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I can calculate the integral on $[0,\pi/2]$ interval, but that solution cant be applied here because it uses $\sin(\pi/2 - x)=\cos x$ and the rule $$\int_{a}^b f(x)dx=\int_{a}^b f(a+b-x) dx$$ which here would turn into $\sin(1-x)$ which is useless. I'd be grateful if anyone could help. (Also i saw some general solutions involving i, i don't know complex integration so please don't go down that route)

J.G.
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mida
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    From How to ask a good question: Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title. – jjagmath Dec 31 '21 at 10:15
  • Do you have any reason to believe this integral can be calculated in terms of elementary functions? That's usually not the case... – jjagmath Dec 31 '21 at 10:19
  • CAS says: $-\ln (2)-\frac{1}{2} \Im\left(\text{Li}_2\left(e^{2 i}\right)\right)$ – Mariusz Iwaniuk Dec 31 '21 at 10:24
  • Use series expansion and termwise integration. – Claude Leibovici Dec 31 '21 at 10:27
  • Where did you encounter a need to evaluate this? If you were asked to, there's a surprisingly good chance it's a misprint for $\int_0^1\ln\sin(\pi x)dx$. – J.G. Dec 31 '21 at 10:29
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    i asked a teacher and he said i could estimate it , saying it is less than $pi(1-ln(sin(1)))$ . i tried to look for a function f that is bigger than ln(sin(x)) and we can integrate easily , but couldnt come up with something that didnt mess up the bound. – mida Dec 31 '21 at 10:32
  • If you just want to know why $\int_0^1\ln\sin xdx<\pi(1-\ln\sin 1)$, it's enough to note the LHS is negative. – J.G. Dec 31 '21 at 11:03
  • i made a slight mistake , sorry. i meant $$- \int_{0}^1 ln(sin(x)) \leq \pi (1-ln sin 1)$$. – mida Dec 31 '21 at 12:08
  • In fact you can drop that factor of $\pi$, because $1.06<1.17$. – J.G. Jan 01 '22 at 10:01
  • I've updated my answer to prove the desired inequality. – J.G. Jan 01 '22 at 10:19

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Let me introduce the polylogarithm$$\operatorname{Li}_s(z):=\sum_{n\ge1}z^nn^{-s}$$(technically we analytically continue the RHS, but I don't think you want that kind of complex-analytic pedantry right now). In particular, $\frac{d}{dz}\operatorname{Li}_2(z)=-z^{-1}\ln(1-z)$ so$$\frac{d}{dx}\operatorname{Li}_2(e^{2ix})=-2i\ln(1-e^{2ix})=-2i\ln\sin x+2x-\color{blue}{2i\ln(-2i)}.$$Applying $\frac{-1}{2i}\int dx$,$$\int\ln\sin xdx=\frac{i}{2}\left(\operatorname{Li}_2(e^{2ix})-x^2\right)+\color{blue}{2i\ln(-2i)}x+C.$$I'll leave you to ponder what the blue coefficient should be, i.e. which branch of the complex logarithm it needs.

As @jjagmath noted, it's highly unlikely the $\int_0^1$ result has a closed form, except for what the above indefinite integral implies. Wolfram Alpha can do no better, although at least it will help you test your thoughts about the blue value. The result is in line with @MariuszIwaniuk comment.

Edit: we can prove $\int_0^1(-\ln\sin x)dx<1-\sin1$ or equivalently $\int_0^1\left(-\ln\frac{\sin x}{x}\right)dx<-\sin1$ by noting the latter integrand increases from $0$ at $x=0$ to $-\ln\sin1$ at $x=1$, so the latter integral has upper bound $\int_0^1(-\ln\sin1)dx=-\ln\sin1$.

J.G.
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  • thank you for this, it's nice. but could you think of any function that is less than ln(sin(x)) and can be integrated easily? its integration on the interval [0,1] is $$\pi (ln(sin(1))-1)$$ apparently. – mida Dec 31 '21 at 10:45