Calculating $intfrac{e^{iz}}{z}, dz$ on the semi-circle given by $re^{itheta}$ where $theta:,0topi$
$begingroup$
As a part of an exercise I need to calculate
$$ lim_{rto0}int_{sigma_{r}}frac{e^{iz}}{z}, dz $$
Where $$ sigma_{r}:,[0,pi]tomathbb{C} $$
$$ sigma_{r}(t)=re^{it} $$
I know how to calculate the integral on the full circle ($|z|=r$)
using Cauchy integral formula: $$2pi ie^{iz}|_{z=0}=2pi i$$
I have checked if
$$
(overline{frac{e^{iz}}{z})}=frac{e^{iz}}{z}
$$
so that the integral I want to calculate is $$frac{1}{2}cdot2pi i=pi i$$
but I got that that the equality above does not hold.
Can someone please suggest a way of calculating this integral ?
complex-analysis integration
asked Jul 14 '13 at 14:30
BelgiBelgi
14.6k1054115
$endgroup$
add a comment |
$begingroup$
As a part of an exercise I need to calculate
$$ lim_{rto0}int_{sigma_{r}}frac{e^{iz}}{z}, dz $$
Where $$ sigma_{r}:,[0,pi]tomathbb{C} $$
$$ sigma_{r}(t)=re^{it} $$
I know how to calculate the integral on the full circle ($|z|=r$)
using Cauchy integral formula: $$2pi ie^{iz}|_{z=0}=2pi i$$
I have checked if
$$
(overline{frac{e^{iz}}{z})}=frac{e^{iz}}{z}
$$
so that the integral I want to calculate is $$frac{1}{2}cdot2pi i=pi i$$
but I got that that the equality above does not hold.
Can someone please suggest a way of calculating this integral ?
complex-analysis integration
asked Jul 14 '13 at 14:30
BelgiBelgi
14.6k1054115
$endgroup$
$begingroup$
There might be a big difference in the difficulty between finding the integral and finding the limit. Are you sure you're asking what you want to ask?
$endgroup$
– Git Gud
Jul 14 '13 at 14:33
$begingroup$
@GitGud - Yes (but I think that the answer is independent of $r$ so there isn't actually a difference between the two)
$endgroup$
– Belgi
Jul 14 '13 at 14:35
$begingroup$
@Belgi The answer is not independent of $r$. You can relatively easily find a series expression for the integral by expanding $e^{iz}$ into its Taylor series.
$endgroup$
– Daniel Fischer♦
Jul 14 '13 at 14:46
add a comment |
$begingroup$
As a part of an exercise I need to calculate
$$ lim_{rto0}int_{sigma_{r}}frac{e^{iz}}{z}, dz $$
Where $$ sigma_{r}:,[0,pi]tomathbb{C} $$
$$ sigma_{r}(t)=re^{it} $$
I know how to calculate the integral on the full circle ($|z|=r$)
using Cauchy integral formula: $$2pi ie^{iz}|_{z=0}=2pi i$$
I have checked if
$$
(overline{frac{e^{iz}}{z})}=frac{e^{iz}}{z}
$$
so that the integral I want to calculate is $$frac{1}{2}cdot2pi i=pi i$$
but I got that that the equality above does not hold.
Can someone please suggest a way of calculating this integral ?
complex-analysis integration
asked Jul 14 '13 at 14:30
BelgiBelgi
14.6k1054115
$endgroup$
As a part of an exercise I need to calculate
$$ lim_{rto0}int_{sigma_{r}}frac{e^{iz}}{z}, dz $$
Where $$ sigma_{r}:,[0,pi]tomathbb{C} $$
$$ sigma_{r}(t)=re^{it} $$
I know how to calculate the integral on the full circle ($|z|=r$)
using Cauchy integral formula: $$2pi ie^{iz}|_{z=0}=2pi i$$
I have checked if
$$
(overline{frac{e^{iz}}{z})}=frac{e^{iz}}{z}
$$
so that the integral I want to calculate is $$frac{1}{2}cdot2pi i=pi i$$
but I got that that the equality above does not hold.
Can someone please suggest a way of calculating this integral ?
complex-analysis integration
complex-analysis integration
asked Jul 14 '13 at 14:30
BelgiBelgi
14.6k1054115
asked Jul 14 '13 at 14:30
BelgiBelgi
14.6k1054115
asked Jul 14 '13 at 14:30
BelgiBelgi
14.6k1054115
asked Jul 14 '13 at 14:30
BelgiBelgi
14.6k1054115
asked Jul 14 '13 at 14:30
BelgiBelgi
14.6k1054115
14.6k1054115
$begingroup$
There might be a big difference in the difficulty between finding the integral and finding the limit. Are you sure you're asking what you want to ask?
$endgroup$
– Git Gud
Jul 14 '13 at 14:33
$begingroup$
@GitGud - Yes (but I think that the answer is independent of $r$ so there isn't actually a difference between the two)
$endgroup$
– Belgi
Jul 14 '13 at 14:35
$begingroup$
@Belgi The answer is not independent of $r$. You can relatively easily find a series expression for the integral by expanding $e^{iz}$ into its Taylor series.
$endgroup$
– Daniel Fischer♦
Jul 14 '13 at 14:46
add a comment |
$begingroup$
There might be a big difference in the difficulty between finding the integral and finding the limit. Are you sure you're asking what you want to ask?
$endgroup$
– Git Gud
Jul 14 '13 at 14:33
$begingroup$
@GitGud - Yes (but I think that the answer is independent of $r$ so there isn't actually a difference between the two)
$endgroup$
– Belgi
Jul 14 '13 at 14:35
$begingroup$
@Belgi The answer is not independent of $r$. You can relatively easily find a series expression for the integral by expanding $e^{iz}$ into its Taylor series.
$endgroup$
– Daniel Fischer♦
Jul 14 '13 at 14:46
$begingroup$
There might be a big difference in the difficulty between finding the integral and finding the limit. Are you sure you're asking what you want to ask?
$endgroup$
– Git Gud
Jul 14 '13 at 14:33
$begingroup$
There might be a big difference in the difficulty between finding the integral and finding the limit. Are you sure you're asking what you want to ask?
$endgroup$
– Git Gud
Jul 14 '13 at 14:33
$begingroup$
@GitGud - Yes (but I think that the answer is independent of $r$ so there isn't actually a difference between the two)
$endgroup$
– Belgi
Jul 14 '13 at 14:35
$begingroup$
@GitGud - Yes (but I think that the answer is independent of $r$ so there isn't actually a difference between the two)
$endgroup$
– Belgi
Jul 14 '13 at 14:35
$begingroup$
@Belgi The answer is not independent of $r$. You can relatively easily find a series expression for the integral by expanding $e^{iz}$ into its Taylor series.
$endgroup$
– Daniel Fischer♦
Jul 14 '13 at 14:46
$begingroup$
@Belgi The answer is not independent of $r$. You can relatively easily find a series expression for the integral by expanding $e^{iz}$ into its Taylor series.
$endgroup$
– Daniel Fischer♦
Jul 14 '13 at 14:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that $dfrac{e^{iz}-1}{z}$ is bounded near $z=0$. Since the length of $sigma_r$ tends to $0$, we get
$$
lim_{rto0^+}int_{largesigma_r}frac{e^{iz}-1}{z},mathrm{d}z=0
$$
Thus,
$$
begin{align}
lim_{rto0^+}int_{largesigma_r}frac{e^{iz}}{z},mathrm{d}z
&=lim_{rto0^+}int_{largesigma_r}frac{e^{iz}-1}{z},mathrm{d}z
+lim_{rto0^+}int_{largesigma_r}frac{1}{z},mathrm{d}z\
&=0+lim_{rto0^+}int_0^pifrac{1}{re^{it}},mathrm{d}re^{it}\
&=lim_{rto0^+}int_0^pi i,mathrm{d}t\[9pt]
&=pi i
end{align}
$$
answered Jul 14 '13 at 17:13
robjohn♦robjohn
267k27308632
$endgroup$
add a comment |
$begingroup$
For $k neq -1$, the function $z mapsto z^k$ has a primitive on $mathbb{C}setminus {0}$, namely $z mapsto frac{1}{k+1}z^{k+1}$.
Thus, for $f$ holomorphic in a punctured neighbourhood of $0$, with the Laurent expansion
$$f(z) = sum_{-infty}^infty a_k z^k,$$
you can semi-explicitly evaluate the integral over $sigma_r$ (for small enough $r$) as
$$begin{align}
int_{sigma_r} f(z),dz &= a_{-1}int_{sigma_r}frac{dz}{z} + sum_{kneq -1} a_k int_{sigma_r}z^k\
&= pi i a_{-1} + sum_{k neq -1} frac{a_k r^{k+1}}{k+1} left((-1)^{k+1} - 1right)\
&= pi i a_{-1} -2 sum_{-infty}^infty frac{a_{2k}r^{2k+1}}{2k+1}.
end{align}$$
From that, you can read off that in general the integral depends on $r$, and it converges to $pi i operatorname{res}_0 f$ when $f$ has a simple pole in $0$ (generally, if $f$ has an odd principal part).
Your integrand
$$frac{e^{iz}}{z} = sum_{k=0}^infty frac{i^k}{k!}z^{k-1} = frac1z + i -frac{z}{2} -ifrac{z^2}{6} + frac{z^3}{24} + dotsb$$
has $0$ as a simple pole, hence the above applies. Since not all even coefficients of the Laurent expansion vanish, the integral depends on $r$, but the limit is $pi i$.
answered Jul 14 '13 at 16:02
Daniel Fischer♦Daniel Fischer
174k16167286
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that $dfrac{e^{iz}-1}{z}$ is bounded near $z=0$. Since the length of $sigma_r$ tends to $0$, we get
$$
lim_{rto0^+}int_{largesigma_r}frac{e^{iz}-1}{z},mathrm{d}z=0
$$
Thus,
$$
begin{align}
lim_{rto0^+}int_{largesigma_r}frac{e^{iz}}{z},mathrm{d}z
&=lim_{rto0^+}int_{largesigma_r}frac{e^{iz}-1}{z},mathrm{d}z
+lim_{rto0^+}int_{largesigma_r}frac{1}{z},mathrm{d}z\
&=0+lim_{rto0^+}int_0^pifrac{1}{re^{it}},mathrm{d}re^{it}\
&=lim_{rto0^+}int_0^pi i,mathrm{d}t\[9pt]
&=pi i
end{align}
$$
answered Jul 14 '13 at 17:13
robjohn♦robjohn
267k27308632
$endgroup$
add a comment |
$begingroup$
Note that $dfrac{e^{iz}-1}{z}$ is bounded near $z=0$. Since the length of $sigma_r$ tends to $0$, we get
$$
lim_{rto0^+}int_{largesigma_r}frac{e^{iz}-1}{z},mathrm{d}z=0
$$
Thus,
$$
begin{align}
lim_{rto0^+}int_{largesigma_r}frac{e^{iz}}{z},mathrm{d}z
&=lim_{rto0^+}int_{largesigma_r}frac{e^{iz}-1}{z},mathrm{d}z
+lim_{rto0^+}int_{largesigma_r}frac{1}{z},mathrm{d}z\
&=0+lim_{rto0^+}int_0^pifrac{1}{re^{it}},mathrm{d}re^{it}\
&=lim_{rto0^+}int_0^pi i,mathrm{d}t\[9pt]
&=pi i
end{align}
$$
answered Jul 14 '13 at 17:13
robjohn♦robjohn
267k27308632
$endgroup$
add a comment |
$begingroup$
Note that $dfrac{e^{iz}-1}{z}$ is bounded near $z=0$. Since the length of $sigma_r$ tends to $0$, we get
$$
lim_{rto0^+}int_{largesigma_r}frac{e^{iz}-1}{z},mathrm{d}z=0
$$
Thus,
$$
begin{align}
lim_{rto0^+}int_{largesigma_r}frac{e^{iz}}{z},mathrm{d}z
&=lim_{rto0^+}int_{largesigma_r}frac{e^{iz}-1}{z},mathrm{d}z
+lim_{rto0^+}int_{largesigma_r}frac{1}{z},mathrm{d}z\
&=0+lim_{rto0^+}int_0^pifrac{1}{re^{it}},mathrm{d}re^{it}\
&=lim_{rto0^+}int_0^pi i,mathrm{d}t\[9pt]
&=pi i
end{align}
$$
answered Jul 14 '13 at 17:13
robjohn♦robjohn
267k27308632
$endgroup$
Note that $dfrac{e^{iz}-1}{z}$ is bounded near $z=0$. Since the length of $sigma_r$ tends to $0$, we get
$$
lim_{rto0^+}int_{largesigma_r}frac{e^{iz}-1}{z},mathrm{d}z=0
$$
Thus,
$$
begin{align}
lim_{rto0^+}int_{largesigma_r}frac{e^{iz}}{z},mathrm{d}z
&=lim_{rto0^+}int_{largesigma_r}frac{e^{iz}-1}{z},mathrm{d}z
+lim_{rto0^+}int_{largesigma_r}frac{1}{z},mathrm{d}z\
&=0+lim_{rto0^+}int_0^pifrac{1}{re^{it}},mathrm{d}re^{it}\
&=lim_{rto0^+}int_0^pi i,mathrm{d}t\[9pt]
&=pi i
end{align}
$$
answered Jul 14 '13 at 17:13
robjohn♦robjohn
267k27308632
edited Jul 14 '13 at 17:18
answered Jul 14 '13 at 17:13
robjohn♦robjohn
267k27308632
answered Jul 14 '13 at 17:13
robjohn♦robjohn
267k27308632
answered Jul 14 '13 at 17:13
robjohn♦robjohn
267k27308632
267k27308632
add a comment |
add a comment |
$begingroup$
For $k neq -1$, the function $z mapsto z^k$ has a primitive on $mathbb{C}setminus {0}$, namely $z mapsto frac{1}{k+1}z^{k+1}$.
Thus, for $f$ holomorphic in a punctured neighbourhood of $0$, with the Laurent expansion
$$f(z) = sum_{-infty}^infty a_k z^k,$$
you can semi-explicitly evaluate the integral over $sigma_r$ (for small enough $r$) as
$$begin{align}
int_{sigma_r} f(z),dz &= a_{-1}int_{sigma_r}frac{dz}{z} + sum_{kneq -1} a_k int_{sigma_r}z^k\
&= pi i a_{-1} + sum_{k neq -1} frac{a_k r^{k+1}}{k+1} left((-1)^{k+1} - 1right)\
&= pi i a_{-1} -2 sum_{-infty}^infty frac{a_{2k}r^{2k+1}}{2k+1}.
end{align}$$
From that, you can read off that in general the integral depends on $r$, and it converges to $pi i operatorname{res}_0 f$ when $f$ has a simple pole in $0$ (generally, if $f$ has an odd principal part).
Your integrand
$$frac{e^{iz}}{z} = sum_{k=0}^infty frac{i^k}{k!}z^{k-1} = frac1z + i -frac{z}{2} -ifrac{z^2}{6} + frac{z^3}{24} + dotsb$$
has $0$ as a simple pole, hence the above applies. Since not all even coefficients of the Laurent expansion vanish, the integral depends on $r$, but the limit is $pi i$.
answered Jul 14 '13 at 16:02
Daniel Fischer♦Daniel Fischer
174k16167286
$endgroup$
add a comment |
$begingroup$
For $k neq -1$, the function $z mapsto z^k$ has a primitive on $mathbb{C}setminus {0}$, namely $z mapsto frac{1}{k+1}z^{k+1}$.
Thus, for $f$ holomorphic in a punctured neighbourhood of $0$, with the Laurent expansion
$$f(z) = sum_{-infty}^infty a_k z^k,$$
you can semi-explicitly evaluate the integral over $sigma_r$ (for small enough $r$) as
$$begin{align}
int_{sigma_r} f(z),dz &= a_{-1}int_{sigma_r}frac{dz}{z} + sum_{kneq -1} a_k int_{sigma_r}z^k\
&= pi i a_{-1} + sum_{k neq -1} frac{a_k r^{k+1}}{k+1} left((-1)^{k+1} - 1right)\
&= pi i a_{-1} -2 sum_{-infty}^infty frac{a_{2k}r^{2k+1}}{2k+1}.
end{align}$$
From that, you can read off that in general the integral depends on $r$, and it converges to $pi i operatorname{res}_0 f$ when $f$ has a simple pole in $0$ (generally, if $f$ has an odd principal part).
Your integrand
$$frac{e^{iz}}{z} = sum_{k=0}^infty frac{i^k}{k!}z^{k-1} = frac1z + i -frac{z}{2} -ifrac{z^2}{6} + frac{z^3}{24} + dotsb$$
has $0$ as a simple pole, hence the above applies. Since not all even coefficients of the Laurent expansion vanish, the integral depends on $r$, but the limit is $pi i$.
answered Jul 14 '13 at 16:02
Daniel Fischer♦Daniel Fischer
174k16167286
$endgroup$
add a comment |
$begingroup$
For $k neq -1$, the function $z mapsto z^k$ has a primitive on $mathbb{C}setminus {0}$, namely $z mapsto frac{1}{k+1}z^{k+1}$.
Thus, for $f$ holomorphic in a punctured neighbourhood of $0$, with the Laurent expansion
$$f(z) = sum_{-infty}^infty a_k z^k,$$
you can semi-explicitly evaluate the integral over $sigma_r$ (for small enough $r$) as
$$begin{align}
int_{sigma_r} f(z),dz &= a_{-1}int_{sigma_r}frac{dz}{z} + sum_{kneq -1} a_k int_{sigma_r}z^k\
&= pi i a_{-1} + sum_{k neq -1} frac{a_k r^{k+1}}{k+1} left((-1)^{k+1} - 1right)\
&= pi i a_{-1} -2 sum_{-infty}^infty frac{a_{2k}r^{2k+1}}{2k+1}.
end{align}$$
From that, you can read off that in general the integral depends on $r$, and it converges to $pi i operatorname{res}_0 f$ when $f$ has a simple pole in $0$ (generally, if $f$ has an odd principal part).
Your integrand
$$frac{e^{iz}}{z} = sum_{k=0}^infty frac{i^k}{k!}z^{k-1} = frac1z + i -frac{z}{2} -ifrac{z^2}{6} + frac{z^3}{24} + dotsb$$
has $0$ as a simple pole, hence the above applies. Since not all even coefficients of the Laurent expansion vanish, the integral depends on $r$, but the limit is $pi i$.
answered Jul 14 '13 at 16:02
Daniel Fischer♦Daniel Fischer
174k16167286
$endgroup$
For $k neq -1$, the function $z mapsto z^k$ has a primitive on $mathbb{C}setminus {0}$, namely $z mapsto frac{1}{k+1}z^{k+1}$.
Thus, for $f$ holomorphic in a punctured neighbourhood of $0$, with the Laurent expansion
$$f(z) = sum_{-infty}^infty a_k z^k,$$
you can semi-explicitly evaluate the integral over $sigma_r$ (for small enough $r$) as
$$begin{align}
int_{sigma_r} f(z),dz &= a_{-1}int_{sigma_r}frac{dz}{z} + sum_{kneq -1} a_k int_{sigma_r}z^k\
&= pi i a_{-1} + sum_{k neq -1} frac{a_k r^{k+1}}{k+1} left((-1)^{k+1} - 1right)\
&= pi i a_{-1} -2 sum_{-infty}^infty frac{a_{2k}r^{2k+1}}{2k+1}.
end{align}$$
From that, you can read off that in general the integral depends on $r$, and it converges to $pi i operatorname{res}_0 f$ when $f$ has a simple pole in $0$ (generally, if $f$ has an odd principal part).
Your integrand
$$frac{e^{iz}}{z} = sum_{k=0}^infty frac{i^k}{k!}z^{k-1} = frac1z + i -frac{z}{2} -ifrac{z^2}{6} + frac{z^3}{24} + dotsb$$
has $0$ as a simple pole, hence the above applies. Since not all even coefficients of the Laurent expansion vanish, the integral depends on $r$, but the limit is $pi i$.
answered Jul 14 '13 at 16:02
Daniel Fischer♦Daniel Fischer
174k16167286
answered Jul 14 '13 at 16:02
Daniel Fischer♦Daniel Fischer
174k16167286
answered Jul 14 '13 at 16:02
Daniel Fischer♦Daniel Fischer
174k16167286
answered Jul 14 '13 at 16:02
Daniel Fischer♦Daniel Fischer
174k16167286
174k16167286
add a comment |
add a comment |
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$begingroup$
There might be a big difference in the difficulty between finding the integral and finding the limit. Are you sure you're asking what you want to ask?
$endgroup$
– Git Gud
Jul 14 '13 at 14:33
$begingroup$
@GitGud - Yes (but I think that the answer is independent of $r$ so there isn't actually a difference between the two)
$endgroup$
– Belgi
Jul 14 '13 at 14:35
$begingroup$
@Belgi The answer is not independent of $r$. You can relatively easily find a series expression for the integral by expanding $e^{iz}$ into its Taylor series.
$endgroup$
– Daniel Fischer♦
Jul 14 '13 at 14:46