What is the value of the Mellin transform ?

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zetafunction
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my question is what is the value of

[tex]\int_{0}^{\infty}dtt^{s-1}[/tex]

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


[tex]\int_{0}^{\infty}dtt^{s-1}cos(at)[/tex] fro every real value of 'a' thanks.
 
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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.
 
thanks.. but could you use analytic continuation for values bigger than Re (s) >1 ??

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

[tex]\int_{0}^{\infty}frac(1/t)t^{s-1}dt= -\zeta(s)/s[/tex] for every s except s=1
 
zetafunction said:
my question is what is the value of

[tex]\int_{0}^{\infty}dtt^{s-1}[/tex]

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

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

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

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

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

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