Does convergence of $Gamma(x)$ imply convergence of $Gamma(z)$? Is it generalisable?
up vote
0
down vote
favorite
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 at 11:45
Idea Flux
64
|
show 1 more comment
up vote
0
down vote
favorite
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 at 11:45
Idea Flux
64
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 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 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 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 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
18 hours ago
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 at 11:45
Idea Flux
64
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 at 11:45
Idea Flux
64
asked Nov 21 at 11:45
Idea Flux
64
edited Nov 22 at 16:13
asked Nov 21 at 11:45
Idea Flux
64
asked Nov 21 at 11:45
Idea Flux
64
asked Nov 21 at 11:45
Idea Flux
64
64
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 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 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 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 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
18 hours ago
|
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 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 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 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 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
18 hours ago
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 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 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 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 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 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 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 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 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
18 hours ago
@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
18 hours ago
|
show 1 more comment
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007613%2fdoes-convergence-of-gammax-imply-convergence-of-gammaz-is-it-generali%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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 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 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 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 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
18 hours ago