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. :-)
On
\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}
On
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
$$\begin{align} \underbrace{\int_0^\infty \frac{e^{-x}\sin(x)}{x}\,dx}_{\int_0^\infty f(x)g(x)\,dx}&=\underbrace{\int_0^\infty \frac{1}{(s+1)^2+1}\,ds}_{\int_0^\infty \mathscr{L}\{f\}(s)\,\,\mathscr{L}^{-1}\{g\}(s)\,ds}\\\\ &=\frac{\pi}{4} \end{align}$$
On
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 $$ \begin{aligned} \int_0^1 \frac{\sin (\ln x)}{\ln x} d x & =I(1)-I(0) \\ & =\int_0^1 \frac{1}{a^2+1} d a \\ & =\left[\tan ^{-1} a\right]_0^1 \\ & =\frac{\pi}{4} \end{aligned} $$
On
$$I=\int_0^1\frac{\sin{\ln x}}{\ln x}dx=\Im\int_0^1\frac{x^i-1}{\ln x}dx$$ 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 $$J=\int_0^1\frac{\cos{\ln{x}}-1}{\ln x}dx=\Re[I(i)]=\frac{\ln 2}{2}$$
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}