How to evaluate integrals like $ int_C frac{z}{e^z-1} dz$ without Residue Theorem?
I'm trying to evaluate two integrals:
$$ I_1 = int_{C(0, 1)} frac{z}{e^z-1} dz,quad I_2 = int_{C(1, 3)} frac{sin(iz)}{e^{iz}-1} dz $$
$ C(a, r) $ is a circular contour with radius $r$ centered at $a$, oriented counterclockwise.
In both cases there's a singularity point $ z=0 $. How can I handle it? Can I use some parametrization to get rid of exponential function?
In general, what can I do to evaluate integrals of type:
$$ I_3 = int_{gamma} frac{f(z)}{e^{g(z)}-1} dz$$
Where $f, g$ holomorphic on open, simply connected subset U, $gamma(t) in U$?
complex-analysis exponential-function cauchy-integral-formula
edited Nov 28 '18 at 19:33
Bernard
118k639112
asked Nov 28 '18 at 18:04
Krzysztof AntoniakKrzysztof Antoniak
837
add a comment |
I'm trying to evaluate two integrals:
$$ I_1 = int_{C(0, 1)} frac{z}{e^z-1} dz,quad I_2 = int_{C(1, 3)} frac{sin(iz)}{e^{iz}-1} dz $$
$ C(a, r) $ is a circular contour with radius $r$ centered at $a$, oriented counterclockwise.
In both cases there's a singularity point $ z=0 $. How can I handle it? Can I use some parametrization to get rid of exponential function?
In general, what can I do to evaluate integrals of type:
$$ I_3 = int_{gamma} frac{f(z)}{e^{g(z)}-1} dz$$
Where $f, g$ holomorphic on open, simply connected subset U, $gamma(t) in U$?
complex-analysis exponential-function cauchy-integral-formula
edited Nov 28 '18 at 19:33
Bernard
118k639112
asked Nov 28 '18 at 18:04
Krzysztof AntoniakKrzysztof Antoniak
837
2
What do you mean by $C(0,1)$ or $C(1,3)$? Why don't you want to use residues? Note that the singularities at $0$ are removable.
– Robert Israel
Nov 28 '18 at 18:09
1
By C I meant circular contour, I updated the question. About residues: my course did not reach series expansions yet, and I'm wondering if it's doable otherwise.
– Krzysztof Antoniak
Nov 28 '18 at 18:16
Were removable singularities mentioned? Cauchy integral theorem?
– Robert Israel
Nov 28 '18 at 19:58
add a comment |
I'm trying to evaluate two integrals:
$$ I_1 = int_{C(0, 1)} frac{z}{e^z-1} dz,quad I_2 = int_{C(1, 3)} frac{sin(iz)}{e^{iz}-1} dz $$
$ C(a, r) $ is a circular contour with radius $r$ centered at $a$, oriented counterclockwise.
In both cases there's a singularity point $ z=0 $. How can I handle it? Can I use some parametrization to get rid of exponential function?
In general, what can I do to evaluate integrals of type:
$$ I_3 = int_{gamma} frac{f(z)}{e^{g(z)}-1} dz$$
Where $f, g$ holomorphic on open, simply connected subset U, $gamma(t) in U$?
complex-analysis exponential-function cauchy-integral-formula
edited Nov 28 '18 at 19:33
Bernard
118k639112
asked Nov 28 '18 at 18:04
Krzysztof AntoniakKrzysztof Antoniak
837
I'm trying to evaluate two integrals:
$$ I_1 = int_{C(0, 1)} frac{z}{e^z-1} dz,quad I_2 = int_{C(1, 3)} frac{sin(iz)}{e^{iz}-1} dz $$
$ C(a, r) $ is a circular contour with radius $r$ centered at $a$, oriented counterclockwise.
In both cases there's a singularity point $ z=0 $. How can I handle it? Can I use some parametrization to get rid of exponential function?
In general, what can I do to evaluate integrals of type:
$$ I_3 = int_{gamma} frac{f(z)}{e^{g(z)}-1} dz$$
Where $f, g$ holomorphic on open, simply connected subset U, $gamma(t) in U$?
complex-analysis exponential-function cauchy-integral-formula
complex-analysis exponential-function cauchy-integral-formula
edited Nov 28 '18 at 19:33
Bernard
118k639112
asked Nov 28 '18 at 18:04
Krzysztof AntoniakKrzysztof Antoniak
837
edited Nov 28 '18 at 19:33
Bernard
118k639112
asked Nov 28 '18 at 18:04
Krzysztof AntoniakKrzysztof Antoniak
837
edited Nov 28 '18 at 19:33
Bernard
118k639112
edited Nov 28 '18 at 19:33
Bernard
118k639112
edited Nov 28 '18 at 19:33
Bernard
118k639112
118k639112
asked Nov 28 '18 at 18:04
Krzysztof AntoniakKrzysztof Antoniak
837
asked Nov 28 '18 at 18:04
Krzysztof AntoniakKrzysztof Antoniak
837
asked Nov 28 '18 at 18:04
Krzysztof AntoniakKrzysztof Antoniak
837
837
2
What do you mean by $C(0,1)$ or $C(1,3)$? Why don't you want to use residues? Note that the singularities at $0$ are removable.
– Robert Israel
Nov 28 '18 at 18:09
1
By C I meant circular contour, I updated the question. About residues: my course did not reach series expansions yet, and I'm wondering if it's doable otherwise.
– Krzysztof Antoniak
Nov 28 '18 at 18:16
Were removable singularities mentioned? Cauchy integral theorem?
– Robert Israel
Nov 28 '18 at 19:58
add a comment |
2
What do you mean by $C(0,1)$ or $C(1,3)$? Why don't you want to use residues? Note that the singularities at $0$ are removable.
– Robert Israel
Nov 28 '18 at 18:09
1
By C I meant circular contour, I updated the question. About residues: my course did not reach series expansions yet, and I'm wondering if it's doable otherwise.
– Krzysztof Antoniak
Nov 28 '18 at 18:16
Were removable singularities mentioned? Cauchy integral theorem?
– Robert Israel
Nov 28 '18 at 19:58
2
2
What do you mean by $C(0,1)$ or $C(1,3)$? Why don't you want to use residues? Note that the singularities at $0$ are removable.
– Robert Israel
Nov 28 '18 at 18:09
What do you mean by $C(0,1)$ or $C(1,3)$? Why don't you want to use residues? Note that the singularities at $0$ are removable.
– Robert Israel
Nov 28 '18 at 18:09
1
1
By C I meant circular contour, I updated the question. About residues: my course did not reach series expansions yet, and I'm wondering if it's doable otherwise.
– Krzysztof Antoniak
Nov 28 '18 at 18:16
By C I meant circular contour, I updated the question. About residues: my course did not reach series expansions yet, and I'm wondering if it's doable otherwise.
– Krzysztof Antoniak
Nov 28 '18 at 18:16
Were removable singularities mentioned? Cauchy integral theorem?
– Robert Israel
Nov 28 '18 at 19:58
Were removable singularities mentioned? Cauchy integral theorem?
– Robert Israel
Nov 28 '18 at 19:58
add a comment |
1 Answer
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This is really using the fact that the singularity is removable without explicitly mentioning it.
Somewhat more generally, consider an integral of the form
$$J = oint_Gamma frac{f(z)}{g(z)}; dz$$
where $f$ and $g$ are analytic in a simply connected region $U$ containing the piecewise smooth simple closed contour $Gamma$, $f(p) = g(p) = 0$ for some point $p$ on $Gamma$, and $g$ has no other zeros in or on $Gamma$. So except for the singularity at $p$ we would have $J=0$ by the Cauchy integral theorem.
We might compare this to another integral
$$ widetilde{J} = oint_widetilde{Gamma} frac{f(z)}{g(z)}; dz = 0$$
where $widetilde{Gamma}$ is obtained by modifying $Gamma$ near $p$ so that $p$ is outside $widetilde{Gamma}$.
The difference between them is the integral over a small contour $widetilde{widetilde Gamma}$ near $p$.
Now suppose we know that $f(z)/g(z)$ is bounded for $z$ near $p$, say $|f(z)/g(z)| < B$ for $|z - p| < epsilon$. If $widetilde{widetilde Gamma}$ is all in the region $|z-p|<epsilon$ and has length at most $k epsilon$ for some constant $k$, then we can
estimate
$$ |J |= |J - widetilde{J}| = left|oint_{widetilde{widetilde Gamma}} frac{f(z)}{g(z)}; dz right| le B cdot text{length}(widetilde{widetilde Gamma}) le B k epsilon $$
and taking the limit as $epsilon to 0$ we conclude that $J = 0$.
answered Nov 28 '18 at 20:24
Robert IsraelRobert Israel
319k23208457
add a comment |
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This is really using the fact that the singularity is removable without explicitly mentioning it.
Somewhat more generally, consider an integral of the form
$$J = oint_Gamma frac{f(z)}{g(z)}; dz$$
where $f$ and $g$ are analytic in a simply connected region $U$ containing the piecewise smooth simple closed contour $Gamma$, $f(p) = g(p) = 0$ for some point $p$ on $Gamma$, and $g$ has no other zeros in or on $Gamma$. So except for the singularity at $p$ we would have $J=0$ by the Cauchy integral theorem.
We might compare this to another integral
$$ widetilde{J} = oint_widetilde{Gamma} frac{f(z)}{g(z)}; dz = 0$$
where $widetilde{Gamma}$ is obtained by modifying $Gamma$ near $p$ so that $p$ is outside $widetilde{Gamma}$.
The difference between them is the integral over a small contour $widetilde{widetilde Gamma}$ near $p$.
Now suppose we know that $f(z)/g(z)$ is bounded for $z$ near $p$, say $|f(z)/g(z)| < B$ for $|z - p| < epsilon$. If $widetilde{widetilde Gamma}$ is all in the region $|z-p|<epsilon$ and has length at most $k epsilon$ for some constant $k$, then we can
estimate
$$ |J |= |J - widetilde{J}| = left|oint_{widetilde{widetilde Gamma}} frac{f(z)}{g(z)}; dz right| le B cdot text{length}(widetilde{widetilde Gamma}) le B k epsilon $$
and taking the limit as $epsilon to 0$ we conclude that $J = 0$.
answered Nov 28 '18 at 20:24
Robert IsraelRobert Israel
319k23208457
add a comment |
This is really using the fact that the singularity is removable without explicitly mentioning it.
Somewhat more generally, consider an integral of the form
$$J = oint_Gamma frac{f(z)}{g(z)}; dz$$
where $f$ and $g$ are analytic in a simply connected region $U$ containing the piecewise smooth simple closed contour $Gamma$, $f(p) = g(p) = 0$ for some point $p$ on $Gamma$, and $g$ has no other zeros in or on $Gamma$. So except for the singularity at $p$ we would have $J=0$ by the Cauchy integral theorem.
We might compare this to another integral
$$ widetilde{J} = oint_widetilde{Gamma} frac{f(z)}{g(z)}; dz = 0$$
where $widetilde{Gamma}$ is obtained by modifying $Gamma$ near $p$ so that $p$ is outside $widetilde{Gamma}$.
The difference between them is the integral over a small contour $widetilde{widetilde Gamma}$ near $p$.
Now suppose we know that $f(z)/g(z)$ is bounded for $z$ near $p$, say $|f(z)/g(z)| < B$ for $|z - p| < epsilon$. If $widetilde{widetilde Gamma}$ is all in the region $|z-p|<epsilon$ and has length at most $k epsilon$ for some constant $k$, then we can
estimate
$$ |J |= |J - widetilde{J}| = left|oint_{widetilde{widetilde Gamma}} frac{f(z)}{g(z)}; dz right| le B cdot text{length}(widetilde{widetilde Gamma}) le B k epsilon $$
and taking the limit as $epsilon to 0$ we conclude that $J = 0$.
answered Nov 28 '18 at 20:24
Robert IsraelRobert Israel
319k23208457
add a comment |
This is really using the fact that the singularity is removable without explicitly mentioning it.
Somewhat more generally, consider an integral of the form
$$J = oint_Gamma frac{f(z)}{g(z)}; dz$$
where $f$ and $g$ are analytic in a simply connected region $U$ containing the piecewise smooth simple closed contour $Gamma$, $f(p) = g(p) = 0$ for some point $p$ on $Gamma$, and $g$ has no other zeros in or on $Gamma$. So except for the singularity at $p$ we would have $J=0$ by the Cauchy integral theorem.
We might compare this to another integral
$$ widetilde{J} = oint_widetilde{Gamma} frac{f(z)}{g(z)}; dz = 0$$
where $widetilde{Gamma}$ is obtained by modifying $Gamma$ near $p$ so that $p$ is outside $widetilde{Gamma}$.
The difference between them is the integral over a small contour $widetilde{widetilde Gamma}$ near $p$.
Now suppose we know that $f(z)/g(z)$ is bounded for $z$ near $p$, say $|f(z)/g(z)| < B$ for $|z - p| < epsilon$. If $widetilde{widetilde Gamma}$ is all in the region $|z-p|<epsilon$ and has length at most $k epsilon$ for some constant $k$, then we can
estimate
$$ |J |= |J - widetilde{J}| = left|oint_{widetilde{widetilde Gamma}} frac{f(z)}{g(z)}; dz right| le B cdot text{length}(widetilde{widetilde Gamma}) le B k epsilon $$
and taking the limit as $epsilon to 0$ we conclude that $J = 0$.
answered Nov 28 '18 at 20:24
Robert IsraelRobert Israel
319k23208457
This is really using the fact that the singularity is removable without explicitly mentioning it.
Somewhat more generally, consider an integral of the form
$$J = oint_Gamma frac{f(z)}{g(z)}; dz$$
where $f$ and $g$ are analytic in a simply connected region $U$ containing the piecewise smooth simple closed contour $Gamma$, $f(p) = g(p) = 0$ for some point $p$ on $Gamma$, and $g$ has no other zeros in or on $Gamma$. So except for the singularity at $p$ we would have $J=0$ by the Cauchy integral theorem.
We might compare this to another integral
$$ widetilde{J} = oint_widetilde{Gamma} frac{f(z)}{g(z)}; dz = 0$$
where $widetilde{Gamma}$ is obtained by modifying $Gamma$ near $p$ so that $p$ is outside $widetilde{Gamma}$.
The difference between them is the integral over a small contour $widetilde{widetilde Gamma}$ near $p$.
Now suppose we know that $f(z)/g(z)$ is bounded for $z$ near $p$, say $|f(z)/g(z)| < B$ for $|z - p| < epsilon$. If $widetilde{widetilde Gamma}$ is all in the region $|z-p|<epsilon$ and has length at most $k epsilon$ for some constant $k$, then we can
estimate
$$ |J |= |J - widetilde{J}| = left|oint_{widetilde{widetilde Gamma}} frac{f(z)}{g(z)}; dz right| le B cdot text{length}(widetilde{widetilde Gamma}) le B k epsilon $$
and taking the limit as $epsilon to 0$ we conclude that $J = 0$.
answered Nov 28 '18 at 20:24
Robert IsraelRobert Israel
319k23208457
answered Nov 28 '18 at 20:24
Robert IsraelRobert Israel
319k23208457
answered Nov 28 '18 at 20:24
Robert IsraelRobert Israel
319k23208457
answered Nov 28 '18 at 20:24
Robert IsraelRobert Israel
319k23208457
319k23208457
add a comment |
add a comment |
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2
What do you mean by $C(0,1)$ or $C(1,3)$? Why don't you want to use residues? Note that the singularities at $0$ are removable.
– Robert Israel
Nov 28 '18 at 18:09
1
By C I meant circular contour, I updated the question. About residues: my course did not reach series expansions yet, and I'm wondering if it's doable otherwise.
– Krzysztof Antoniak
Nov 28 '18 at 18:16
Were removable singularities mentioned? Cauchy integral theorem?
– Robert Israel
Nov 28 '18 at 19:58