Mixing
Does convergence of $Gamma(x)$ imply convergence of $Gamma(z)$? Is it generalisable?
If we have the gamma function in integral form $Gamma(x)=int_limits0^infty e^{-t}t^{x-1}dt$ and have proven that it converges for real $x>0$ (to a point), can we then immediately conclude that $Gamma(z)$ converges to a point if Re($z$)$>0$?
I'm asking this because we note that $t^{z-1}$ is complex and thus can be written in polar form $re^{ivarphi}$. The distance between the origin and $t^{z-1}$ is equal to $r>0$. We can now easily associate this distance with the distance between $x=r$ and the origin. We already know that $Gamma(x)$ converges, thus $Gamma(z)$ must converge to a given distance from the origin. However, we have completely ignored the angle $varphi$. Do we know that the angle converges? If so, how?
If $Gamma(z)$ converges, can we somehow generalise the idea?
complex-analysis gamma-function absolute-convergence
asked Nov 21 '18 at 11:45
Idea Flux
86
|
show 1 more comment
If we have the gamma function in integral form $Gamma(x)=int_limits0^infty e^{-t}t^{x-1}dt$ and have proven that it converges for real $x>0$ (to a point), can we then immediately conclude that $Gamma(z)$ converges to a point if Re($z$)$>0$?
I'm asking this because we note that $t^{z-1}$ is complex and thus can be written in polar form $re^{ivarphi}$. The distance between the origin and $t^{z-1}$ is equal to $r>0$. We can now easily associate this distance with the distance between $x=r$ and the origin. We already know that $Gamma(x)$ converges, thus $Gamma(z)$ must converge to a given distance from the origin. However, we have completely ignored the angle $varphi$. Do we know that the angle converges? If so, how?
If $Gamma(z)$ converges, can we somehow generalise the idea?
complex-analysis gamma-function absolute-convergence
asked Nov 21 '18 at 11:45
Idea Flux
86
What exactly do you mean by $int_a^b$. Is this a path integral from $a$ to $b$? Can it be any path? Why do you think that $int_a^b |f(z)|text{d}z$ is positive? What do you mean by "$|a| = infty$"
– Jakobian
Nov 21 '18 at 11:54
I have completely changed my question to 1) make it more accurate and 2) more understable. I hope everything is clear and correctly worded now.
– Idea Flux
Nov 22 '18 at 16:14
For $x =Re(z)$ and $0 < a < b < infty$ then $int_a^b e^{-t} t^{x-1}dt=int_a^b |e^{-t} t^{z-1}|dt$ so $Gamma(z,a,b)=int_a^b e^{-t} t^{z-1}dt$ converges. Since $e^{-u} u^{z-1}$ is analytic for $Re(u) > 0$ it implies that $Gamma(z,a,b)=oint_a^b e^{-t} t^{z-1}dt$ for any finite length path $ato b$. Letting $b to +infty$ you have a result for $Gamma(z) = lim_{b to +infty} Gamma(z,1/b,b)$.
– reuns
Nov 22 '18 at 20:31
Now a stronger result is that $Gamma(z) = lim_{Re(b) to +infty} Gamma(z,0,b)$ (for example $Gamma(z) = lim_{c to +infty} Gamma(z,0,c(1+i))$) and this is because $e^{-u}u^{z-1}$ decreases very fast as $Re(u) to +infty$ so that for any $Re(s) > 0$ then $lim_{c to +infty} oint_c^{cs} e^{-u} u^{z-1}du= 0$. The limiting case $s =pm i$ is interesting too.
– reuns
Nov 22 '18 at 20:31
@reuns Okay, so $x=$Re$(z)$ follows from $|e^{-t}t^{z-1}|=e^{-t}sqrt{t^{x+iy-1}cdot t^{x-iy-1}}$, then the equality between the integrals follows indeed, but why then "so $Gamma(z,a,b)=intlimits_a^b e^{-t}t^{z-1}dt$ converges"?
– Idea Flux
Nov 27 '18 at 12:22
|
show 1 more comment
If we have the gamma function in integral form $Gamma(x)=int_limits0^infty e^{-t}t^{x-1}dt$ and have proven that it converges for real $x>0$ (to a point), can we then immediately conclude that $Gamma(z)$ converges to a point if Re($z$)$>0$?
I'm asking this because we note that $t^{z-1}$ is complex and thus can be written in polar form $re^{ivarphi}$. The distance between the origin and $t^{z-1}$ is equal to $r>0$. We can now easily associate this distance with the distance between $x=r$ and the origin. We already know that $Gamma(x)$ converges, thus $Gamma(z)$ must converge to a given distance from the origin. However, we have completely ignored the angle $varphi$. Do we know that the angle converges? If so, how?
If $Gamma(z)$ converges, can we somehow generalise the idea?
complex-analysis gamma-function absolute-convergence
asked Nov 21 '18 at 11:45
Idea Flux
86
If we have the gamma function in integral form $Gamma(x)=int_limits0^infty e^{-t}t^{x-1}dt$ and have proven that it converges for real $x>0$ (to a point), can we then immediately conclude that $Gamma(z)$ converges to a point if Re($z$)$>0$?
I'm asking this because we note that $t^{z-1}$ is complex and thus can be written in polar form $re^{ivarphi}$. The distance between the origin and $t^{z-1}$ is equal to $r>0$. We can now easily associate this distance with the distance between $x=r$ and the origin. We already know that $Gamma(x)$ converges, thus $Gamma(z)$ must converge to a given distance from the origin. However, we have completely ignored the angle $varphi$. Do we know that the angle converges? If so, how?
If $Gamma(z)$ converges, can we somehow generalise the idea?
complex-analysis gamma-function absolute-convergence
complex-analysis gamma-function absolute-convergence
asked Nov 21 '18 at 11:45
Idea Flux
86
asked Nov 21 '18 at 11:45
Idea Flux
86
edited Nov 22 '18 at 16:13
asked Nov 21 '18 at 11:45
Idea Flux
86
asked Nov 21 '18 at 11:45
Idea Flux
86
asked Nov 21 '18 at 11:45
Idea Flux
86
86
What exactly do you mean by $int_a^b$. Is this a path integral from $a$ to $b$? Can it be any path? Why do you think that $int_a^b |f(z)|text{d}z$ is positive? What do you mean by "$|a| = infty$"
– Jakobian
Nov 21 '18 at 11:54
I have completely changed my question to 1) make it more accurate and 2) more understable. I hope everything is clear and correctly worded now.
– Idea Flux
Nov 22 '18 at 16:14
For $x =Re(z)$ and $0 < a < b < infty$ then $int_a^b e^{-t} t^{x-1}dt=int_a^b |e^{-t} t^{z-1}|dt$ so $Gamma(z,a,b)=int_a^b e^{-t} t^{z-1}dt$ converges. Since $e^{-u} u^{z-1}$ is analytic for $Re(u) > 0$ it implies that $Gamma(z,a,b)=oint_a^b e^{-t} t^{z-1}dt$ for any finite length path $ato b$. Letting $b to +infty$ you have a result for $Gamma(z) = lim_{b to +infty} Gamma(z,1/b,b)$.
– reuns
Nov 22 '18 at 20:31
Now a stronger result is that $Gamma(z) = lim_{Re(b) to +infty} Gamma(z,0,b)$ (for example $Gamma(z) = lim_{c to +infty} Gamma(z,0,c(1+i))$) and this is because $e^{-u}u^{z-1}$ decreases very fast as $Re(u) to +infty$ so that for any $Re(s) > 0$ then $lim_{c to +infty} oint_c^{cs} e^{-u} u^{z-1}du= 0$. The limiting case $s =pm i$ is interesting too.
– reuns
Nov 22 '18 at 20:31
@reuns Okay, so $x=$Re$(z)$ follows from $|e^{-t}t^{z-1}|=e^{-t}sqrt{t^{x+iy-1}cdot t^{x-iy-1}}$, then the equality between the integrals follows indeed, but why then "so $Gamma(z,a,b)=intlimits_a^b e^{-t}t^{z-1}dt$ converges"?
– Idea Flux
Nov 27 '18 at 12:22
|
show 1 more comment
What exactly do you mean by $int_a^b$. Is this a path integral from $a$ to $b$? Can it be any path? Why do you think that $int_a^b |f(z)|text{d}z$ is positive? What do you mean by "$|a| = infty$"
– Jakobian
Nov 21 '18 at 11:54
I have completely changed my question to 1) make it more accurate and 2) more understable. I hope everything is clear and correctly worded now.
– Idea Flux
Nov 22 '18 at 16:14
For $x =Re(z)$ and $0 < a < b < infty$ then $int_a^b e^{-t} t^{x-1}dt=int_a^b |e^{-t} t^{z-1}|dt$ so $Gamma(z,a,b)=int_a^b e^{-t} t^{z-1}dt$ converges. Since $e^{-u} u^{z-1}$ is analytic for $Re(u) > 0$ it implies that $Gamma(z,a,b)=oint_a^b e^{-t} t^{z-1}dt$ for any finite length path $ato b$. Letting $b to +infty$ you have a result for $Gamma(z) = lim_{b to +infty} Gamma(z,1/b,b)$.
– reuns
Nov 22 '18 at 20:31
Now a stronger result is that $Gamma(z) = lim_{Re(b) to +infty} Gamma(z,0,b)$ (for example $Gamma(z) = lim_{c to +infty} Gamma(z,0,c(1+i))$) and this is because $e^{-u}u^{z-1}$ decreases very fast as $Re(u) to +infty$ so that for any $Re(s) > 0$ then $lim_{c to +infty} oint_c^{cs} e^{-u} u^{z-1}du= 0$. The limiting case $s =pm i$ is interesting too.
– reuns
Nov 22 '18 at 20:31
@reuns Okay, so $x=$Re$(z)$ follows from $|e^{-t}t^{z-1}|=e^{-t}sqrt{t^{x+iy-1}cdot t^{x-iy-1}}$, then the equality between the integrals follows indeed, but why then "so $Gamma(z,a,b)=intlimits_a^b e^{-t}t^{z-1}dt$ converges"?
– Idea Flux
Nov 27 '18 at 12:22
What exactly do you mean by $int_a^b$. Is this a path integral from $a$ to $b$? Can it be any path? Why do you think that $int_a^b |f(z)|text{d}z$ is positive? What do you mean by "$|a| = infty$"
– Jakobian
Nov 21 '18 at 11:54
What exactly do you mean by $int_a^b$. Is this a path integral from $a$ to $b$? Can it be any path? Why do you think that $int_a^b |f(z)|text{d}z$ is positive? What do you mean by "$|a| = infty$"
– Jakobian
Nov 21 '18 at 11:54
I have completely changed my question to 1) make it more accurate and 2) more understable. I hope everything is clear and correctly worded now.
– Idea Flux
Nov 22 '18 at 16:14
I have completely changed my question to 1) make it more accurate and 2) more understable. I hope everything is clear and correctly worded now.
– Idea Flux
Nov 22 '18 at 16:14
For $x =Re(z)$ and $0 < a < b < infty$ then $int_a^b e^{-t} t^{x-1}dt=int_a^b |e^{-t} t^{z-1}|dt$ so $Gamma(z,a,b)=int_a^b e^{-t} t^{z-1}dt$ converges. Since $e^{-u} u^{z-1}$ is analytic for $Re(u) > 0$ it implies that $Gamma(z,a,b)=oint_a^b e^{-t} t^{z-1}dt$ for any finite length path $ato b$. Letting $b to +infty$ you have a result for $Gamma(z) = lim_{b to +infty} Gamma(z,1/b,b)$.
– reuns
Nov 22 '18 at 20:31
For $x =Re(z)$ and $0 < a < b < infty$ then $int_a^b e^{-t} t^{x-1}dt=int_a^b |e^{-t} t^{z-1}|dt$ so $Gamma(z,a,b)=int_a^b e^{-t} t^{z-1}dt$ converges. Since $e^{-u} u^{z-1}$ is analytic for $Re(u) > 0$ it implies that $Gamma(z,a,b)=oint_a^b e^{-t} t^{z-1}dt$ for any finite length path $ato b$. Letting $b to +infty$ you have a result for $Gamma(z) = lim_{b to +infty} Gamma(z,1/b,b)$.
– reuns
Nov 22 '18 at 20:31
Now a stronger result is that $Gamma(z) = lim_{Re(b) to +infty} Gamma(z,0,b)$ (for example $Gamma(z) = lim_{c to +infty} Gamma(z,0,c(1+i))$) and this is because $e^{-u}u^{z-1}$ decreases very fast as $Re(u) to +infty$ so that for any $Re(s) > 0$ then $lim_{c to +infty} oint_c^{cs} e^{-u} u^{z-1}du= 0$. The limiting case $s =pm i$ is interesting too.
– reuns
Nov 22 '18 at 20:31
Now a stronger result is that $Gamma(z) = lim_{Re(b) to +infty} Gamma(z,0,b)$ (for example $Gamma(z) = lim_{c to +infty} Gamma(z,0,c(1+i))$) and this is because $e^{-u}u^{z-1}$ decreases very fast as $Re(u) to +infty$ so that for any $Re(s) > 0$ then $lim_{c to +infty} oint_c^{cs} e^{-u} u^{z-1}du= 0$. The limiting case $s =pm i$ is interesting too.
– reuns
Nov 22 '18 at 20:31
@reuns Okay, so $x=$Re$(z)$ follows from $|e^{-t}t^{z-1}|=e^{-t}sqrt{t^{x+iy-1}cdot t^{x-iy-1}}$, then the equality between the integrals follows indeed, but why then "so $Gamma(z,a,b)=intlimits_a^b e^{-t}t^{z-1}dt$ converges"?
– Idea Flux
Nov 27 '18 at 12:22
@reuns Okay, so $x=$Re$(z)$ follows from $|e^{-t}t^{z-1}|=e^{-t}sqrt{t^{x+iy-1}cdot t^{x-iy-1}}$, then the equality between the integrals follows indeed, but why then "so $Gamma(z,a,b)=intlimits_a^b e^{-t}t^{z-1}dt$ converges"?
– Idea Flux
Nov 27 '18 at 12:22
|
show 1 more comment
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What exactly do you mean by $int_a^b$. Is this a path integral from $a$ to $b$? Can it be any path? Why do you think that $int_a^b |f(z)|text{d}z$ is positive? What do you mean by "$|a| = infty$"
– Jakobian
Nov 21 '18 at 11:54
I have completely changed my question to 1) make it more accurate and 2) more understable. I hope everything is clear and correctly worded now.
– Idea Flux
Nov 22 '18 at 16:14
For $x =Re(z)$ and $0 < a < b < infty$ then $int_a^b e^{-t} t^{x-1}dt=int_a^b |e^{-t} t^{z-1}|dt$ so $Gamma(z,a,b)=int_a^b e^{-t} t^{z-1}dt$ converges. Since $e^{-u} u^{z-1}$ is analytic for $Re(u) > 0$ it implies that $Gamma(z,a,b)=oint_a^b e^{-t} t^{z-1}dt$ for any finite length path $ato b$. Letting $b to +infty$ you have a result for $Gamma(z) = lim_{b to +infty} Gamma(z,1/b,b)$.
– reuns
Nov 22 '18 at 20:31
Now a stronger result is that $Gamma(z) = lim_{Re(b) to +infty} Gamma(z,0,b)$ (for example $Gamma(z) = lim_{c to +infty} Gamma(z,0,c(1+i))$) and this is because $e^{-u}u^{z-1}$ decreases very fast as $Re(u) to +infty$ so that for any $Re(s) > 0$ then $lim_{c to +infty} oint_c^{cs} e^{-u} u^{z-1}du= 0$. The limiting case $s =pm i$ is interesting too.
– reuns
Nov 22 '18 at 20:31
@reuns Okay, so $x=$Re$(z)$ follows from $|e^{-t}t^{z-1}|=e^{-t}sqrt{t^{x+iy-1}cdot t^{x-iy-1}}$, then the equality between the integrals follows indeed, but why then "so $Gamma(z,a,b)=intlimits_a^b e^{-t}t^{z-1}dt$ converges"?
– Idea Flux
Nov 27 '18 at 12:22