1. Oct 2, 2009

zetafunction

given [x] the 'integer function' would be the following bound valid ??

$$\int_{0}^{\infty}dt [g(x/t)]f(t) \le \int_{0}^{\infty}dt g(x/t)f(t)$$

for a given functions f(t) and g(u) u=x/t , here g is a non-decreasing positive function for positive arguments (and real) of parameter u=x/t

2. Oct 2, 2009

vineethbs

Hi,

if f(t) >= 0 for t >=0, then using (g(x/t) + 1) would be a better idea ?
Now if
$$f(t) = f^{+}(t) - f^{-}(t)$$,
where $$f^{+}(t) and f^{-}(t)$$ denotes max(f(t),0) and max(-f(t),0) respectively
then one can use g(x/t) + 1 for f+(t) and g(x/t) - 1 for f-(t) to obtain a bound.