Mixing
Lower bound for $Gamma(x+i y)$ where $x>0$ involving only the real part
$begingroup$
I am trying to have some inequalities involving the special Gamma function. I am able to get an upper bound for $Gamma(x+i y)$, for $x>0$,
$$
begin{align}
|Gamma(x+iy)|
&=left|int_0^infty e^{-t},t^{x+iy-1};mathrm{d}tright|\
&=left|int_0^infty e^{-t},t^{x-1},e^{iylog(t)};mathrm{d}tright|\
&leint_0^inftyleft|e^{-t},t^{x-1},e^{iylog(t)}right|;mathrm{d}t\
&=int_0^infty e^{-t},t^{x-1};mathrm{d}t\
&=Gamma(x)\
&=|Gamma(x)|tag{1}.
end{align}
$$
However, I did not succeed to get a lower bound for $Gamma(x+i y)$ such that I get rid of the imaginary part. Any help in this direction?
Using the suggestion of @reuns and the Euler's reflection formula, one can shows that
begin{align} left|Gamma(x+i y)right| &= frac{pi}{left|Gamma(1-(x+i y)right| left|sin(pi (x+i y))right|} \
&ge frac{pi}{left|Gamma(1-x)right| left|sin(pi (x+i y))right|}
end{align}
But I can not still get a lower bound for the complex sine term such that I get rid of the imaginary part! Any suggestion?
complex-analysis inequality special-functions gamma-function
asked Jan 31 at 7:18
user144209user144209
42314
$endgroup$
add a comment |
$begingroup$
I am trying to have some inequalities involving the special Gamma function. I am able to get an upper bound for $Gamma(x+i y)$, for $x>0$,
$$
begin{align}
|Gamma(x+iy)|
&=left|int_0^infty e^{-t},t^{x+iy-1};mathrm{d}tright|\
&=left|int_0^infty e^{-t},t^{x-1},e^{iylog(t)};mathrm{d}tright|\
&leint_0^inftyleft|e^{-t},t^{x-1},e^{iylog(t)}right|;mathrm{d}t\
&=int_0^infty e^{-t},t^{x-1};mathrm{d}t\
&=Gamma(x)\
&=|Gamma(x)|tag{1}.
end{align}
$$
However, I did not succeed to get a lower bound for $Gamma(x+i y)$ such that I get rid of the imaginary part. Any help in this direction?
Using the suggestion of @reuns and the Euler's reflection formula, one can shows that
begin{align} left|Gamma(x+i y)right| &= frac{pi}{left|Gamma(1-(x+i y)right| left|sin(pi (x+i y))right|} \
&ge frac{pi}{left|Gamma(1-x)right| left|sin(pi (x+i y))right|}
end{align}
But I can not still get a lower bound for the complex sine term such that I get rid of the imaginary part! Any suggestion?
complex-analysis inequality special-functions gamma-function
asked Jan 31 at 7:18
user144209user144209
42314
$endgroup$
1
$begingroup$
You won't obtain anything directly from the integral, you need to show the explicit product formula for $Gamma(s)$, or things like the reflection formula.
$endgroup$
– reuns
Jan 31 at 8:21
$begingroup$
Thank you reuns for your comment. Can you please provide more details or a reference about that. Thanks.
$endgroup$
– user144209
Jan 31 at 9:55
1
$begingroup$
$Gamma(s)Gamma(1-s) = pi/sin(pi s)$ gives the exact decay on $Re(s) in mathbb{Z}/2$. For other $Re(s)$ the easiest is to look at the (complex) Stirling approximation, a direct consequence of the product formula en.wikipedia.org/wiki/…
$endgroup$
– reuns
Jan 31 at 10:02
add a comment |
$begingroup$
I am trying to have some inequalities involving the special Gamma function. I am able to get an upper bound for $Gamma(x+i y)$, for $x>0$,
$$
begin{align}
|Gamma(x+iy)|
&=left|int_0^infty e^{-t},t^{x+iy-1};mathrm{d}tright|\
&=left|int_0^infty e^{-t},t^{x-1},e^{iylog(t)};mathrm{d}tright|\
&leint_0^inftyleft|e^{-t},t^{x-1},e^{iylog(t)}right|;mathrm{d}t\
&=int_0^infty e^{-t},t^{x-1};mathrm{d}t\
&=Gamma(x)\
&=|Gamma(x)|tag{1}.
end{align}
$$
However, I did not succeed to get a lower bound for $Gamma(x+i y)$ such that I get rid of the imaginary part. Any help in this direction?
Using the suggestion of @reuns and the Euler's reflection formula, one can shows that
begin{align} left|Gamma(x+i y)right| &= frac{pi}{left|Gamma(1-(x+i y)right| left|sin(pi (x+i y))right|} \
&ge frac{pi}{left|Gamma(1-x)right| left|sin(pi (x+i y))right|}
end{align}
But I can not still get a lower bound for the complex sine term such that I get rid of the imaginary part! Any suggestion?
complex-analysis inequality special-functions gamma-function
asked Jan 31 at 7:18
user144209user144209
42314
$endgroup$
I am trying to have some inequalities involving the special Gamma function. I am able to get an upper bound for $Gamma(x+i y)$, for $x>0$,
$$
begin{align}
|Gamma(x+iy)|
&=left|int_0^infty e^{-t},t^{x+iy-1};mathrm{d}tright|\
&=left|int_0^infty e^{-t},t^{x-1},e^{iylog(t)};mathrm{d}tright|\
&leint_0^inftyleft|e^{-t},t^{x-1},e^{iylog(t)}right|;mathrm{d}t\
&=int_0^infty e^{-t},t^{x-1};mathrm{d}t\
&=Gamma(x)\
&=|Gamma(x)|tag{1}.
end{align}
$$
However, I did not succeed to get a lower bound for $Gamma(x+i y)$ such that I get rid of the imaginary part. Any help in this direction?
Using the suggestion of @reuns and the Euler's reflection formula, one can shows that
begin{align} left|Gamma(x+i y)right| &= frac{pi}{left|Gamma(1-(x+i y)right| left|sin(pi (x+i y))right|} \
&ge frac{pi}{left|Gamma(1-x)right| left|sin(pi (x+i y))right|}
end{align}
But I can not still get a lower bound for the complex sine term such that I get rid of the imaginary part! Any suggestion?
complex-analysis inequality special-functions gamma-function
complex-analysis inequality special-functions gamma-function
asked Jan 31 at 7:18
user144209user144209
42314
asked Jan 31 at 7:18
user144209user144209
42314
edited Feb 15 at 7:43
user144209
asked Jan 31 at 7:18
user144209user144209
42314
asked Jan 31 at 7:18
user144209user144209
42314
asked Jan 31 at 7:18
user144209user144209
42314
42314
1
$begingroup$
You won't obtain anything directly from the integral, you need to show the explicit product formula for $Gamma(s)$, or things like the reflection formula.
$endgroup$
– reuns
Jan 31 at 8:21
$begingroup$
Thank you reuns for your comment. Can you please provide more details or a reference about that. Thanks.
$endgroup$
– user144209
Jan 31 at 9:55
1
$begingroup$
$Gamma(s)Gamma(1-s) = pi/sin(pi s)$ gives the exact decay on $Re(s) in mathbb{Z}/2$. For other $Re(s)$ the easiest is to look at the (complex) Stirling approximation, a direct consequence of the product formula en.wikipedia.org/wiki/…
$endgroup$
– reuns
Jan 31 at 10:02
add a comment |
1
$begingroup$
You won't obtain anything directly from the integral, you need to show the explicit product formula for $Gamma(s)$, or things like the reflection formula.
$endgroup$
– reuns
Jan 31 at 8:21
$begingroup$
Thank you reuns for your comment. Can you please provide more details or a reference about that. Thanks.
$endgroup$
– user144209
Jan 31 at 9:55
1
$begingroup$
$Gamma(s)Gamma(1-s) = pi/sin(pi s)$ gives the exact decay on $Re(s) in mathbb{Z}/2$. For other $Re(s)$ the easiest is to look at the (complex) Stirling approximation, a direct consequence of the product formula en.wikipedia.org/wiki/…
$endgroup$
– reuns
Jan 31 at 10:02
1
1
$begingroup$
You won't obtain anything directly from the integral, you need to show the explicit product formula for $Gamma(s)$, or things like the reflection formula.
$endgroup$
– reuns
Jan 31 at 8:21
$begingroup$
You won't obtain anything directly from the integral, you need to show the explicit product formula for $Gamma(s)$, or things like the reflection formula.
$endgroup$
– reuns
Jan 31 at 8:21
$begingroup$
Thank you reuns for your comment. Can you please provide more details or a reference about that. Thanks.
$endgroup$
– user144209
Jan 31 at 9:55
$begingroup$
Thank you reuns for your comment. Can you please provide more details or a reference about that. Thanks.
$endgroup$
– user144209
Jan 31 at 9:55
1
1
$begingroup$
$Gamma(s)Gamma(1-s) = pi/sin(pi s)$ gives the exact decay on $Re(s) in mathbb{Z}/2$. For other $Re(s)$ the easiest is to look at the (complex) Stirling approximation, a direct consequence of the product formula en.wikipedia.org/wiki/…
$endgroup$
– reuns
Jan 31 at 10:02
$begingroup$
$Gamma(s)Gamma(1-s) = pi/sin(pi s)$ gives the exact decay on $Re(s) in mathbb{Z}/2$. For other $Re(s)$ the easiest is to look at the (complex) Stirling approximation, a direct consequence of the product formula en.wikipedia.org/wiki/…
$endgroup$
– reuns
Jan 31 at 10:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
According to the NIST Handbook, edition of 2010, formula 5.6.7, there is the estimate
$$
|Gamma(x + i y)| ge frac{ Gamma(x)}{sqrt{cosh (pi y)}}
$$
for $x ge frac{1}{2}$. By the comment above, this is sharp for $x = frac{1}{2}$. This seems to say that you can't get rid of the imaginary part.
answered Feb 14 at 15:41
Hans EnglerHans Engler
10.7k11836
$endgroup$
1
$begingroup$
$|Gamma|$ goes very fast to zero on any fixed vertical line as the imaginary part goes to infinity (both plus and minus for that matter) so obviously any lower bound must involve the imaginary part...
$endgroup$
– Conrad
Feb 14 at 15:59
$begingroup$
well I can get a better bound from my derivation if I am not worried about the imaginary part using the fact that $|sin(x+iy)| le sinh(sqrt{x^2+y^2}) $ but my aim is to get rid of the imaginary part.
$endgroup$
– user144209
Feb 15 at 7:42
add a comment |
Your Answer
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1 Answer
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active
oldest
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1 Answer
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active
oldest
votes
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oldest
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active
oldest
votes
$begingroup$
According to the NIST Handbook, edition of 2010, formula 5.6.7, there is the estimate
$$
|Gamma(x + i y)| ge frac{ Gamma(x)}{sqrt{cosh (pi y)}}
$$
for $x ge frac{1}{2}$. By the comment above, this is sharp for $x = frac{1}{2}$. This seems to say that you can't get rid of the imaginary part.
answered Feb 14 at 15:41
Hans EnglerHans Engler
10.7k11836
$endgroup$
1
$begingroup$
$|Gamma|$ goes very fast to zero on any fixed vertical line as the imaginary part goes to infinity (both plus and minus for that matter) so obviously any lower bound must involve the imaginary part...
$endgroup$
– Conrad
Feb 14 at 15:59
$begingroup$
well I can get a better bound from my derivation if I am not worried about the imaginary part using the fact that $|sin(x+iy)| le sinh(sqrt{x^2+y^2}) $ but my aim is to get rid of the imaginary part.
$endgroup$
– user144209
Feb 15 at 7:42
add a comment |
$begingroup$
According to the NIST Handbook, edition of 2010, formula 5.6.7, there is the estimate
$$
|Gamma(x + i y)| ge frac{ Gamma(x)}{sqrt{cosh (pi y)}}
$$
for $x ge frac{1}{2}$. By the comment above, this is sharp for $x = frac{1}{2}$. This seems to say that you can't get rid of the imaginary part.
answered Feb 14 at 15:41
Hans EnglerHans Engler
10.7k11836
$endgroup$
1
$begingroup$
$|Gamma|$ goes very fast to zero on any fixed vertical line as the imaginary part goes to infinity (both plus and minus for that matter) so obviously any lower bound must involve the imaginary part...
$endgroup$
– Conrad
Feb 14 at 15:59
$begingroup$
well I can get a better bound from my derivation if I am not worried about the imaginary part using the fact that $|sin(x+iy)| le sinh(sqrt{x^2+y^2}) $ but my aim is to get rid of the imaginary part.
$endgroup$
– user144209
Feb 15 at 7:42
add a comment |
$begingroup$
According to the NIST Handbook, edition of 2010, formula 5.6.7, there is the estimate
$$
|Gamma(x + i y)| ge frac{ Gamma(x)}{sqrt{cosh (pi y)}}
$$
for $x ge frac{1}{2}$. By the comment above, this is sharp for $x = frac{1}{2}$. This seems to say that you can't get rid of the imaginary part.
answered Feb 14 at 15:41
Hans EnglerHans Engler
10.7k11836
$endgroup$
According to the NIST Handbook, edition of 2010, formula 5.6.7, there is the estimate
$$
|Gamma(x + i y)| ge frac{ Gamma(x)}{sqrt{cosh (pi y)}}
$$
for $x ge frac{1}{2}$. By the comment above, this is sharp for $x = frac{1}{2}$. This seems to say that you can't get rid of the imaginary part.
answered Feb 14 at 15:41
Hans EnglerHans Engler
10.7k11836
edited Feb 14 at 15:50
answered Feb 14 at 15:41
Hans EnglerHans Engler
10.7k11836
answered Feb 14 at 15:41
Hans EnglerHans Engler
10.7k11836
answered Feb 14 at 15:41
Hans EnglerHans Engler
10.7k11836
10.7k11836
1
$begingroup$
$|Gamma|$ goes very fast to zero on any fixed vertical line as the imaginary part goes to infinity (both plus and minus for that matter) so obviously any lower bound must involve the imaginary part...
$endgroup$
– Conrad
Feb 14 at 15:59
$begingroup$
well I can get a better bound from my derivation if I am not worried about the imaginary part using the fact that $|sin(x+iy)| le sinh(sqrt{x^2+y^2}) $ but my aim is to get rid of the imaginary part.
$endgroup$
– user144209
Feb 15 at 7:42
add a comment |
1
$begingroup$
$|Gamma|$ goes very fast to zero on any fixed vertical line as the imaginary part goes to infinity (both plus and minus for that matter) so obviously any lower bound must involve the imaginary part...
$endgroup$
– Conrad
Feb 14 at 15:59
$begingroup$
well I can get a better bound from my derivation if I am not worried about the imaginary part using the fact that $|sin(x+iy)| le sinh(sqrt{x^2+y^2}) $ but my aim is to get rid of the imaginary part.
$endgroup$
– user144209
Feb 15 at 7:42
1
1
$begingroup$
$|Gamma|$ goes very fast to zero on any fixed vertical line as the imaginary part goes to infinity (both plus and minus for that matter) so obviously any lower bound must involve the imaginary part...
$endgroup$
– Conrad
Feb 14 at 15:59
$begingroup$
$|Gamma|$ goes very fast to zero on any fixed vertical line as the imaginary part goes to infinity (both plus and minus for that matter) so obviously any lower bound must involve the imaginary part...
$endgroup$
– Conrad
Feb 14 at 15:59
$begingroup$
well I can get a better bound from my derivation if I am not worried about the imaginary part using the fact that $|sin(x+iy)| le sinh(sqrt{x^2+y^2}) $ but my aim is to get rid of the imaginary part.
$endgroup$
– user144209
Feb 15 at 7:42
$begingroup$
well I can get a better bound from my derivation if I am not worried about the imaginary part using the fact that $|sin(x+iy)| le sinh(sqrt{x^2+y^2}) $ but my aim is to get rid of the imaginary part.
$endgroup$
– user144209
Feb 15 at 7:42
add a comment |
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$begingroup$
You won't obtain anything directly from the integral, you need to show the explicit product formula for $Gamma(s)$, or things like the reflection formula.
$endgroup$
– reuns
Jan 31 at 8:21
$begingroup$
Thank you reuns for your comment. Can you please provide more details or a reference about that. Thanks.
$endgroup$
– user144209
Jan 31 at 9:55
1
$begingroup$
$Gamma(s)Gamma(1-s) = pi/sin(pi s)$ gives the exact decay on $Re(s) in mathbb{Z}/2$. For other $Re(s)$ the easiest is to look at the (complex) Stirling approximation, a direct consequence of the product formula en.wikipedia.org/wiki/…
$endgroup$
– reuns
Jan 31 at 10:02