{n=0}^\infty\frac{(-1)^n(2n)!}{(2n+1)!}\\
&=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\\
&=\arctan(1)\\
&=\boxed{\frac{\pi}{4}}\\
\end{align}
NB:
i assume $\displaystyle \int_0^1 \ln^{2n} x\,dx=(2n)!$ and $\displaystyle |x|\leq 1,\arctan(x)=\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}$, and, $\displaystyle x\in\mathbb{R},\sin(x)=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}$
PS:
$n\geq 0$, integer.
\begin{align}K_n&=\int_0^1\ln^{n} x\,dx\\
K_0&=\int_0^1\ln^{0} x\,dx\\
&=1\\
K_1&=\Big[x(\ln x-1)\Big]_0^1\\
&=-1\\
K_{n+2}&=\Big[x(\ln x-1)\ln^{n+1} x\Big]_0^1-(n+1)\int_0^1 (\ln x-1)\ln^{n} x\,dx\\
&=(n+1)K_{n}-(n+1)K_{n+1}\\
\end{align}
Proof by induction:
For $\displaystyle 0\leq m\leq n+1, K_m=(-1)^m\times m!$
\begin{align}K_{n+2}&=(n+1)K_{n}-(n+1)K_{n+1}\\
&=(n+1)\times(-1)^{n}\times n!-(n+1)\times (-1)^{n+1}\times (n+1)! \\
&=(-1)^n(n+2)!\left(\frac{1}{n+2}+\frac{n+1}{n+2}\right)\\
&=(-1)^n(n+2)!\\
&=\boxed{(-1)^{n+2}(n+2)!}
\end{align}
A: Using Feynman’s Integration Technique, we first let
$$
I(a)=\int_0^1 \frac{\sin (a\ln x)}{\ln x} d x 
$$
Then differentiating $I(a)$ w.r.t. $a$ yields
$$
I^{\prime}(a)=\int_0^1 \cos (a \ln x) d x
$$
Letting $y=-a\ln x$ transforms
$$
 I^{\prime}(a) =\frac{1}{a} \int_0^{\infty} e^{\frac{y}{a}} \cos y d y =\frac{1}{a^2+1}
$$
Integrating back from $a=0$ to $1$ gives
$$
\begin{aligned}
\int_0^1 \frac{\sin (\ln x)}{\ln x} d x & =I(1)-I(0) \\
& =\int_0^1 \frac{1}{a^2+1} d a \\
& =\left[\tan ^{-1} a\right]_0^1 \\
& =\frac{\pi}{4}
\end{aligned}
$$
A: $$I=\int_0^1\frac{\sin{\ln x}}{\ln x}dx=\Im\int_0^1\frac{x^i-1}{\ln x}dx$$
This is a classic integral to solve with Feynman's trick
$$I(a)=\int_0^1\frac{x^a-1}{\ln x}dx, I(0)=0$$
$$I'(a)=\int_0^1\frac{x^a\ln x}{\ln x}dx=\int_0^1x^adx=\frac{x^{a+1}}{a+1}\Bigg|_0^1=\frac{1}{a+1}$$
$$I(a)=\ln(a+1)+C$$
$$I(0)=\ln(1)+C=C=0$$
$$I(a)=\ln(a+1)$$
Going back to $I$
$$I=\Im\int_0^1\frac{x^i-1}{\ln x}dx=\Im [I(i)]=\Im[\ln(1+i)]=\frac{\pi}{4}$$
We also get another integral for free
$$J=\int_0^1\frac{\cos{\ln{x}}-1}{\ln x}dx=\Re[I(i)]=\frac{\ln 2}{2}$$
