- Thread starter
- #1
DreamWeaver
Well-known member
- Sep 16, 2013
- 337
There's a unified pattern at work here...
Problem 1:
[tex]\int_0^{\infty}\frac{\log(1-\sin x)}{x}\,dx=-\frac{\pi^2}{6}=-\zeta(2)[/tex]
Problem 2:
[tex]\int_0^{\infty}\frac{ \text{Li}_m(\sin x) }{x}\,dx=\text{Li}_{m+1}(1)=\zeta (m+1)[/tex]
Problem 1:
[tex]\int_0^{\infty}\frac{\log(1-\sin x)}{x}\,dx=-\frac{\pi^2}{6}=-\zeta(2)[/tex]
Problem 2:
[tex]\int_0^{\infty}\frac{ \text{Li}_m(\sin x) }{x}\,dx=\text{Li}_{m+1}(1)=\zeta (m+1)[/tex]