Ln-sine integral

∫01xnln⁡(sin⁡πx) dx\int _{ 0 }^{ 1 }{ { x }^{ n }\ln { (\sin { \pi x } ) } \, dx }

Find a closed form of the integral above, where nn is a positive integer.

This is a part of the set Formidable Series and Integrals

#Calculus

Note by Hamza A
5 years ago

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Comments

the trick is to write the Fourier series ln⁡(sin⁡(πx))=−ln⁡(2)−∑k=1∞cos⁡(2πxk)k\ln(\sin(\pi x))=-\ln(2)-\sum_{k=1}^\infty \dfrac{\cos(2\pi x k)}{k} it follows that the integral is then: ∫01xnln⁡(sin⁡(πx))dx=−ln⁡(2)n+1−∑k=1∞1k∫01xncos⁡(2πxk)dx=−ln⁡(2)n+1−∑k=1∞1(2π)n+1kn+2∫02πkxncos⁡(x)dx\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 In=∫02πkxncos⁡(x)dx=−n∫02πkxn−1sin⁡(x)dx=n(2πk)n−1+n(n−1)In−2I_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 In=∑j<n,j=oddn!(n−j)!(2πk)n−jI_n=\sum_{j<n , j=odd} \dfrac{n!}{(n-j)!} (2\pi k)^{n-j} and our original integral is now −ln⁡(2)n+1−∑k=1∞∑j<n,j=oddn!(n−j)!(2πk)n−j(2π)n+1kn+2=−ln⁡(2)n+1−∑j<n,j=oddn!ζ(j+2)(n−j)!(2π)j+1-\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 - 4 years ago
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