Is the residue theorem the correct approach for this integral?
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I want to calculate
$$int_{gamma} frac{sin z}{z^2 + 1}dz$$ where $gamma$ is the upper-half circle of radius 2 centered at the origin starting at 2. I know that since $gamma$ is not a closed curve, Cauchy's theorem will be of no use, but I believe that the integrand has a singularity at i, which is between the curve and the real axis. Can I use the residue theorem here?
complex-analysis contour-integration residue-calculus
asked Dec 12 '18 at 18:36
Richard VillalobosRichard Villalobos
1757
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add a comment |
$begingroup$
I want to calculate
$$int_{gamma} frac{sin z}{z^2 + 1}dz$$ where $gamma$ is the upper-half circle of radius 2 centered at the origin starting at 2. I know that since $gamma$ is not a closed curve, Cauchy's theorem will be of no use, but I believe that the integrand has a singularity at i, which is between the curve and the real axis. Can I use the residue theorem here?
complex-analysis contour-integration residue-calculus
asked Dec 12 '18 at 18:36
Richard VillalobosRichard Villalobos
1757
$endgroup$
add a comment |
$begingroup$
I want to calculate
$$int_{gamma} frac{sin z}{z^2 + 1}dz$$ where $gamma$ is the upper-half circle of radius 2 centered at the origin starting at 2. I know that since $gamma$ is not a closed curve, Cauchy's theorem will be of no use, but I believe that the integrand has a singularity at i, which is between the curve and the real axis. Can I use the residue theorem here?
complex-analysis contour-integration residue-calculus
asked Dec 12 '18 at 18:36
Richard VillalobosRichard Villalobos
1757
$endgroup$
I want to calculate
$$int_{gamma} frac{sin z}{z^2 + 1}dz$$ where $gamma$ is the upper-half circle of radius 2 centered at the origin starting at 2. I know that since $gamma$ is not a closed curve, Cauchy's theorem will be of no use, but I believe that the integrand has a singularity at i, which is between the curve and the real axis. Can I use the residue theorem here?
complex-analysis contour-integration residue-calculus
complex-analysis contour-integration residue-calculus
asked Dec 12 '18 at 18:36
Richard VillalobosRichard Villalobos
1757
asked Dec 12 '18 at 18:36
Richard VillalobosRichard Villalobos
1757
edited Dec 12 '18 at 18:53
Richard Villalobos
asked Dec 12 '18 at 18:36
Richard VillalobosRichard Villalobos
1757
asked Dec 12 '18 at 18:36
Richard VillalobosRichard Villalobos
1757
asked Dec 12 '18 at 18:36
Richard VillalobosRichard Villalobos
1757
1757
add a comment |
add a comment |
1 Answer
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Residue theorem is the right way to go. Note that since the integrand is odd on $mathbb R$,
$$ int_{gamma} frac{sin z}{z^2 + 1}dz = int_{gamma} frac{sin z}{z^2 + 1}dz + int_{-2}^2 frac{sin z}{z^2 + 1}dz = 2pi i operatorname{Res}_{z=i}left( frac{sin z}{z^2 + 1}right)$$
answered Dec 12 '18 at 18:40
Calvin KhorCalvin Khor
12.1k21438
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$begingroup$
I think you mean integral from $-2$ to $2$, looking at OP's definition of $gamma$.
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– SvanN
Dec 12 '18 at 18:42
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@S.vanNigtevecht yeah sorry, already edited
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– Calvin Khor
Dec 12 '18 at 18:43
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
Residue theorem is the right way to go. Note that since the integrand is odd on $mathbb R$,
$$ int_{gamma} frac{sin z}{z^2 + 1}dz = int_{gamma} frac{sin z}{z^2 + 1}dz + int_{-2}^2 frac{sin z}{z^2 + 1}dz = 2pi i operatorname{Res}_{z=i}left( frac{sin z}{z^2 + 1}right)$$
answered Dec 12 '18 at 18:40
Calvin KhorCalvin Khor
12.1k21438
$endgroup$
$begingroup$
I think you mean integral from $-2$ to $2$, looking at OP's definition of $gamma$.
$endgroup$
– SvanN
Dec 12 '18 at 18:42
$begingroup$
@S.vanNigtevecht yeah sorry, already edited
$endgroup$
– Calvin Khor
Dec 12 '18 at 18:43
add a comment |
$begingroup$
Residue theorem is the right way to go. Note that since the integrand is odd on $mathbb R$,
$$ int_{gamma} frac{sin z}{z^2 + 1}dz = int_{gamma} frac{sin z}{z^2 + 1}dz + int_{-2}^2 frac{sin z}{z^2 + 1}dz = 2pi i operatorname{Res}_{z=i}left( frac{sin z}{z^2 + 1}right)$$
answered Dec 12 '18 at 18:40
Calvin KhorCalvin Khor
12.1k21438
$endgroup$
$begingroup$
I think you mean integral from $-2$ to $2$, looking at OP's definition of $gamma$.
$endgroup$
– SvanN
Dec 12 '18 at 18:42
$begingroup$
@S.vanNigtevecht yeah sorry, already edited
$endgroup$
– Calvin Khor
Dec 12 '18 at 18:43
add a comment |
$begingroup$
Residue theorem is the right way to go. Note that since the integrand is odd on $mathbb R$,
$$ int_{gamma} frac{sin z}{z^2 + 1}dz = int_{gamma} frac{sin z}{z^2 + 1}dz + int_{-2}^2 frac{sin z}{z^2 + 1}dz = 2pi i operatorname{Res}_{z=i}left( frac{sin z}{z^2 + 1}right)$$
answered Dec 12 '18 at 18:40
Calvin KhorCalvin Khor
12.1k21438
$endgroup$
Residue theorem is the right way to go. Note that since the integrand is odd on $mathbb R$,
$$ int_{gamma} frac{sin z}{z^2 + 1}dz = int_{gamma} frac{sin z}{z^2 + 1}dz + int_{-2}^2 frac{sin z}{z^2 + 1}dz = 2pi i operatorname{Res}_{z=i}left( frac{sin z}{z^2 + 1}right)$$
answered Dec 12 '18 at 18:40
Calvin KhorCalvin Khor
12.1k21438
answered Dec 12 '18 at 18:40
Calvin KhorCalvin Khor
12.1k21438
answered Dec 12 '18 at 18:40
Calvin KhorCalvin Khor
12.1k21438
answered Dec 12 '18 at 18:40
Calvin KhorCalvin Khor
12.1k21438
12.1k21438
$begingroup$
I think you mean integral from $-2$ to $2$, looking at OP's definition of $gamma$.
$endgroup$
– SvanN
Dec 12 '18 at 18:42
$begingroup$
@S.vanNigtevecht yeah sorry, already edited
$endgroup$
– Calvin Khor
Dec 12 '18 at 18:43
add a comment |
$begingroup$
I think you mean integral from $-2$ to $2$, looking at OP's definition of $gamma$.
$endgroup$
– SvanN
Dec 12 '18 at 18:42
$begingroup$
@S.vanNigtevecht yeah sorry, already edited
$endgroup$
– Calvin Khor
Dec 12 '18 at 18:43
$begingroup$
I think you mean integral from $-2$ to $2$, looking at OP's definition of $gamma$.
$endgroup$
– SvanN
Dec 12 '18 at 18:42
$begingroup$
I think you mean integral from $-2$ to $2$, looking at OP's definition of $gamma$.
$endgroup$
– SvanN
Dec 12 '18 at 18:42
$begingroup$
@S.vanNigtevecht yeah sorry, already edited
$endgroup$
– Calvin Khor
Dec 12 '18 at 18:43
$begingroup$
@S.vanNigtevecht yeah sorry, already edited
$endgroup$
– Calvin Khor
Dec 12 '18 at 18:43
add a comment |
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