# What is the value of the Mellin transform ?

1. Feb 3, 2010

### zetafunction

my question is what is the value of

$$\int_{0}^{\infty}dtt^{s-1}$$

if s >1 the integral is divergent as x -->oo

if s<1 the integral would become divergent as x-->0

then what is the value of this integral, or the value (regularized i presume) of

$$\int_{0}^{\infty}dtt^{s-1}cos(at)$$ fro every real value of 'a' thanks.

2. Feb 3, 2010

### mathman

It has meaning only for 0<s<1. For s = 1/2, the value is √(π/2a). Since cos is even, a >0 can be assumed.

3. Feb 4, 2010

### zetafunction

thanks.. but could you use analytic continuation for values bigger than Re (s) >1 ??

for example if we consider in a regularized sense that $$\int_{0}^{\infty}t^{s-1}dt=0$$ for every 's' then

$$\int_{0}^{\infty}frac(1/t)t^{s-1}dt= -\zeta(s)/s$$ for every s except s=1

4. Feb 4, 2010

### jostpuur

It is divergent on the limit $x\to\infty$ when $s\geq 0$, and on the limit $x\to 0^+$ when $s\leq 0$. The whole integral is thus divergent for any $s$.

What do you mean by this "regularization". You have given value 0 to a divergent integral! :surprised