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zeros of imaginary part of $log(Gamma(z))$
Quotient of gamma functions?Inequality $Gamma(alpha+x)Gamma(beta+y)leq C(alpha; beta)Gamma(x+y-1)$Log of 1-reguralized incomplete gamma function (upper)Evaluating $int_0^1 int_0^1log (Gamma(s,x)) ,ds ,dx $Prove that log Gamma is real analyticCan you generalise the Gamma function?Imaginary Part of log(f(z))An integration related to incomplete gamma functionAn equation with Gamma Euler function in critical strip$2 cdot int_0^1 logbig(Gamma(x)big) cdot sin(2 pi n x) dx = fracgamma + log(2 pi) + log(n)n pi$
$begingroup$
I want to discover if there exist a solution of the following equation:
$$Im(log(Gamma(iy)) - ybeta = 0 $$
for $yinmathbbRsetminus 0$ and $beta >0$
I already proved that if there is a solution $y^*<0$ then also $-y^*$ is a solution, but this do not help me with the general proof.
Thanks a lot
complex-analysis special-functions gamma-function
asked Jan 30 at 14:54
Guido MazzucaGuido Mazzuca
33
$endgroup$
add a comment |
$begingroup$
I want to discover if there exist a solution of the following equation:
$$Im(log(Gamma(iy)) - ybeta = 0 $$
for $yinmathbbRsetminus 0$ and $beta >0$
I already proved that if there is a solution $y^*<0$ then also $-y^*$ is a solution, but this do not help me with the general proof.
Thanks a lot
complex-analysis special-functions gamma-function
asked Jan 30 at 14:54
Guido MazzucaGuido Mazzuca
33
$endgroup$
$begingroup$
Which definition of logarithm are you using?
$endgroup$
– Szeto
Jan 30 at 22:17
$begingroup$
The branch cut is on the negative real axis, so Arg(z) is is $(-pi,pi)$
$endgroup$
– Guido Mazzuca
Jan 31 at 7:51
add a comment |
$begingroup$
I want to discover if there exist a solution of the following equation:
$$Im(log(Gamma(iy)) - ybeta = 0 $$
for $yinmathbbRsetminus 0$ and $beta >0$
I already proved that if there is a solution $y^*<0$ then also $-y^*$ is a solution, but this do not help me with the general proof.
Thanks a lot
complex-analysis special-functions gamma-function
asked Jan 30 at 14:54
Guido MazzucaGuido Mazzuca
33
$endgroup$
I want to discover if there exist a solution of the following equation:
$$Im(log(Gamma(iy)) - ybeta = 0 $$
for $yinmathbbRsetminus 0$ and $beta >0$
I already proved that if there is a solution $y^*<0$ then also $-y^*$ is a solution, but this do not help me with the general proof.
Thanks a lot
complex-analysis special-functions gamma-function
complex-analysis special-functions gamma-function
asked Jan 30 at 14:54
Guido MazzucaGuido Mazzuca
33
asked Jan 30 at 14:54
Guido MazzucaGuido Mazzuca
33
asked Jan 30 at 14:54
Guido MazzucaGuido Mazzuca
33
asked Jan 30 at 14:54
Guido MazzucaGuido Mazzuca
33
asked Jan 30 at 14:54
Guido MazzucaGuido Mazzuca
33
33
$begingroup$
Which definition of logarithm are you using?
$endgroup$
– Szeto
Jan 30 at 22:17
$begingroup$
The branch cut is on the negative real axis, so Arg(z) is is $(-pi,pi)$
$endgroup$
– Guido Mazzuca
Jan 31 at 7:51
add a comment |
$begingroup$
Which definition of logarithm are you using?
$endgroup$
– Szeto
Jan 30 at 22:17
$begingroup$
The branch cut is on the negative real axis, so Arg(z) is is $(-pi,pi)$
$endgroup$
– Guido Mazzuca
Jan 31 at 7:51
$begingroup$
Which definition of logarithm are you using?
$endgroup$
– Szeto
Jan 30 at 22:17
$begingroup$
Which definition of logarithm are you using?
$endgroup$
– Szeto
Jan 30 at 22:17
$begingroup$
The branch cut is on the negative real axis, so Arg(z) is is $(-pi,pi)$
$endgroup$
– Guido Mazzuca
Jan 31 at 7:51
$begingroup$
The branch cut is on the negative real axis, so Arg(z) is is $(-pi,pi)$
$endgroup$
– Guido Mazzuca
Jan 31 at 7:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If the function is $ln(Gamma(i y))$, a solution may not exist, because the imaginary part is negative for small positive $y$ and cannot be greater than $pi$.
If $operatornamelnGamma$ is defined as the analytic continuation of $ln(Gamma(z))$ from the positive real axis to $mathbb C backslash (-infty, 0]$, then $operatornameIm operatornamelnGamma(+i 0) = -pi/2$; $operatornameIm operatornamelnGamma(i y) sim y ln y$ at infinity; $operatornamelnGamma(i y)$ is continuous on $(0, infty)$.
Therefore $y = y(beta)$ s.t. $operatornameIm operatornamelnGamma(i y) = beta y, , y > 0$ exists for any $beta > 0$.
answered Mar 21 at 14:57
MaximMaxim
6,1681221
$endgroup$
add a comment |
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$begingroup$
If the function is $ln(Gamma(i y))$, a solution may not exist, because the imaginary part is negative for small positive $y$ and cannot be greater than $pi$.
If $operatornamelnGamma$ is defined as the analytic continuation of $ln(Gamma(z))$ from the positive real axis to $mathbb C backslash (-infty, 0]$, then $operatornameIm operatornamelnGamma(+i 0) = -pi/2$; $operatornameIm operatornamelnGamma(i y) sim y ln y$ at infinity; $operatornamelnGamma(i y)$ is continuous on $(0, infty)$.
Therefore $y = y(beta)$ s.t. $operatornameIm operatornamelnGamma(i y) = beta y, , y > 0$ exists for any $beta > 0$.
answered Mar 21 at 14:57
MaximMaxim
6,1681221
$endgroup$
add a comment |
$begingroup$
If the function is $ln(Gamma(i y))$, a solution may not exist, because the imaginary part is negative for small positive $y$ and cannot be greater than $pi$.
If $operatornamelnGamma$ is defined as the analytic continuation of $ln(Gamma(z))$ from the positive real axis to $mathbb C backslash (-infty, 0]$, then $operatornameIm operatornamelnGamma(+i 0) = -pi/2$; $operatornameIm operatornamelnGamma(i y) sim y ln y$ at infinity; $operatornamelnGamma(i y)$ is continuous on $(0, infty)$.
Therefore $y = y(beta)$ s.t. $operatornameIm operatornamelnGamma(i y) = beta y, , y > 0$ exists for any $beta > 0$.
answered Mar 21 at 14:57
MaximMaxim
6,1681221
$endgroup$
add a comment |
$begingroup$
If the function is $ln(Gamma(i y))$, a solution may not exist, because the imaginary part is negative for small positive $y$ and cannot be greater than $pi$.
If $operatornamelnGamma$ is defined as the analytic continuation of $ln(Gamma(z))$ from the positive real axis to $mathbb C backslash (-infty, 0]$, then $operatornameIm operatornamelnGamma(+i 0) = -pi/2$; $operatornameIm operatornamelnGamma(i y) sim y ln y$ at infinity; $operatornamelnGamma(i y)$ is continuous on $(0, infty)$.
Therefore $y = y(beta)$ s.t. $operatornameIm operatornamelnGamma(i y) = beta y, , y > 0$ exists for any $beta > 0$.
answered Mar 21 at 14:57
MaximMaxim
6,1681221
$endgroup$
If the function is $ln(Gamma(i y))$, a solution may not exist, because the imaginary part is negative for small positive $y$ and cannot be greater than $pi$.
If $operatornamelnGamma$ is defined as the analytic continuation of $ln(Gamma(z))$ from the positive real axis to $mathbb C backslash (-infty, 0]$, then $operatornameIm operatornamelnGamma(+i 0) = -pi/2$; $operatornameIm operatornamelnGamma(i y) sim y ln y$ at infinity; $operatornamelnGamma(i y)$ is continuous on $(0, infty)$.
Therefore $y = y(beta)$ s.t. $operatornameIm operatornamelnGamma(i y) = beta y, , y > 0$ exists for any $beta > 0$.
answered Mar 21 at 14:57
MaximMaxim
6,1681221
answered Mar 21 at 14:57
MaximMaxim
6,1681221
answered Mar 21 at 14:57
MaximMaxim
6,1681221
answered Mar 21 at 14:57
MaximMaxim
6,1681221
6,1681221
add a comment |
add a comment |
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$begingroup$
Which definition of logarithm are you using?
$endgroup$
– Szeto
Jan 30 at 22:17
$begingroup$
The branch cut is on the negative real axis, so Arg(z) is is $(-pi,pi)$
$endgroup$
– Guido Mazzuca
Jan 31 at 7:51