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Why does $|e^-tt^s-1| = e^-tt^sigma-1$, where $sigma = Re(s)$?
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I am going over Stein and Shakarchi's section on the gamma function, and most of it makes sense except for this little technicality. I've also read a little bit of Titchmarsh and he makes an identical claim, but I still don't know why this is true.
complex-analysis complex-numbers
asked Jul 16 at 14:04
Josh
477
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up vote
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favorite
I am going over Stein and Shakarchi's section on the gamma function, and most of it makes sense except for this little technicality. I've also read a little bit of Titchmarsh and he makes an identical claim, but I still don't know why this is true.
complex-analysis complex-numbers
asked Jul 16 at 14:04
Josh
477
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am going over Stein and Shakarchi's section on the gamma function, and most of it makes sense except for this little technicality. I've also read a little bit of Titchmarsh and he makes an identical claim, but I still don't know why this is true.
complex-analysis complex-numbers
asked Jul 16 at 14:04
Josh
477
I am going over Stein and Shakarchi's section on the gamma function, and most of it makes sense except for this little technicality. I've also read a little bit of Titchmarsh and he makes an identical claim, but I still don't know why this is true.
complex-analysis complex-numbers
asked Jul 16 at 14:04
Josh
477
asked Jul 16 at 14:04
Josh
477
asked Jul 16 at 14:04
Josh
477
asked Jul 16 at 14:04
Josh
477
477
add a comment |Â
add a comment |Â
2 Answers
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Because, if we set $s=sigma+itau$, $sigma,tauin bf R$, we have
$$bigl|t^s-1bigr|=bigl|t^sigma-1+itaubigr|=bigl|t^sigma-1bigr|cdotbigl|t^itaubigr|$$
and $t^itau$ has modulus $1$ since it is
$$t^itau=rm e^itauln t.$$
answered Jul 16 at 14:16
Bernard
110k635103
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$|e^-tt^s-1| = |e^-te^ln(t)(sigma + jomega - 1)| =|e^-t t^sigma - 1e^jomega ln(t)| = e^-tt^sigma-1$, where $sigma = Re(s)$
answered Jul 16 at 14:14
user1949350
977
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Because, if we set $s=sigma+itau$, $sigma,tauin bf R$, we have
$$bigl|t^s-1bigr|=bigl|t^sigma-1+itaubigr|=bigl|t^sigma-1bigr|cdotbigl|t^itaubigr|$$
and $t^itau$ has modulus $1$ since it is
$$t^itau=rm e^itauln t.$$
answered Jul 16 at 14:16
Bernard
110k635103
add a comment |Â
up vote
0
down vote
accepted
Because, if we set $s=sigma+itau$, $sigma,tauin bf R$, we have
$$bigl|t^s-1bigr|=bigl|t^sigma-1+itaubigr|=bigl|t^sigma-1bigr|cdotbigl|t^itaubigr|$$
and $t^itau$ has modulus $1$ since it is
$$t^itau=rm e^itauln t.$$
answered Jul 16 at 14:16
Bernard
110k635103
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Because, if we set $s=sigma+itau$, $sigma,tauin bf R$, we have
$$bigl|t^s-1bigr|=bigl|t^sigma-1+itaubigr|=bigl|t^sigma-1bigr|cdotbigl|t^itaubigr|$$
and $t^itau$ has modulus $1$ since it is
$$t^itau=rm e^itauln t.$$
answered Jul 16 at 14:16
Bernard
110k635103
Because, if we set $s=sigma+itau$, $sigma,tauin bf R$, we have
$$bigl|t^s-1bigr|=bigl|t^sigma-1+itaubigr|=bigl|t^sigma-1bigr|cdotbigl|t^itaubigr|$$
and $t^itau$ has modulus $1$ since it is
$$t^itau=rm e^itauln t.$$
answered Jul 16 at 14:16
Bernard
110k635103
answered Jul 16 at 14:16
Bernard
110k635103
answered Jul 16 at 14:16
Bernard
110k635103
answered Jul 16 at 14:16
Bernard
110k635103
110k635103
add a comment |Â
add a comment |Â
up vote
0
down vote
$|e^-tt^s-1| = |e^-te^ln(t)(sigma + jomega - 1)| =|e^-t t^sigma - 1e^jomega ln(t)| = e^-tt^sigma-1$, where $sigma = Re(s)$
answered Jul 16 at 14:14
user1949350
977
add a comment |Â
up vote
0
down vote
$|e^-tt^s-1| = |e^-te^ln(t)(sigma + jomega - 1)| =|e^-t t^sigma - 1e^jomega ln(t)| = e^-tt^sigma-1$, where $sigma = Re(s)$
answered Jul 16 at 14:14
user1949350
977
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$|e^-tt^s-1| = |e^-te^ln(t)(sigma + jomega - 1)| =|e^-t t^sigma - 1e^jomega ln(t)| = e^-tt^sigma-1$, where $sigma = Re(s)$
answered Jul 16 at 14:14
user1949350
977
$|e^-tt^s-1| = |e^-te^ln(t)(sigma + jomega - 1)| =|e^-t t^sigma - 1e^jomega ln(t)| = e^-tt^sigma-1$, where $sigma = Re(s)$
answered Jul 16 at 14:14
user1949350
977
answered Jul 16 at 14:14
user1949350
977
answered Jul 16 at 14:14
user1949350
977
answered Jul 16 at 14:14
user1949350
977
977
add a comment |Â
add a comment |Â
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