Mixing
Evaluating an integral using residue theorem
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I'm evaluating $frac{1}{2pi i}oint_C frac{e^{zt}}{z(z^2+1)}dz$, $t > 0$ where $C$ is defined by the vertices $2+2i$, $-2+2i$, $-2-2i$, and $2-2i$. I'm pretty sure this is done by finding the poles in $C$ and the residues at the poles and then adding them together, but the answer is supposed to be $1-cos t$. Here is what I've done:
$$frac{1}{2pi i}oint_Cf(z)dz=a_{-1}$$
$$a_{-1}=lim_{z to a}frac{1}{(m-1)!}frac{d^{m-1}}{dz^{m-1}}{(z-a)^mf(z)}$$
The poles are found where $z(z^2+1)$ would be zero, which are $z=0,i,-i$. I need to take the above limit as $z$ goes to each pole.
$$lim_{z to 0}frac{1}{0!}frac{d^0}{dz^0}{(z-0)^1frac{e^{zt}}{z(z^2+1)}}=lim_{z to 0}frac{e^{zt}}{z^2+1}=1$$
$$lim_{z to i}frac{1}{0!}frac{d^0}{dz^0}{(z-i)^1frac{e^z}{z(z^2+1)}}=lim_{z to i}frac{e^{zt}+(z-i)e^{zt}t}{2z}=frac{e^{it}}{2i}$$
$$lim_{z to -i}frac{1}{0!}frac{d^0}{dz^0}{(z+i)^1frac{e^z}{z(z^2+1)}}=lim_{z to -i}frac{e^{zt}+(z+i)e^{zt}t}{2z}=frac{e^{-it}}{-2i}$$
If I were correct, the answer would be $1+frac{e^{it}}{2i}-frac{e^{-it}}{2i}$.
complex-analysis proof-verification residue-calculus
asked Nov 18 at 1:33
AdamK
1138
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1
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I'm evaluating $frac{1}{2pi i}oint_C frac{e^{zt}}{z(z^2+1)}dz$, $t > 0$ where $C$ is defined by the vertices $2+2i$, $-2+2i$, $-2-2i$, and $2-2i$. I'm pretty sure this is done by finding the poles in $C$ and the residues at the poles and then adding them together, but the answer is supposed to be $1-cos t$. Here is what I've done:
$$frac{1}{2pi i}oint_Cf(z)dz=a_{-1}$$
$$a_{-1}=lim_{z to a}frac{1}{(m-1)!}frac{d^{m-1}}{dz^{m-1}}{(z-a)^mf(z)}$$
The poles are found where $z(z^2+1)$ would be zero, which are $z=0,i,-i$. I need to take the above limit as $z$ goes to each pole.
$$lim_{z to 0}frac{1}{0!}frac{d^0}{dz^0}{(z-0)^1frac{e^{zt}}{z(z^2+1)}}=lim_{z to 0}frac{e^{zt}}{z^2+1}=1$$
$$lim_{z to i}frac{1}{0!}frac{d^0}{dz^0}{(z-i)^1frac{e^z}{z(z^2+1)}}=lim_{z to i}frac{e^{zt}+(z-i)e^{zt}t}{2z}=frac{e^{it}}{2i}$$
$$lim_{z to -i}frac{1}{0!}frac{d^0}{dz^0}{(z+i)^1frac{e^z}{z(z^2+1)}}=lim_{z to -i}frac{e^{zt}+(z+i)e^{zt}t}{2z}=frac{e^{-it}}{-2i}$$
If I were correct, the answer would be $1+frac{e^{it}}{2i}-frac{e^{-it}}{2i}$.
complex-analysis proof-verification residue-calculus
asked Nov 18 at 1:33
AdamK
1138
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm evaluating $frac{1}{2pi i}oint_C frac{e^{zt}}{z(z^2+1)}dz$, $t > 0$ where $C$ is defined by the vertices $2+2i$, $-2+2i$, $-2-2i$, and $2-2i$. I'm pretty sure this is done by finding the poles in $C$ and the residues at the poles and then adding them together, but the answer is supposed to be $1-cos t$. Here is what I've done:
$$frac{1}{2pi i}oint_Cf(z)dz=a_{-1}$$
$$a_{-1}=lim_{z to a}frac{1}{(m-1)!}frac{d^{m-1}}{dz^{m-1}}{(z-a)^mf(z)}$$
The poles are found where $z(z^2+1)$ would be zero, which are $z=0,i,-i$. I need to take the above limit as $z$ goes to each pole.
$$lim_{z to 0}frac{1}{0!}frac{d^0}{dz^0}{(z-0)^1frac{e^{zt}}{z(z^2+1)}}=lim_{z to 0}frac{e^{zt}}{z^2+1}=1$$
$$lim_{z to i}frac{1}{0!}frac{d^0}{dz^0}{(z-i)^1frac{e^z}{z(z^2+1)}}=lim_{z to i}frac{e^{zt}+(z-i)e^{zt}t}{2z}=frac{e^{it}}{2i}$$
$$lim_{z to -i}frac{1}{0!}frac{d^0}{dz^0}{(z+i)^1frac{e^z}{z(z^2+1)}}=lim_{z to -i}frac{e^{zt}+(z+i)e^{zt}t}{2z}=frac{e^{-it}}{-2i}$$
If I were correct, the answer would be $1+frac{e^{it}}{2i}-frac{e^{-it}}{2i}$.
complex-analysis proof-verification residue-calculus
asked Nov 18 at 1:33
AdamK
1138
I'm evaluating $frac{1}{2pi i}oint_C frac{e^{zt}}{z(z^2+1)}dz$, $t > 0$ where $C$ is defined by the vertices $2+2i$, $-2+2i$, $-2-2i$, and $2-2i$. I'm pretty sure this is done by finding the poles in $C$ and the residues at the poles and then adding them together, but the answer is supposed to be $1-cos t$. Here is what I've done:
$$frac{1}{2pi i}oint_Cf(z)dz=a_{-1}$$
$$a_{-1}=lim_{z to a}frac{1}{(m-1)!}frac{d^{m-1}}{dz^{m-1}}{(z-a)^mf(z)}$$
The poles are found where $z(z^2+1)$ would be zero, which are $z=0,i,-i$. I need to take the above limit as $z$ goes to each pole.
$$lim_{z to 0}frac{1}{0!}frac{d^0}{dz^0}{(z-0)^1frac{e^{zt}}{z(z^2+1)}}=lim_{z to 0}frac{e^{zt}}{z^2+1}=1$$
$$lim_{z to i}frac{1}{0!}frac{d^0}{dz^0}{(z-i)^1frac{e^z}{z(z^2+1)}}=lim_{z to i}frac{e^{zt}+(z-i)e^{zt}t}{2z}=frac{e^{it}}{2i}$$
$$lim_{z to -i}frac{1}{0!}frac{d^0}{dz^0}{(z+i)^1frac{e^z}{z(z^2+1)}}=lim_{z to -i}frac{e^{zt}+(z+i)e^{zt}t}{2z}=frac{e^{-it}}{-2i}$$
If I were correct, the answer would be $1+frac{e^{it}}{2i}-frac{e^{-it}}{2i}$.
complex-analysis proof-verification residue-calculus
complex-analysis proof-verification residue-calculus
asked Nov 18 at 1:33
AdamK
1138
asked Nov 18 at 1:33
AdamK
1138
edited yesterday
asked Nov 18 at 1:33
AdamK
1138
asked Nov 18 at 1:33
AdamK
1138
asked Nov 18 at 1:33
AdamK
1138
1138
add a comment |
add a comment |
1 Answer
1
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up vote
1
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I think that you made a simple mistake in evaluating the second limit, which can be rewritten as:
$lim_{zto i}frac{(z-i)e^{zt}}{z(z+i)(z-i)} = -frac{1}{2}e^{it}$ so you missed a minus sign
answered yesterday
Sorin Tirc
5039
Yes, thank you. I was just giving it another try, and realized there were a few mistakes and typos, including taking a derivative incorrectly when applying L'Hopital's rule. So the sum of residues should be $1-frac{e^{it}}{2}-frac{e^{-it}}{2}$ according to what I have done. I suspect (hope) this can be worked around to the correct answer, but am currently struggling with this.
– AdamK
yesterday
1
You may have forgotten: $$frac{e^{it}+ e^{-it}}{2}=cos t.$$
– zhw.
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I think that you made a simple mistake in evaluating the second limit, which can be rewritten as:
$lim_{zto i}frac{(z-i)e^{zt}}{z(z+i)(z-i)} = -frac{1}{2}e^{it}$ so you missed a minus sign
answered yesterday
Sorin Tirc
5039
Yes, thank you. I was just giving it another try, and realized there were a few mistakes and typos, including taking a derivative incorrectly when applying L'Hopital's rule. So the sum of residues should be $1-frac{e^{it}}{2}-frac{e^{-it}}{2}$ according to what I have done. I suspect (hope) this can be worked around to the correct answer, but am currently struggling with this.
– AdamK
yesterday
1
You may have forgotten: $$frac{e^{it}+ e^{-it}}{2}=cos t.$$
– zhw.
yesterday
add a comment |
up vote
1
down vote
I think that you made a simple mistake in evaluating the second limit, which can be rewritten as:
$lim_{zto i}frac{(z-i)e^{zt}}{z(z+i)(z-i)} = -frac{1}{2}e^{it}$ so you missed a minus sign
answered yesterday
Sorin Tirc
5039
Yes, thank you. I was just giving it another try, and realized there were a few mistakes and typos, including taking a derivative incorrectly when applying L'Hopital's rule. So the sum of residues should be $1-frac{e^{it}}{2}-frac{e^{-it}}{2}$ according to what I have done. I suspect (hope) this can be worked around to the correct answer, but am currently struggling with this.
– AdamK
yesterday
1
You may have forgotten: $$frac{e^{it}+ e^{-it}}{2}=cos t.$$
– zhw.
yesterday
add a comment |
up vote
1
down vote
up vote
1
down vote
I think that you made a simple mistake in evaluating the second limit, which can be rewritten as:
$lim_{zto i}frac{(z-i)e^{zt}}{z(z+i)(z-i)} = -frac{1}{2}e^{it}$ so you missed a minus sign
answered yesterday
Sorin Tirc
5039
I think that you made a simple mistake in evaluating the second limit, which can be rewritten as:
$lim_{zto i}frac{(z-i)e^{zt}}{z(z+i)(z-i)} = -frac{1}{2}e^{it}$ so you missed a minus sign
answered yesterday
Sorin Tirc
5039
answered yesterday
Sorin Tirc
5039
answered yesterday
Sorin Tirc
5039
answered yesterday
Sorin Tirc
5039
5039
Yes, thank you. I was just giving it another try, and realized there were a few mistakes and typos, including taking a derivative incorrectly when applying L'Hopital's rule. So the sum of residues should be $1-frac{e^{it}}{2}-frac{e^{-it}}{2}$ according to what I have done. I suspect (hope) this can be worked around to the correct answer, but am currently struggling with this.
– AdamK
yesterday
1
You may have forgotten: $$frac{e^{it}+ e^{-it}}{2}=cos t.$$
– zhw.
yesterday
add a comment |
Yes, thank you. I was just giving it another try, and realized there were a few mistakes and typos, including taking a derivative incorrectly when applying L'Hopital's rule. So the sum of residues should be $1-frac{e^{it}}{2}-frac{e^{-it}}{2}$ according to what I have done. I suspect (hope) this can be worked around to the correct answer, but am currently struggling with this.
– AdamK
yesterday
1
You may have forgotten: $$frac{e^{it}+ e^{-it}}{2}=cos t.$$
– zhw.
yesterday
Yes, thank you. I was just giving it another try, and realized there were a few mistakes and typos, including taking a derivative incorrectly when applying L'Hopital's rule. So the sum of residues should be $1-frac{e^{it}}{2}-frac{e^{-it}}{2}$ according to what I have done. I suspect (hope) this can be worked around to the correct answer, but am currently struggling with this.
– AdamK
yesterday
Yes, thank you. I was just giving it another try, and realized there were a few mistakes and typos, including taking a derivative incorrectly when applying L'Hopital's rule. So the sum of residues should be $1-frac{e^{it}}{2}-frac{e^{-it}}{2}$ according to what I have done. I suspect (hope) this can be worked around to the correct answer, but am currently struggling with this.
– AdamK
yesterday
1
1
You may have forgotten: $$frac{e^{it}+ e^{-it}}{2}=cos t.$$
– zhw.
yesterday
You may have forgotten: $$frac{e^{it}+ e^{-it}}{2}=cos t.$$
– zhw.
yesterday
add a comment |
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