benorin
Feb15-10, 01:10 PM
Induction shows that
\zeta(n) = \lim_{x\to 1^{ - }}\int_{0}^{x}\int_{0}^{1}\cdots\int_{0}^{1}\frac {dy_{n - 1}\cdots dy_{1}dy_{0}}{1 - y_{0}y_{1}\cdots y_{n - 1}} \qquard \mbox{ for }n = 2,3,4,\ldots
= \lim_{x\to 1^{ - }}\int_{0}^{x}\int_{0}^{x}\cdots\int_{0}^{x}\frac {dy_{n - 1}\cdots dy_{1}dy_{0}}{1 - y_{0}y_{1}\cdots y_{n - 1}} \qquard \mbox{ for }n = 2,3,4,\ldots
where the second n-d multiple integral for the zeta function is equivalent to the first and only given in hope that it might be somehow useful...
:idea: Problem: devise a change of variables that transforms either of the above n-d multiple integrals into an n-fold iterated integral of the form
\lim_{x\to 1^{ - }}\int_{0}^{x}\int_{0}^{t_{0}}\cdots\int_{0}^{t_{n - 2}}F(t_{n})dt_{n - 1}\cdots dt_{1}dt_{0}
which is readily converted into a fractional integral.
Thanks,
-Ben
\zeta(n) = \lim_{x\to 1^{ - }}\int_{0}^{x}\int_{0}^{1}\cdots\int_{0}^{1}\frac {dy_{n - 1}\cdots dy_{1}dy_{0}}{1 - y_{0}y_{1}\cdots y_{n - 1}} \qquard \mbox{ for }n = 2,3,4,\ldots
= \lim_{x\to 1^{ - }}\int_{0}^{x}\int_{0}^{x}\cdots\int_{0}^{x}\frac {dy_{n - 1}\cdots dy_{1}dy_{0}}{1 - y_{0}y_{1}\cdots y_{n - 1}} \qquard \mbox{ for }n = 2,3,4,\ldots
where the second n-d multiple integral for the zeta function is equivalent to the first and only given in hope that it might be somehow useful...
:idea: Problem: devise a change of variables that transforms either of the above n-d multiple integrals into an n-fold iterated integral of the form
\lim_{x\to 1^{ - }}\int_{0}^{x}\int_{0}^{t_{0}}\cdots\int_{0}^{t_{n - 2}}F(t_{n})dt_{n - 1}\cdots dt_{1}dt_{0}
which is readily converted into a fractional integral.
Thanks,
-Ben