I'm trying to find the following integral for fun:

$$\int \frac{\sin x}{\ln x}\,\mathrm dx.$$

I simply tried to convert the nominator into something log-friendly using:

$$\sin x = \frac{e^{ix}-e^{-ix}}{2i} $$

Then, I tried to change integration variable using $y = e^{ix}$ or $y = \ln x$, but neither seems to be helpful.

What is the right way to analytically wrestle with this integral?

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    $\begingroup$ I doubt this can be done as an indefinite integral. $\endgroup$ – John Doe Apr 25 '18 at 4:32
  • $\begingroup$ @JohnDoe: Thanks. Can you explain where this observation comes from? $\endgroup$ – Roboticist Apr 25 '18 at 4:33
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    $\begingroup$ Just from the fact that these are two very different kinds of functions (so you are unlikely to find a substitution that simplifies both terms) and that you've divided one by the other (this usually makes things difficult). Edit: also, it appears that wolfram alpha fails to solve it $\endgroup$ – John Doe Apr 25 '18 at 4:34
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    $\begingroup$ $\int\frac{\sin x}{\log x}\,dx$ is not an elementary function, but $$\int_{0}^{1}\frac{\sin(\pi x)}{\log x}\,dx$$ might have some interesting alternative representations since $\frac{1}{\log x}=\int_{1}^{+\infty}\frac{da}{x^{a-1}}$ and $\int_{0}^{+\infty}\frac{\sin(x)}{x^{a-1}}\,dx$ can be computed through the Laplace transform or contour integrals. $\endgroup$ – Jack D'Aurizio Apr 25 '18 at 14:35

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