Argument Principle-like complex integral involving logarithm
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Let $g$ be a holomorphic function on $C$ and its interior, where $C$ is a circle with a radius $r$ around the origin. Furthermore let's assume that $g$ doesn't assume a zero value on $C$ and its interior. Evaluate the integral:
$$oint_C log z frac{g'(z)}{g(z)} dz$$
I tried parametrizing the circle as $re^{it}$ and using the line $t = pi$ as a branch cut for the logarithm. The most I could get is:
$$oint_C log z frac{g'(z)}{g(z)} dz = -r int_{-pi}^{pi}te^{it}frac{g'(re^{it})}{g(re^{it})} dt = 2pi i log(g(-r)) - iint_{-pi}^{pi}log(g(re^{it}))dt$$
where in the last part I used integration by parts. However I can't evaluate the last integral. If I apply integral by parts again to it I return back to the initial problem.
On the other side the integral resembles the Argument Principle, but this time $g$ is holomorphic on the region, so I don't how we can take advantage of it.
Additionally if I'm not mistaken we have that:
$$lim_{r to 0}oint_C log z frac{g'(z)}{g(z)} dz = 0$$
I proved this by using the fact that $left|frac{g'(z)}{g(z)}right|$ can be bounded by some $M$, not dependent on $r$ and then:
$$left|oint_C log z frac{g'(z)}{g(z)} dzright|leoint_C left|log z frac{g'(z)}{g(z)}right| dz le 2pi r (ln r + pi)M$$
Upon taking limit from both sides we get the answer.
This somehow does the job for me, but I'm still interested in whether the integral can be evaluated explicitly for any $r>0$
integration complex-analysis logarithms complex-integration
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0
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Let $g$ be a holomorphic function on $C$ and its interior, where $C$ is a circle with a radius $r$ around the origin. Furthermore let's assume that $g$ doesn't assume a zero value on $C$ and its interior. Evaluate the integral:
$$oint_C log z frac{g'(z)}{g(z)} dz$$
I tried parametrizing the circle as $re^{it}$ and using the line $t = pi$ as a branch cut for the logarithm. The most I could get is:
$$oint_C log z frac{g'(z)}{g(z)} dz = -r int_{-pi}^{pi}te^{it}frac{g'(re^{it})}{g(re^{it})} dt = 2pi i log(g(-r)) - iint_{-pi}^{pi}log(g(re^{it}))dt$$
where in the last part I used integration by parts. However I can't evaluate the last integral. If I apply integral by parts again to it I return back to the initial problem.
On the other side the integral resembles the Argument Principle, but this time $g$ is holomorphic on the region, so I don't how we can take advantage of it.
Additionally if I'm not mistaken we have that:
$$lim_{r to 0}oint_C log z frac{g'(z)}{g(z)} dz = 0$$
I proved this by using the fact that $left|frac{g'(z)}{g(z)}right|$ can be bounded by some $M$, not dependent on $r$ and then:
$$left|oint_C log z frac{g'(z)}{g(z)} dzright|leoint_C left|log z frac{g'(z)}{g(z)}right| dz le 2pi r (ln r + pi)M$$
Upon taking limit from both sides we get the answer.
This somehow does the job for me, but I'm still interested in whether the integral can be evaluated explicitly for any $r>0$
integration complex-analysis logarithms complex-integration
If $arg zin[a,a+2pi)$, the integral should equal $-2pi i(ln(g(re^{ia}))-ln(g(0)))$.
– Szeto
Nov 18 at 14:10
@Szeto Could you please elaborate more
– Stefan4024
Nov 18 at 14:28
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $g$ be a holomorphic function on $C$ and its interior, where $C$ is a circle with a radius $r$ around the origin. Furthermore let's assume that $g$ doesn't assume a zero value on $C$ and its interior. Evaluate the integral:
$$oint_C log z frac{g'(z)}{g(z)} dz$$
I tried parametrizing the circle as $re^{it}$ and using the line $t = pi$ as a branch cut for the logarithm. The most I could get is:
$$oint_C log z frac{g'(z)}{g(z)} dz = -r int_{-pi}^{pi}te^{it}frac{g'(re^{it})}{g(re^{it})} dt = 2pi i log(g(-r)) - iint_{-pi}^{pi}log(g(re^{it}))dt$$
where in the last part I used integration by parts. However I can't evaluate the last integral. If I apply integral by parts again to it I return back to the initial problem.
On the other side the integral resembles the Argument Principle, but this time $g$ is holomorphic on the region, so I don't how we can take advantage of it.
Additionally if I'm not mistaken we have that:
$$lim_{r to 0}oint_C log z frac{g'(z)}{g(z)} dz = 0$$
I proved this by using the fact that $left|frac{g'(z)}{g(z)}right|$ can be bounded by some $M$, not dependent on $r$ and then:
$$left|oint_C log z frac{g'(z)}{g(z)} dzright|leoint_C left|log z frac{g'(z)}{g(z)}right| dz le 2pi r (ln r + pi)M$$
Upon taking limit from both sides we get the answer.
This somehow does the job for me, but I'm still interested in whether the integral can be evaluated explicitly for any $r>0$
integration complex-analysis logarithms complex-integration
Let $g$ be a holomorphic function on $C$ and its interior, where $C$ is a circle with a radius $r$ around the origin. Furthermore let's assume that $g$ doesn't assume a zero value on $C$ and its interior. Evaluate the integral:
$$oint_C log z frac{g'(z)}{g(z)} dz$$
I tried parametrizing the circle as $re^{it}$ and using the line $t = pi$ as a branch cut for the logarithm. The most I could get is:
$$oint_C log z frac{g'(z)}{g(z)} dz = -r int_{-pi}^{pi}te^{it}frac{g'(re^{it})}{g(re^{it})} dt = 2pi i log(g(-r)) - iint_{-pi}^{pi}log(g(re^{it}))dt$$
where in the last part I used integration by parts. However I can't evaluate the last integral. If I apply integral by parts again to it I return back to the initial problem.
On the other side the integral resembles the Argument Principle, but this time $g$ is holomorphic on the region, so I don't how we can take advantage of it.
Additionally if I'm not mistaken we have that:
$$lim_{r to 0}oint_C log z frac{g'(z)}{g(z)} dz = 0$$
I proved this by using the fact that $left|frac{g'(z)}{g(z)}right|$ can be bounded by some $M$, not dependent on $r$ and then:
$$left|oint_C log z frac{g'(z)}{g(z)} dzright|leoint_C left|log z frac{g'(z)}{g(z)}right| dz le 2pi r (ln r + pi)M$$
Upon taking limit from both sides we get the answer.
This somehow does the job for me, but I'm still interested in whether the integral can be evaluated explicitly for any $r>0$
integration complex-analysis logarithms complex-integration
integration complex-analysis logarithms complex-integration
asked Nov 18 at 13:56
Stefan4024
30k63377
30k63377
If $arg zin[a,a+2pi)$, the integral should equal $-2pi i(ln(g(re^{ia}))-ln(g(0)))$.
– Szeto
Nov 18 at 14:10
@Szeto Could you please elaborate more
– Stefan4024
Nov 18 at 14:28
add a comment |
If $arg zin[a,a+2pi)$, the integral should equal $-2pi i(ln(g(re^{ia}))-ln(g(0)))$.
– Szeto
Nov 18 at 14:10
@Szeto Could you please elaborate more
– Stefan4024
Nov 18 at 14:28
If $arg zin[a,a+2pi)$, the integral should equal $-2pi i(ln(g(re^{ia}))-ln(g(0)))$.
– Szeto
Nov 18 at 14:10
If $arg zin[a,a+2pi)$, the integral should equal $-2pi i(ln(g(re^{ia}))-ln(g(0)))$.
– Szeto
Nov 18 at 14:10
@Szeto Could you please elaborate more
– Stefan4024
Nov 18 at 14:28
@Szeto Could you please elaborate more
– Stefan4024
Nov 18 at 14:28
add a comment |
1 Answer
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Let $f(z)=g’/g$.
Your aim is to find $$int_{gamma_1}f(z)log z~dz$$
By Cauchy’s integral theorem,
$$int_{gamma_1}=-left(int_{gamma_2}+ int_{gamma_3}+ int_{gamma_4}right)$$
You have proved $int_{gamma_3}=0$.
By the theorem I wrote here,
$$left(int_{gamma_2}+ int_{gamma_4}right)f(z)log z~dz=-2pi iint^{re^{ia}}_0f(z)dz$$
So, indeed,
$$int_{gamma_1}frac{g’(z)}{g(z)}log z~dz=2pi i[ln g(re^{ia})-ln g(0)]$$
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
accepted
Let $f(z)=g’/g$.
Your aim is to find $$int_{gamma_1}f(z)log z~dz$$
By Cauchy’s integral theorem,
$$int_{gamma_1}=-left(int_{gamma_2}+ int_{gamma_3}+ int_{gamma_4}right)$$
You have proved $int_{gamma_3}=0$.
By the theorem I wrote here,
$$left(int_{gamma_2}+ int_{gamma_4}right)f(z)log z~dz=-2pi iint^{re^{ia}}_0f(z)dz$$
So, indeed,
$$int_{gamma_1}frac{g’(z)}{g(z)}log z~dz=2pi i[ln g(re^{ia})-ln g(0)]$$
add a comment |
up vote
1
down vote
accepted
Let $f(z)=g’/g$.
Your aim is to find $$int_{gamma_1}f(z)log z~dz$$
By Cauchy’s integral theorem,
$$int_{gamma_1}=-left(int_{gamma_2}+ int_{gamma_3}+ int_{gamma_4}right)$$
You have proved $int_{gamma_3}=0$.
By the theorem I wrote here,
$$left(int_{gamma_2}+ int_{gamma_4}right)f(z)log z~dz=-2pi iint^{re^{ia}}_0f(z)dz$$
So, indeed,
$$int_{gamma_1}frac{g’(z)}{g(z)}log z~dz=2pi i[ln g(re^{ia})-ln g(0)]$$
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $f(z)=g’/g$.
Your aim is to find $$int_{gamma_1}f(z)log z~dz$$
By Cauchy’s integral theorem,
$$int_{gamma_1}=-left(int_{gamma_2}+ int_{gamma_3}+ int_{gamma_4}right)$$
You have proved $int_{gamma_3}=0$.
By the theorem I wrote here,
$$left(int_{gamma_2}+ int_{gamma_4}right)f(z)log z~dz=-2pi iint^{re^{ia}}_0f(z)dz$$
So, indeed,
$$int_{gamma_1}frac{g’(z)}{g(z)}log z~dz=2pi i[ln g(re^{ia})-ln g(0)]$$
Let $f(z)=g’/g$.
Your aim is to find $$int_{gamma_1}f(z)log z~dz$$
By Cauchy’s integral theorem,
$$int_{gamma_1}=-left(int_{gamma_2}+ int_{gamma_3}+ int_{gamma_4}right)$$
You have proved $int_{gamma_3}=0$.
By the theorem I wrote here,
$$left(int_{gamma_2}+ int_{gamma_4}right)f(z)log z~dz=-2pi iint^{re^{ia}}_0f(z)dz$$
So, indeed,
$$int_{gamma_1}frac{g’(z)}{g(z)}log z~dz=2pi i[ln g(re^{ia})-ln g(0)]$$
edited Nov 19 at 0:09
answered Nov 19 at 0:02
Szeto
6,2792726
6,2792726
add a comment |
add a comment |
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If $arg zin[a,a+2pi)$, the integral should equal $-2pi i(ln(g(re^{ia}))-ln(g(0)))$.
– Szeto
Nov 18 at 14:10
@Szeto Could you please elaborate more
– Stefan4024
Nov 18 at 14:28