Why does convergence of $Gamma(x)$ imply convergence of $Gamma(z)$?
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A while ago I asked a question (link at the bottom) about the gamma function. It was answered with:
'$intlimits_a^b e^{-t} t^{x-1}dt=intlimits_a^b |e^{-t} t^{z-1}|dt$ with $xinmathbb{R}_{>0}$ and $zinmathbb{C}_{Re(z)>0}$. Since the LHS converges, we have that $intlimits_a^b e^t t^{z-1} dt$ converges.'
Why can we draw this conclusion? In my mind we can only draw conclusions about $int|f(z)|dt$ and not about $int f(z) dt$.
.
.
Link: Does convergence of $Gamma(x)$ imply convergence of $Gamma(z)$? Is it generalisable?
complex-analysis convergence
asked Dec 15 '18 at 14:00
Idea FluxIdea Flux
86
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add a comment |
$begingroup$
A while ago I asked a question (link at the bottom) about the gamma function. It was answered with:
'$intlimits_a^b e^{-t} t^{x-1}dt=intlimits_a^b |e^{-t} t^{z-1}|dt$ with $xinmathbb{R}_{>0}$ and $zinmathbb{C}_{Re(z)>0}$. Since the LHS converges, we have that $intlimits_a^b e^t t^{z-1} dt$ converges.'
Why can we draw this conclusion? In my mind we can only draw conclusions about $int|f(z)|dt$ and not about $int f(z) dt$.
.
.
Link: Does convergence of $Gamma(x)$ imply convergence of $Gamma(z)$? Is it generalisable?
complex-analysis convergence
asked Dec 15 '18 at 14:00
Idea FluxIdea Flux
86
$endgroup$
$begingroup$
If $int_a^b|f(u)|,du$ is finite (and $f$ measurable) then $int_a^bf(u),du$ exists (and is finite).
$endgroup$
– Did
Dec 15 '18 at 14:48
$begingroup$
@Did But if $b=infty$, mayhaps $intlimits_a^b f(u)du$ becomes finite yet not convergent. In context of the Gamma function: we expect to get $int f = cinmathbb{C}$ and thus can write $c=re^{ivarphi}$. Can't it be that as we take the limit $btoinfty$ we have that $|c|$ converges, but $varphi$ does not which means that c becomes finite yet does not converge?
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– Idea Flux
Dec 15 '18 at 15:13
$begingroup$
@Did I think I cannot express myself more clearly, but the idea is that the "answer" to the integral could converge to the line $d(0,z) = 10$, but not a specific point on this line. This due to the fact that $|c|=|re^{ivarphi}|=r$ converges but $c=re^{ivarphi}$ does not since $varphi$ does not converge as $btoinfty$
$endgroup$
– Idea Flux
Dec 15 '18 at 15:25
$begingroup$
Possible duplicate of Absolutely improper integrability implies improper integrability
$endgroup$
– user10354138
Dec 16 '18 at 8:20
add a comment |
$begingroup$
A while ago I asked a question (link at the bottom) about the gamma function. It was answered with:
'$intlimits_a^b e^{-t} t^{x-1}dt=intlimits_a^b |e^{-t} t^{z-1}|dt$ with $xinmathbb{R}_{>0}$ and $zinmathbb{C}_{Re(z)>0}$. Since the LHS converges, we have that $intlimits_a^b e^t t^{z-1} dt$ converges.'
Why can we draw this conclusion? In my mind we can only draw conclusions about $int|f(z)|dt$ and not about $int f(z) dt$.
.
.
Link: Does convergence of $Gamma(x)$ imply convergence of $Gamma(z)$? Is it generalisable?
complex-analysis convergence
asked Dec 15 '18 at 14:00
Idea FluxIdea Flux
86
$endgroup$
A while ago I asked a question (link at the bottom) about the gamma function. It was answered with:
'$intlimits_a^b e^{-t} t^{x-1}dt=intlimits_a^b |e^{-t} t^{z-1}|dt$ with $xinmathbb{R}_{>0}$ and $zinmathbb{C}_{Re(z)>0}$. Since the LHS converges, we have that $intlimits_a^b e^t t^{z-1} dt$ converges.'
Why can we draw this conclusion? In my mind we can only draw conclusions about $int|f(z)|dt$ and not about $int f(z) dt$.
.
.
Link: Does convergence of $Gamma(x)$ imply convergence of $Gamma(z)$? Is it generalisable?
complex-analysis convergence
complex-analysis convergence
asked Dec 15 '18 at 14:00
Idea FluxIdea Flux
86
asked Dec 15 '18 at 14:00
Idea FluxIdea Flux
86
edited Dec 15 '18 at 14:05
Idea Flux
asked Dec 15 '18 at 14:00
Idea FluxIdea Flux
86
asked Dec 15 '18 at 14:00
Idea FluxIdea Flux
86
asked Dec 15 '18 at 14:00
Idea FluxIdea Flux
86
86
$begingroup$
If $int_a^b|f(u)|,du$ is finite (and $f$ measurable) then $int_a^bf(u),du$ exists (and is finite).
$endgroup$
– Did
Dec 15 '18 at 14:48
$begingroup$
@Did But if $b=infty$, mayhaps $intlimits_a^b f(u)du$ becomes finite yet not convergent. In context of the Gamma function: we expect to get $int f = cinmathbb{C}$ and thus can write $c=re^{ivarphi}$. Can't it be that as we take the limit $btoinfty$ we have that $|c|$ converges, but $varphi$ does not which means that c becomes finite yet does not converge?
$endgroup$
– Idea Flux
Dec 15 '18 at 15:13
$begingroup$
@Did I think I cannot express myself more clearly, but the idea is that the "answer" to the integral could converge to the line $d(0,z) = 10$, but not a specific point on this line. This due to the fact that $|c|=|re^{ivarphi}|=r$ converges but $c=re^{ivarphi}$ does not since $varphi$ does not converge as $btoinfty$
$endgroup$
– Idea Flux
Dec 15 '18 at 15:25
$begingroup$
Possible duplicate of Absolutely improper integrability implies improper integrability
$endgroup$
– user10354138
Dec 16 '18 at 8:20
add a comment |
$begingroup$
If $int_a^b|f(u)|,du$ is finite (and $f$ measurable) then $int_a^bf(u),du$ exists (and is finite).
$endgroup$
– Did
Dec 15 '18 at 14:48
$begingroup$
@Did But if $b=infty$, mayhaps $intlimits_a^b f(u)du$ becomes finite yet not convergent. In context of the Gamma function: we expect to get $int f = cinmathbb{C}$ and thus can write $c=re^{ivarphi}$. Can't it be that as we take the limit $btoinfty$ we have that $|c|$ converges, but $varphi$ does not which means that c becomes finite yet does not converge?
$endgroup$
– Idea Flux
Dec 15 '18 at 15:13
$begingroup$
@Did I think I cannot express myself more clearly, but the idea is that the "answer" to the integral could converge to the line $d(0,z) = 10$, but not a specific point on this line. This due to the fact that $|c|=|re^{ivarphi}|=r$ converges but $c=re^{ivarphi}$ does not since $varphi$ does not converge as $btoinfty$
$endgroup$
– Idea Flux
Dec 15 '18 at 15:25
$begingroup$
Possible duplicate of Absolutely improper integrability implies improper integrability
$endgroup$
– user10354138
Dec 16 '18 at 8:20
$begingroup$
If $int_a^b|f(u)|,du$ is finite (and $f$ measurable) then $int_a^bf(u),du$ exists (and is finite).
$endgroup$
– Did
Dec 15 '18 at 14:48
$begingroup$
If $int_a^b|f(u)|,du$ is finite (and $f$ measurable) then $int_a^bf(u),du$ exists (and is finite).
$endgroup$
– Did
Dec 15 '18 at 14:48
$begingroup$
@Did But if $b=infty$, mayhaps $intlimits_a^b f(u)du$ becomes finite yet not convergent. In context of the Gamma function: we expect to get $int f = cinmathbb{C}$ and thus can write $c=re^{ivarphi}$. Can't it be that as we take the limit $btoinfty$ we have that $|c|$ converges, but $varphi$ does not which means that c becomes finite yet does not converge?
$endgroup$
– Idea Flux
Dec 15 '18 at 15:13
$begingroup$
@Did But if $b=infty$, mayhaps $intlimits_a^b f(u)du$ becomes finite yet not convergent. In context of the Gamma function: we expect to get $int f = cinmathbb{C}$ and thus can write $c=re^{ivarphi}$. Can't it be that as we take the limit $btoinfty$ we have that $|c|$ converges, but $varphi$ does not which means that c becomes finite yet does not converge?
$endgroup$
– Idea Flux
Dec 15 '18 at 15:13
$begingroup$
@Did I think I cannot express myself more clearly, but the idea is that the "answer" to the integral could converge to the line $d(0,z) = 10$, but not a specific point on this line. This due to the fact that $|c|=|re^{ivarphi}|=r$ converges but $c=re^{ivarphi}$ does not since $varphi$ does not converge as $btoinfty$
$endgroup$
– Idea Flux
Dec 15 '18 at 15:25
$begingroup$
@Did I think I cannot express myself more clearly, but the idea is that the "answer" to the integral could converge to the line $d(0,z) = 10$, but not a specific point on this line. This due to the fact that $|c|=|re^{ivarphi}|=r$ converges but $c=re^{ivarphi}$ does not since $varphi$ does not converge as $btoinfty$
$endgroup$
– Idea Flux
Dec 15 '18 at 15:25
$begingroup$
Possible duplicate of Absolutely improper integrability implies improper integrability
$endgroup$
– user10354138
Dec 16 '18 at 8:20
$begingroup$
Possible duplicate of Absolutely improper integrability implies improper integrability
$endgroup$
– user10354138
Dec 16 '18 at 8:20
add a comment |
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$begingroup$
If $int_a^b|f(u)|,du$ is finite (and $f$ measurable) then $int_a^bf(u),du$ exists (and is finite).
$endgroup$
– Did
Dec 15 '18 at 14:48
$begingroup$
@Did But if $b=infty$, mayhaps $intlimits_a^b f(u)du$ becomes finite yet not convergent. In context of the Gamma function: we expect to get $int f = cinmathbb{C}$ and thus can write $c=re^{ivarphi}$. Can't it be that as we take the limit $btoinfty$ we have that $|c|$ converges, but $varphi$ does not which means that c becomes finite yet does not converge?
$endgroup$
– Idea Flux
Dec 15 '18 at 15:13
$begingroup$
@Did I think I cannot express myself more clearly, but the idea is that the "answer" to the integral could converge to the line $d(0,z) = 10$, but not a specific point on this line. This due to the fact that $|c|=|re^{ivarphi}|=r$ converges but $c=re^{ivarphi}$ does not since $varphi$ does not converge as $btoinfty$
$endgroup$
– Idea Flux
Dec 15 '18 at 15:25
$begingroup$
Possible duplicate of Absolutely improper integrability implies improper integrability
$endgroup$
– user10354138
Dec 16 '18 at 8:20