12/09/2021, 11:48 PM
Consider the following (double) integral where h is a positive infinitesimal
\( B(s)=\int_h^s \int_h^\infty \frac{cos(zt)dtdz}{t^t} \)
This integral is intended as an analogue for erf(s) but which is suppose to go - C for Re(s) << -1 and + C for Re(s) >> 1 ( independant of the imaginary part ! ).
Where C is a (probably nonzero and positive ) real constant.
Assuming that indeed 0 < C we continue :
\( tb(s)=\frac{1+\frac{B(s)}{C}}{2} \)
Now consider
\( f(s)=\exp(tb(s) f(s-1)) \)
And finally we get lim n to +oo ;
\( tet_{tb}(s)=ln^{[n]}f(s+n) \)
I call it tommy beta method , hence "tb"
This ofcourse requires more research.
regards
tommy1729
Tom Marcel Raes
\( B(s)=\int_h^s \int_h^\infty \frac{cos(zt)dtdz}{t^t} \)
This integral is intended as an analogue for erf(s) but which is suppose to go - C for Re(s) << -1 and + C for Re(s) >> 1 ( independant of the imaginary part ! ).
Where C is a (probably nonzero and positive ) real constant.
Assuming that indeed 0 < C we continue :
\( tb(s)=\frac{1+\frac{B(s)}{C}}{2} \)
Now consider
\( f(s)=\exp(tb(s) f(s-1)) \)
And finally we get lim n to +oo ;
\( tet_{tb}(s)=ln^{[n]}f(s+n) \)
I call it tommy beta method , hence "tb"
This ofcourse requires more research.
regards
tommy1729
Tom Marcel Raes