\[\int _{ 0 }^{ 1 }{ { x }^{ n }\ln { (\sin { \pi x } ) } \, dx } \]

Find a closed form of the integral above, where \(n\) is a positive integer.

This is a part of the set Formidable Series and Integrals

\[\int _{ 0 }^{ 1 }{ { x }^{ n }\ln { (\sin { \pi x } ) } \, dx } \]

Find a closed form of the integral above, where \(n\) is a positive integer.

This is a part of the set Formidable Series and Integrals

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TopNewestthe trick is to write the Fourier series \[\ln(\sin(\pi x))=-\ln(2)-\sum_{k=1}^\infty \dfrac{\cos(2\pi x k)}{k}\] it follows that the integral is then: \[\int_0^1 x^n \ln(\sin(\pi x)) dx= -\dfrac{\ln(2)}{n+1}-\sum_{k=1}^\infty \dfrac{1}{k} \int_0^1 x^n \cos(2\pi x k) dx= -\dfrac{\ln(2)}{n+1}-\sum_{k=1}^\infty \dfrac{1}{(2\pi)^{n+1}k^{n+2}} \int_0^{2\pi k} x^n \cos(x) dx\] we can see that if \[I_n=\int_0^{2\pi k} x^n \cos(x) dx=-n\int_0^{2\pi k} x^{n-1} \sin(x) dx=n(2\pi k)^{n-1}+n(n-1)I_{n-2}\] we see that \[I_n=\sum_{j<n , j=odd} \dfrac{n!}{(n-j)!} (2\pi k)^{n-j}\] and our original integral is now \[-\dfrac{\ln(2)}{n+1}-\sum_{k=1}^\infty \dfrac{\sum_{j<n , j=odd} \dfrac{n!}{(n-j)!} (2\pi k)^{n-j}}{(2\pi)^{n+1}k^{n+2}}=-\dfrac{\ln(2)}{n+1} -\sum_{j<n , j=odd}\dfrac{n!\zeta(j+2)}{(n-j)! (2\pi)^{j+1}} \] – Aareyan Manzoor · 3 months, 2 weeks ago

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