On $sum_k=1^infty1/(k!k^s)$.Is this formula for $ sum_n (n^2+z^2)^-s $ correct?How to get the floor function as a Mellin inverse of the Hadamard product of the Riemann zeta function?Wrong proof of the functional equation for $ zeta (s) $ but why is the result correct?Using The Riemann Zeta Functional EquationHow do you use the Riemann Zeta Function?What's about $sum_n=1^infty e^-p_n u$, where $p_n$ is the nth-prime number?Calculate $zeta(s) = frac1Gamma(s) sum_n=0^infty fracB_nn! int_0^infty x^s + n - 2 dx$Xi Function on Critical Strip - Mellin TransformRelating the Dirichlet $beta$-function to the $zeta$-function through the Parseval theorem. Is this theorem reversible?On the series expansion of $fracoperatornameLi_3(-x)1+x$ and its usage
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On $sum_k=1^infty1/(k!k^s)$.
Is this formula for $ sum_n (n^2+z^2)^-s $ correct?How to get the floor function as a Mellin inverse of the Hadamard product of the Riemann zeta function?Wrong proof of the functional equation for $ zeta (s) $ but why is the result correct?Using The Riemann Zeta Functional EquationHow do you use the Riemann Zeta Function?What's about $sum_n=1^infty e^-p_n u$, where $p_n$ is the nth-prime number?Calculate $zeta(s) = frac1Gamma(s) sum_n=0^infty fracB_nn! int_0^infty x^s + n - 2 dx$Xi Function on Critical Strip - Mellin TransformRelating the Dirichlet $beta$-function to the $zeta$-function through the Parseval theorem. Is this theorem reversible?On the series expansion of $fracoperatornameLi_3(-x)1+x$ and its usage
$begingroup$
Has the following $zeta$-like function been studied before?
$$f(z;s)=sum_k=1^inftyfracz^kk!k^s.$$
I believe this is an entire function since using the ratio test,
$$lim_ntoinftyfraca_n+1a_n=lim_ntoinftyfracfrac1(n+1)!(n+1)^sfrac1n!n^s=lim_ntoinftyfracn!n^s(n+1)!(n+1)^s=lim_ntoinftyfrac1n+1left(1+frac1nright)^-s=0,$$
so $R=infty$.
The polylogarithm,
$$textLi_s(z)=sum_k=1^inftyfracz^kk^s,$$
"comes close," but does anyone know anything about $f(z;s)$, or similar function, in the literature?
@reuns comment reminded me of Ramanujan's Master Theorem. If we consider the slightly different function,
$$f_1(x;s)=sum_k=0^inftyfrac1k!(k+1)^s(-x)^k,$$
then applying Ramanujan's Master Theorem with $phi(k)=1/(1+k)^s$,
$$int_0^infty x^s-1f_1(x;s)dx=fracGamma(s)(1-s)^s.$$
Applying the inverse Mellin transform then gives
$$f_1(x;s)=frac12pi iint_c-iinfty^c+iinftyx^-sfracGamma(s)(1-s)^sds,$$
assuming everything converges.
reference-request riemann-zeta polylogarithm
asked yesterday
AntinousAntinous
5,81942452
$endgroup$
add a comment |
$begingroup$
Has the following $zeta$-like function been studied before?
$$f(z;s)=sum_k=1^inftyfracz^kk!k^s.$$
I believe this is an entire function since using the ratio test,
$$lim_ntoinftyfraca_n+1a_n=lim_ntoinftyfracfrac1(n+1)!(n+1)^sfrac1n!n^s=lim_ntoinftyfracn!n^s(n+1)!(n+1)^s=lim_ntoinftyfrac1n+1left(1+frac1nright)^-s=0,$$
so $R=infty$.
The polylogarithm,
$$textLi_s(z)=sum_k=1^inftyfracz^kk^s,$$
"comes close," but does anyone know anything about $f(z;s)$, or similar function, in the literature?
@reuns comment reminded me of Ramanujan's Master Theorem. If we consider the slightly different function,
$$f_1(x;s)=sum_k=0^inftyfrac1k!(k+1)^s(-x)^k,$$
then applying Ramanujan's Master Theorem with $phi(k)=1/(1+k)^s$,
$$int_0^infty x^s-1f_1(x;s)dx=fracGamma(s)(1-s)^s.$$
Applying the inverse Mellin transform then gives
$$f_1(x;s)=frac12pi iint_c-iinfty^c+iinftyx^-sfracGamma(s)(1-s)^sds,$$
assuming everything converges.
reference-request riemann-zeta polylogarithm
asked yesterday
AntinousAntinous
5,81942452
$endgroup$
1
$begingroup$
For $|z| < 1$, $int_0^infty e^-x f(zx,s)dx = Li_s(z)$ the LHS is a (Mellin) convolution. With $g(z) = sum_k=1^infty fracz^kk!^2$ (the Bessel function $J_0$) $int_0^infty x^s-1 g(z e^-x) dx = Gamma(s) f(z,s)$.
$endgroup$
– reuns
yesterday
add a comment |
$begingroup$
Has the following $zeta$-like function been studied before?
$$f(z;s)=sum_k=1^inftyfracz^kk!k^s.$$
I believe this is an entire function since using the ratio test,
$$lim_ntoinftyfraca_n+1a_n=lim_ntoinftyfracfrac1(n+1)!(n+1)^sfrac1n!n^s=lim_ntoinftyfracn!n^s(n+1)!(n+1)^s=lim_ntoinftyfrac1n+1left(1+frac1nright)^-s=0,$$
so $R=infty$.
The polylogarithm,
$$textLi_s(z)=sum_k=1^inftyfracz^kk^s,$$
"comes close," but does anyone know anything about $f(z;s)$, or similar function, in the literature?
@reuns comment reminded me of Ramanujan's Master Theorem. If we consider the slightly different function,
$$f_1(x;s)=sum_k=0^inftyfrac1k!(k+1)^s(-x)^k,$$
then applying Ramanujan's Master Theorem with $phi(k)=1/(1+k)^s$,
$$int_0^infty x^s-1f_1(x;s)dx=fracGamma(s)(1-s)^s.$$
Applying the inverse Mellin transform then gives
$$f_1(x;s)=frac12pi iint_c-iinfty^c+iinftyx^-sfracGamma(s)(1-s)^sds,$$
assuming everything converges.
reference-request riemann-zeta polylogarithm
asked yesterday
AntinousAntinous
5,81942452
$endgroup$
Has the following $zeta$-like function been studied before?
$$f(z;s)=sum_k=1^inftyfracz^kk!k^s.$$
I believe this is an entire function since using the ratio test,
$$lim_ntoinftyfraca_n+1a_n=lim_ntoinftyfracfrac1(n+1)!(n+1)^sfrac1n!n^s=lim_ntoinftyfracn!n^s(n+1)!(n+1)^s=lim_ntoinftyfrac1n+1left(1+frac1nright)^-s=0,$$
so $R=infty$.
The polylogarithm,
$$textLi_s(z)=sum_k=1^inftyfracz^kk^s,$$
"comes close," but does anyone know anything about $f(z;s)$, or similar function, in the literature?
@reuns comment reminded me of Ramanujan's Master Theorem. If we consider the slightly different function,
$$f_1(x;s)=sum_k=0^inftyfrac1k!(k+1)^s(-x)^k,$$
then applying Ramanujan's Master Theorem with $phi(k)=1/(1+k)^s$,
$$int_0^infty x^s-1f_1(x;s)dx=fracGamma(s)(1-s)^s.$$
Applying the inverse Mellin transform then gives
$$f_1(x;s)=frac12pi iint_c-iinfty^c+iinftyx^-sfracGamma(s)(1-s)^sds,$$
assuming everything converges.
reference-request riemann-zeta polylogarithm
reference-request riemann-zeta polylogarithm
asked yesterday
AntinousAntinous
5,81942452
asked yesterday
AntinousAntinous
5,81942452
edited 22 hours ago
Antinous
asked yesterday
AntinousAntinous
5,81942452
asked yesterday
AntinousAntinous
5,81942452
asked yesterday
AntinousAntinous
5,81942452
5,81942452
1
$begingroup$
For $|z| < 1$, $int_0^infty e^-x f(zx,s)dx = Li_s(z)$ the LHS is a (Mellin) convolution. With $g(z) = sum_k=1^infty fracz^kk!^2$ (the Bessel function $J_0$) $int_0^infty x^s-1 g(z e^-x) dx = Gamma(s) f(z,s)$.
$endgroup$
– reuns
yesterday
add a comment |
1
$begingroup$
For $|z| < 1$, $int_0^infty e^-x f(zx,s)dx = Li_s(z)$ the LHS is a (Mellin) convolution. With $g(z) = sum_k=1^infty fracz^kk!^2$ (the Bessel function $J_0$) $int_0^infty x^s-1 g(z e^-x) dx = Gamma(s) f(z,s)$.
$endgroup$
– reuns
yesterday
1
1
$begingroup$
For $|z| < 1$, $int_0^infty e^-x f(zx,s)dx = Li_s(z)$ the LHS is a (Mellin) convolution. With $g(z) = sum_k=1^infty fracz^kk!^2$ (the Bessel function $J_0$) $int_0^infty x^s-1 g(z e^-x) dx = Gamma(s) f(z,s)$.
$endgroup$
– reuns
yesterday
$begingroup$
For $|z| < 1$, $int_0^infty e^-x f(zx,s)dx = Li_s(z)$ the LHS is a (Mellin) convolution. With $g(z) = sum_k=1^infty fracz^kk!^2$ (the Bessel function $J_0$) $int_0^infty x^s-1 g(z e^-x) dx = Gamma(s) f(z,s)$.
$endgroup$
– reuns
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$f(z;s)=sum_k=1^inftyfracz^kk!k^s.$$
$$f(z;0)=e^z-1$$
$$f(z;1)=Ei(z)-ln(z)-gamma$$
$$f(z;2)=z :_3F_3(1,1,1;2,2,2;z)$$
In case of $s=n=$integer the closed form is a generalized hypergeometric function :
$$f(z;n)=z :_n+1F_n+1(1,...1;2,...,2;z)$$
$$f(z;-1)= ze^z$$
$$f(z;-2)= z(z+1)e^z$$
$$f(z;-3)= z(z^2+3z+1)e^z$$
In case of $-s=n=$integer the closed form is on the form of the next equation where $P(z)$ is a polynomial of degree $n-1$.
$$f(z;-n)= zP(z)e^z$$
In the general case ($s$ real) it is doubtful that a closed form exists.
answered 22 hours ago
JJacquelinJJacquelin
44.5k21854
$endgroup$
add a comment |
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$begingroup$
$$f(z;s)=sum_k=1^inftyfracz^kk!k^s.$$
$$f(z;0)=e^z-1$$
$$f(z;1)=Ei(z)-ln(z)-gamma$$
$$f(z;2)=z :_3F_3(1,1,1;2,2,2;z)$$
In case of $s=n=$integer the closed form is a generalized hypergeometric function :
$$f(z;n)=z :_n+1F_n+1(1,...1;2,...,2;z)$$
$$f(z;-1)= ze^z$$
$$f(z;-2)= z(z+1)e^z$$
$$f(z;-3)= z(z^2+3z+1)e^z$$
In case of $-s=n=$integer the closed form is on the form of the next equation where $P(z)$ is a polynomial of degree $n-1$.
$$f(z;-n)= zP(z)e^z$$
In the general case ($s$ real) it is doubtful that a closed form exists.
answered 22 hours ago
JJacquelinJJacquelin
44.5k21854
$endgroup$
add a comment |
$begingroup$
$$f(z;s)=sum_k=1^inftyfracz^kk!k^s.$$
$$f(z;0)=e^z-1$$
$$f(z;1)=Ei(z)-ln(z)-gamma$$
$$f(z;2)=z :_3F_3(1,1,1;2,2,2;z)$$
In case of $s=n=$integer the closed form is a generalized hypergeometric function :
$$f(z;n)=z :_n+1F_n+1(1,...1;2,...,2;z)$$
$$f(z;-1)= ze^z$$
$$f(z;-2)= z(z+1)e^z$$
$$f(z;-3)= z(z^2+3z+1)e^z$$
In case of $-s=n=$integer the closed form is on the form of the next equation where $P(z)$ is a polynomial of degree $n-1$.
$$f(z;-n)= zP(z)e^z$$
In the general case ($s$ real) it is doubtful that a closed form exists.
answered 22 hours ago
JJacquelinJJacquelin
44.5k21854
$endgroup$
add a comment |
$begingroup$
$$f(z;s)=sum_k=1^inftyfracz^kk!k^s.$$
$$f(z;0)=e^z-1$$
$$f(z;1)=Ei(z)-ln(z)-gamma$$
$$f(z;2)=z :_3F_3(1,1,1;2,2,2;z)$$
In case of $s=n=$integer the closed form is a generalized hypergeometric function :
$$f(z;n)=z :_n+1F_n+1(1,...1;2,...,2;z)$$
$$f(z;-1)= ze^z$$
$$f(z;-2)= z(z+1)e^z$$
$$f(z;-3)= z(z^2+3z+1)e^z$$
In case of $-s=n=$integer the closed form is on the form of the next equation where $P(z)$ is a polynomial of degree $n-1$.
$$f(z;-n)= zP(z)e^z$$
In the general case ($s$ real) it is doubtful that a closed form exists.
answered 22 hours ago
JJacquelinJJacquelin
44.5k21854
$endgroup$
$$f(z;s)=sum_k=1^inftyfracz^kk!k^s.$$
$$f(z;0)=e^z-1$$
$$f(z;1)=Ei(z)-ln(z)-gamma$$
$$f(z;2)=z :_3F_3(1,1,1;2,2,2;z)$$
In case of $s=n=$integer the closed form is a generalized hypergeometric function :
$$f(z;n)=z :_n+1F_n+1(1,...1;2,...,2;z)$$
$$f(z;-1)= ze^z$$
$$f(z;-2)= z(z+1)e^z$$
$$f(z;-3)= z(z^2+3z+1)e^z$$
In case of $-s=n=$integer the closed form is on the form of the next equation where $P(z)$ is a polynomial of degree $n-1$.
$$f(z;-n)= zP(z)e^z$$
In the general case ($s$ real) it is doubtful that a closed form exists.
answered 22 hours ago
JJacquelinJJacquelin
44.5k21854
answered 22 hours ago
JJacquelinJJacquelin
44.5k21854
answered 22 hours ago
JJacquelinJJacquelin
44.5k21854
answered 22 hours ago
JJacquelinJJacquelin
44.5k21854
44.5k21854
add a comment |
add a comment |
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$begingroup$
For $|z| < 1$, $int_0^infty e^-x f(zx,s)dx = Li_s(z)$ the LHS is a (Mellin) convolution. With $g(z) = sum_k=1^infty fracz^kk!^2$ (the Bessel function $J_0$) $int_0^infty x^s-1 g(z e^-x) dx = Gamma(s) f(z,s)$.
$endgroup$
– reuns
yesterday