Cauchy's integral formula with special contour 4
up vote
0
down vote
favorite
suppose $gamma: [a,b] rightarrow mathbb{C} $ is a path of integration with $ gamma(a)=0, gamma(b)=1 and pm i notingamma([a,b]) $
Show that,
$$ int_{gamma} frac{1}{1+z^2} = frac{pi}{4} + k pi $$
I would try to apply Cauchy's integral formula. Therefore i can split the integrand in partial fractions with singularities at $ pm i$
How do i have to choose my contour such that it fullfilles the conditions at the beginning.
integration cauchy-integral-formula
asked Nov 24 at 9:37
Steven33
144
add a comment |
up vote
0
down vote
favorite
suppose $gamma: [a,b] rightarrow mathbb{C} $ is a path of integration with $ gamma(a)=0, gamma(b)=1 and pm i notingamma([a,b]) $
Show that,
$$ int_{gamma} frac{1}{1+z^2} = frac{pi}{4} + k pi $$
I would try to apply Cauchy's integral formula. Therefore i can split the integrand in partial fractions with singularities at $ pm i$
How do i have to choose my contour such that it fullfilles the conditions at the beginning.
integration cauchy-integral-formula
asked Nov 24 at 9:37
Steven33
144
Maybe i have to construct a closed path of integration. out of $ gamma(a)=0, gamma(b)=1$ ?
– Steven33
Nov 24 at 9:51
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
suppose $gamma: [a,b] rightarrow mathbb{C} $ is a path of integration with $ gamma(a)=0, gamma(b)=1 and pm i notingamma([a,b]) $
Show that,
$$ int_{gamma} frac{1}{1+z^2} = frac{pi}{4} + k pi $$
I would try to apply Cauchy's integral formula. Therefore i can split the integrand in partial fractions with singularities at $ pm i$
How do i have to choose my contour such that it fullfilles the conditions at the beginning.
integration cauchy-integral-formula
asked Nov 24 at 9:37
Steven33
144
suppose $gamma: [a,b] rightarrow mathbb{C} $ is a path of integration with $ gamma(a)=0, gamma(b)=1 and pm i notingamma([a,b]) $
Show that,
$$ int_{gamma} frac{1}{1+z^2} = frac{pi}{4} + k pi $$
I would try to apply Cauchy's integral formula. Therefore i can split the integrand in partial fractions with singularities at $ pm i$
How do i have to choose my contour such that it fullfilles the conditions at the beginning.
integration cauchy-integral-formula
integration cauchy-integral-formula
asked Nov 24 at 9:37
Steven33
144
asked Nov 24 at 9:37
Steven33
144
asked Nov 24 at 9:37
Steven33
144
asked Nov 24 at 9:37
Steven33
144
asked Nov 24 at 9:37
Steven33
144
144
Maybe i have to construct a closed path of integration. out of $ gamma(a)=0, gamma(b)=1$ ?
– Steven33
Nov 24 at 9:51
add a comment |
Maybe i have to construct a closed path of integration. out of $ gamma(a)=0, gamma(b)=1$ ?
– Steven33
Nov 24 at 9:51
Maybe i have to construct a closed path of integration. out of $ gamma(a)=0, gamma(b)=1$ ?
– Steven33
Nov 24 at 9:51
Maybe i have to construct a closed path of integration. out of $ gamma(a)=0, gamma(b)=1$ ?
– Steven33
Nov 24 at 9:51
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
Let $Gamma$ be another path: a straight line connecting from $1$ to $0$.
By residue theorem (let $n$ and $m$ be the winding numbers around $z=+i$ and $z=-i$ respectively),
$$
begin{align}
oint_{gamma+Gamma}frac{dz}{1+z^2}
&=2pi i left(noperatorname*{Res}_{z=i}frac1{1+z^2}+ moperatorname*{Res}_{z=i}frac1{1+z^2}right) \
&=pi(-n+m) \
&=pi(m-n)
end{align}
$$
Since
$$int_{Gamma}frac{dz}{1+z^2}=int^0_1frac{dt}{1+t^2}=operatorname{arctan}tbiggvert^{t=0}_{t=1}=-frac{pi}4$$
Thus,
$$int_{gamma}frac{dz}{1+z^2}=frac{pi}4+(m-n)pi$$
Doing without residue theorem:
Firstly, recognize that
$$intfrac{dz}{1+z^2}=frac1{2i}intleft(frac1{z+i}-frac1{z-i}right)dz=frac{ln(z+i)-ln(z-i)}{2i}$$
Thus,
$$intfrac{dz}{1+z^2}= frac{ln(z+i)-ln(z-i)}{2i} biggvert^{z=1}_{z=0}$$
Considering the multi-value-ness of $ln$ you would obtain the desired result.
In essence,
$$begin{align}
frac{ln(z+i)-ln(z-i)}{2i}biggvert^{z=1}_{z=0}
&=frac{ln(1+i)-ln(1-i)+ln(i)-ln(-i)}{2i}\
&=frac{ln(sqrt2e^{i(pi/4+2mpi)})-ln(sqrt2e^{i(-pi/4+2npi)})}{2i}\
&~~~~+frac{ln(e^{i(pi/2+2ppi)})-ln(e^{i(-pi/2+2qpi)})}{2i}\
&=ifrac{pi/4+pi/4+pi/2+pi/2+2Npi}{2i}\
&=frac{pi}4+N’pi\
end{align}
$$
where $N’$ is an arbitrary integer.
answered Nov 24 at 10:21
Szeto
6,3192826
I am not allowed to uses the residue theorem. Therefore i wanted to split the integrand into partial fractions. How do u choose $ gamma_1, gamma_2 $?
– Steven33
Nov 24 at 10:26
Dont i have to enclose my singularities with $ Gamma$
– Steven33
Nov 24 at 10:39
@Steven33 Please see my edited answer.
– Szeto
Nov 24 at 12:03
Thank you:) Do i have to parameterize the straight line connecting from 0 to 1 and put that into my definition of the contour integral, meaning: z=t $ forall t in [0,1]$. Ok i see that makes not such a big difference.:)
– Steven33
Nov 24 at 12:42
Thank you very much:) You made my day. Thank you:) I made a horrible mistake.
– Steven33
Nov 24 at 13:00
|
show 2 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Let $Gamma$ be another path: a straight line connecting from $1$ to $0$.
By residue theorem (let $n$ and $m$ be the winding numbers around $z=+i$ and $z=-i$ respectively),
$$
begin{align}
oint_{gamma+Gamma}frac{dz}{1+z^2}
&=2pi i left(noperatorname*{Res}_{z=i}frac1{1+z^2}+ moperatorname*{Res}_{z=i}frac1{1+z^2}right) \
&=pi(-n+m) \
&=pi(m-n)
end{align}
$$
Since
$$int_{Gamma}frac{dz}{1+z^2}=int^0_1frac{dt}{1+t^2}=operatorname{arctan}tbiggvert^{t=0}_{t=1}=-frac{pi}4$$
Thus,
$$int_{gamma}frac{dz}{1+z^2}=frac{pi}4+(m-n)pi$$
Doing without residue theorem:
Firstly, recognize that
$$intfrac{dz}{1+z^2}=frac1{2i}intleft(frac1{z+i}-frac1{z-i}right)dz=frac{ln(z+i)-ln(z-i)}{2i}$$
Thus,
$$intfrac{dz}{1+z^2}= frac{ln(z+i)-ln(z-i)}{2i} biggvert^{z=1}_{z=0}$$
Considering the multi-value-ness of $ln$ you would obtain the desired result.
In essence,
$$begin{align}
frac{ln(z+i)-ln(z-i)}{2i}biggvert^{z=1}_{z=0}
&=frac{ln(1+i)-ln(1-i)+ln(i)-ln(-i)}{2i}\
&=frac{ln(sqrt2e^{i(pi/4+2mpi)})-ln(sqrt2e^{i(-pi/4+2npi)})}{2i}\
&~~~~+frac{ln(e^{i(pi/2+2ppi)})-ln(e^{i(-pi/2+2qpi)})}{2i}\
&=ifrac{pi/4+pi/4+pi/2+pi/2+2Npi}{2i}\
&=frac{pi}4+N’pi\
end{align}
$$
where $N’$ is an arbitrary integer.
answered Nov 24 at 10:21
Szeto
6,3192826
I am not allowed to uses the residue theorem. Therefore i wanted to split the integrand into partial fractions. How do u choose $ gamma_1, gamma_2 $?
– Steven33
Nov 24 at 10:26
Dont i have to enclose my singularities with $ Gamma$
– Steven33
Nov 24 at 10:39
@Steven33 Please see my edited answer.
– Szeto
Nov 24 at 12:03
Thank you:) Do i have to parameterize the straight line connecting from 0 to 1 and put that into my definition of the contour integral, meaning: z=t $ forall t in [0,1]$. Ok i see that makes not such a big difference.:)
– Steven33
Nov 24 at 12:42
Thank you very much:) You made my day. Thank you:) I made a horrible mistake.
– Steven33
Nov 24 at 13:00
|
show 2 more comments
up vote
0
down vote
accepted
Let $Gamma$ be another path: a straight line connecting from $1$ to $0$.
By residue theorem (let $n$ and $m$ be the winding numbers around $z=+i$ and $z=-i$ respectively),
$$
begin{align}
oint_{gamma+Gamma}frac{dz}{1+z^2}
&=2pi i left(noperatorname*{Res}_{z=i}frac1{1+z^2}+ moperatorname*{Res}_{z=i}frac1{1+z^2}right) \
&=pi(-n+m) \
&=pi(m-n)
end{align}
$$
Since
$$int_{Gamma}frac{dz}{1+z^2}=int^0_1frac{dt}{1+t^2}=operatorname{arctan}tbiggvert^{t=0}_{t=1}=-frac{pi}4$$
Thus,
$$int_{gamma}frac{dz}{1+z^2}=frac{pi}4+(m-n)pi$$
Doing without residue theorem:
Firstly, recognize that
$$intfrac{dz}{1+z^2}=frac1{2i}intleft(frac1{z+i}-frac1{z-i}right)dz=frac{ln(z+i)-ln(z-i)}{2i}$$
Thus,
$$intfrac{dz}{1+z^2}= frac{ln(z+i)-ln(z-i)}{2i} biggvert^{z=1}_{z=0}$$
Considering the multi-value-ness of $ln$ you would obtain the desired result.
In essence,
$$begin{align}
frac{ln(z+i)-ln(z-i)}{2i}biggvert^{z=1}_{z=0}
&=frac{ln(1+i)-ln(1-i)+ln(i)-ln(-i)}{2i}\
&=frac{ln(sqrt2e^{i(pi/4+2mpi)})-ln(sqrt2e^{i(-pi/4+2npi)})}{2i}\
&~~~~+frac{ln(e^{i(pi/2+2ppi)})-ln(e^{i(-pi/2+2qpi)})}{2i}\
&=ifrac{pi/4+pi/4+pi/2+pi/2+2Npi}{2i}\
&=frac{pi}4+N’pi\
end{align}
$$
where $N’$ is an arbitrary integer.
answered Nov 24 at 10:21
Szeto
6,3192826
I am not allowed to uses the residue theorem. Therefore i wanted to split the integrand into partial fractions. How do u choose $ gamma_1, gamma_2 $?
– Steven33
Nov 24 at 10:26
Dont i have to enclose my singularities with $ Gamma$
– Steven33
Nov 24 at 10:39
@Steven33 Please see my edited answer.
– Szeto
Nov 24 at 12:03
Thank you:) Do i have to parameterize the straight line connecting from 0 to 1 and put that into my definition of the contour integral, meaning: z=t $ forall t in [0,1]$. Ok i see that makes not such a big difference.:)
– Steven33
Nov 24 at 12:42
Thank you very much:) You made my day. Thank you:) I made a horrible mistake.
– Steven33
Nov 24 at 13:00
|
show 2 more comments
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Let $Gamma$ be another path: a straight line connecting from $1$ to $0$.
By residue theorem (let $n$ and $m$ be the winding numbers around $z=+i$ and $z=-i$ respectively),
$$
begin{align}
oint_{gamma+Gamma}frac{dz}{1+z^2}
&=2pi i left(noperatorname*{Res}_{z=i}frac1{1+z^2}+ moperatorname*{Res}_{z=i}frac1{1+z^2}right) \
&=pi(-n+m) \
&=pi(m-n)
end{align}
$$
Since
$$int_{Gamma}frac{dz}{1+z^2}=int^0_1frac{dt}{1+t^2}=operatorname{arctan}tbiggvert^{t=0}_{t=1}=-frac{pi}4$$
Thus,
$$int_{gamma}frac{dz}{1+z^2}=frac{pi}4+(m-n)pi$$
Doing without residue theorem:
Firstly, recognize that
$$intfrac{dz}{1+z^2}=frac1{2i}intleft(frac1{z+i}-frac1{z-i}right)dz=frac{ln(z+i)-ln(z-i)}{2i}$$
Thus,
$$intfrac{dz}{1+z^2}= frac{ln(z+i)-ln(z-i)}{2i} biggvert^{z=1}_{z=0}$$
Considering the multi-value-ness of $ln$ you would obtain the desired result.
In essence,
$$begin{align}
frac{ln(z+i)-ln(z-i)}{2i}biggvert^{z=1}_{z=0}
&=frac{ln(1+i)-ln(1-i)+ln(i)-ln(-i)}{2i}\
&=frac{ln(sqrt2e^{i(pi/4+2mpi)})-ln(sqrt2e^{i(-pi/4+2npi)})}{2i}\
&~~~~+frac{ln(e^{i(pi/2+2ppi)})-ln(e^{i(-pi/2+2qpi)})}{2i}\
&=ifrac{pi/4+pi/4+pi/2+pi/2+2Npi}{2i}\
&=frac{pi}4+N’pi\
end{align}
$$
where $N’$ is an arbitrary integer.
answered Nov 24 at 10:21
Szeto
6,3192826
Let $Gamma$ be another path: a straight line connecting from $1$ to $0$.
By residue theorem (let $n$ and $m$ be the winding numbers around $z=+i$ and $z=-i$ respectively),
$$
begin{align}
oint_{gamma+Gamma}frac{dz}{1+z^2}
&=2pi i left(noperatorname*{Res}_{z=i}frac1{1+z^2}+ moperatorname*{Res}_{z=i}frac1{1+z^2}right) \
&=pi(-n+m) \
&=pi(m-n)
end{align}
$$
Since
$$int_{Gamma}frac{dz}{1+z^2}=int^0_1frac{dt}{1+t^2}=operatorname{arctan}tbiggvert^{t=0}_{t=1}=-frac{pi}4$$
Thus,
$$int_{gamma}frac{dz}{1+z^2}=frac{pi}4+(m-n)pi$$
Doing without residue theorem:
Firstly, recognize that
$$intfrac{dz}{1+z^2}=frac1{2i}intleft(frac1{z+i}-frac1{z-i}right)dz=frac{ln(z+i)-ln(z-i)}{2i}$$
Thus,
$$intfrac{dz}{1+z^2}= frac{ln(z+i)-ln(z-i)}{2i} biggvert^{z=1}_{z=0}$$
Considering the multi-value-ness of $ln$ you would obtain the desired result.
In essence,
$$begin{align}
frac{ln(z+i)-ln(z-i)}{2i}biggvert^{z=1}_{z=0}
&=frac{ln(1+i)-ln(1-i)+ln(i)-ln(-i)}{2i}\
&=frac{ln(sqrt2e^{i(pi/4+2mpi)})-ln(sqrt2e^{i(-pi/4+2npi)})}{2i}\
&~~~~+frac{ln(e^{i(pi/2+2ppi)})-ln(e^{i(-pi/2+2qpi)})}{2i}\
&=ifrac{pi/4+pi/4+pi/2+pi/2+2Npi}{2i}\
&=frac{pi}4+N’pi\
end{align}
$$
where $N’$ is an arbitrary integer.
answered Nov 24 at 10:21
Szeto
6,3192826
edited Nov 24 at 13:02
answered Nov 24 at 10:21
Szeto
6,3192826
answered Nov 24 at 10:21
Szeto
6,3192826
answered Nov 24 at 10:21
Szeto
6,3192826
6,3192826
I am not allowed to uses the residue theorem. Therefore i wanted to split the integrand into partial fractions. How do u choose $ gamma_1, gamma_2 $?
– Steven33
Nov 24 at 10:26
Dont i have to enclose my singularities with $ Gamma$
– Steven33
Nov 24 at 10:39
@Steven33 Please see my edited answer.
– Szeto
Nov 24 at 12:03
Thank you:) Do i have to parameterize the straight line connecting from 0 to 1 and put that into my definition of the contour integral, meaning: z=t $ forall t in [0,1]$. Ok i see that makes not such a big difference.:)
– Steven33
Nov 24 at 12:42
Thank you very much:) You made my day. Thank you:) I made a horrible mistake.
– Steven33
Nov 24 at 13:00
|
show 2 more comments
I am not allowed to uses the residue theorem. Therefore i wanted to split the integrand into partial fractions. How do u choose $ gamma_1, gamma_2 $?
– Steven33
Nov 24 at 10:26
Dont i have to enclose my singularities with $ Gamma$
– Steven33
Nov 24 at 10:39
@Steven33 Please see my edited answer.
– Szeto
Nov 24 at 12:03
Thank you:) Do i have to parameterize the straight line connecting from 0 to 1 and put that into my definition of the contour integral, meaning: z=t $ forall t in [0,1]$. Ok i see that makes not such a big difference.:)
– Steven33
Nov 24 at 12:42
Thank you very much:) You made my day. Thank you:) I made a horrible mistake.
– Steven33
Nov 24 at 13:00
I am not allowed to uses the residue theorem. Therefore i wanted to split the integrand into partial fractions. How do u choose $ gamma_1, gamma_2 $?
– Steven33
Nov 24 at 10:26
I am not allowed to uses the residue theorem. Therefore i wanted to split the integrand into partial fractions. How do u choose $ gamma_1, gamma_2 $?
– Steven33
Nov 24 at 10:26
Dont i have to enclose my singularities with $ Gamma$
– Steven33
Nov 24 at 10:39
Dont i have to enclose my singularities with $ Gamma$
– Steven33
Nov 24 at 10:39
@Steven33 Please see my edited answer.
– Szeto
Nov 24 at 12:03
@Steven33 Please see my edited answer.
– Szeto
Nov 24 at 12:03
Thank you:) Do i have to parameterize the straight line connecting from 0 to 1 and put that into my definition of the contour integral, meaning: z=t $ forall t in [0,1]$. Ok i see that makes not such a big difference.:)
– Steven33
Nov 24 at 12:42
Thank you:) Do i have to parameterize the straight line connecting from 0 to 1 and put that into my definition of the contour integral, meaning: z=t $ forall t in [0,1]$. Ok i see that makes not such a big difference.:)
– Steven33
Nov 24 at 12:42
Thank you very much:) You made my day. Thank you:) I made a horrible mistake.
– Steven33
Nov 24 at 13:00
Thank you very much:) You made my day. Thank you:) I made a horrible mistake.
– Steven33
Nov 24 at 13:00
|
show 2 more comments
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Maybe i have to construct a closed path of integration. out of $ gamma(a)=0, gamma(b)=1$ ?
– Steven33
Nov 24 at 9:51