How to calculate $intfrac{e^{iz}}{z}, dz$ on the semi-circle given by $re^{itheta}$ where $theta:,0topi$












0












$begingroup$


I was reading this post and in the comments someone said that the difficult in calculating the limit as $r$ goes to $0$ is a lot different than calculating the limit. I tried to calculate the integral and don't seems that difficult but a lot of algebraic work, are there a trick to easily calculate the integral?.






complex-analysis line-integrals







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    We can make a keyhole contour, use Cauchy residue theorem and the fact that we know what the only poles inside are. I don't know which contour would be easiest. I guess either full circle and use some symmetricity or or half circle.
    $endgroup$
    – mathreadler
    Nov 30 '18 at 16:50










  • $begingroup$
    @mathreadler Symmetry would be hard to come by, as the integrand takes much larger values on the lower half-plane than on the upper.
    $endgroup$
    – Connor Harris
    Nov 30 '18 at 16:57
















0












$begingroup$


I was reading this post and in the comments someone said that the difficult in calculating the limit as $r$ goes to $0$ is a lot different than calculating the limit. I tried to calculate the integral and don't seems that difficult but a lot of algebraic work, are there a trick to easily calculate the integral?.






complex-analysis line-integrals







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    We can make a keyhole contour, use Cauchy residue theorem and the fact that we know what the only poles inside are. I don't know which contour would be easiest. I guess either full circle and use some symmetricity or or half circle.
    $endgroup$
    – mathreadler
    Nov 30 '18 at 16:50










  • $begingroup$
    @mathreadler Symmetry would be hard to come by, as the integrand takes much larger values on the lower half-plane than on the upper.
    $endgroup$
    – Connor Harris
    Nov 30 '18 at 16:57














0












0








0





$begingroup$


I was reading this post and in the comments someone said that the difficult in calculating the limit as $r$ goes to $0$ is a lot different than calculating the limit. I tried to calculate the integral and don't seems that difficult but a lot of algebraic work, are there a trick to easily calculate the integral?.






complex-analysis line-integrals







share|cite|improve this question









$endgroup$




I was reading this post and in the comments someone said that the difficult in calculating the limit as $r$ goes to $0$ is a lot different than calculating the limit. I tried to calculate the integral and don't seems that difficult but a lot of algebraic work, are there a trick to easily calculate the integral?.






complex-analysis line-integrals




complex-analysis line-integrals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 30 '18 at 16:27









Maria GuthierMaria Guthier

867




867








  • 1




    $begingroup$
    We can make a keyhole contour, use Cauchy residue theorem and the fact that we know what the only poles inside are. I don't know which contour would be easiest. I guess either full circle and use some symmetricity or or half circle.
    $endgroup$
    – mathreadler
    Nov 30 '18 at 16:50










  • $begingroup$
    @mathreadler Symmetry would be hard to come by, as the integrand takes much larger values on the lower half-plane than on the upper.
    $endgroup$
    – Connor Harris
    Nov 30 '18 at 16:57














  • 1




    $begingroup$
    We can make a keyhole contour, use Cauchy residue theorem and the fact that we know what the only poles inside are. I don't know which contour would be easiest. I guess either full circle and use some symmetricity or or half circle.
    $endgroup$
    – mathreadler
    Nov 30 '18 at 16:50










  • $begingroup$
    @mathreadler Symmetry would be hard to come by, as the integrand takes much larger values on the lower half-plane than on the upper.
    $endgroup$
    – Connor Harris
    Nov 30 '18 at 16:57








1




1




$begingroup$
We can make a keyhole contour, use Cauchy residue theorem and the fact that we know what the only poles inside are. I don't know which contour would be easiest. I guess either full circle and use some symmetricity or or half circle.
$endgroup$
– mathreadler
Nov 30 '18 at 16:50




$begingroup$
We can make a keyhole contour, use Cauchy residue theorem and the fact that we know what the only poles inside are. I don't know which contour would be easiest. I guess either full circle and use some symmetricity or or half circle.
$endgroup$
– mathreadler
Nov 30 '18 at 16:50












$begingroup$
@mathreadler Symmetry would be hard to come by, as the integrand takes much larger values on the lower half-plane than on the upper.
$endgroup$
– Connor Harris
Nov 30 '18 at 16:57




$begingroup$
@mathreadler Symmetry would be hard to come by, as the integrand takes much larger values on the lower half-plane than on the upper.
$endgroup$
– Connor Harris
Nov 30 '18 at 16:57










2 Answers
2






active

oldest

votes


















1












$begingroup$

There's no closed form. Take a counterclockwise contour comprising a large semicircle ${r e^{i theta} | 0 leq theta leq pi}$, a small semicircle ${epsilon e^{i theta} | 0 leq theta leq pi}$, and real intervals $[-r, -epsilon]$ and $[epsilon, r]$. This looks like a semicircle with a dimple cut out around the origin. The integrand has no poles in this region, so the integral around the whole contour is zero. If $x$ is real, then $$frac{e^{ix}}{x} = frac{i sin x}{x} = frac{i sin (-x)}{-x} = frac{e^{-ix}}{-x}$$ so the integrals over the real intervals are the same and equal $i int_epsilon^r frac{sin x}{x}, dx$, which has no closed form; the special function $operatorname{Si}(x) = int_0^x frac{sin t}{t}, dt$ is sometimes used.



Finally, as shown in the question you linked, at the $epsilon to 0$ limit, the integral over the small semicircle is $-pi i$ (with the negative sign because the contour traverses this small semicircle in the wrong direction, left-to-right). Thus the integral over the outer semicircle is $$pi i - 2i operatorname{Si}(r).$$



It's possible to show that the $r to infty$ limit of this expression is zero.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But the integral of just the arc! is then $pi i$
    $endgroup$
    – Maria Guthier
    Nov 30 '18 at 21:04



















1












$begingroup$

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$




By closing the contour along the real axis while an indent $ds{braces{z mid z = epsilonexpo{ictheta}, r > epsilon > 0, theta in bracks{0,pi} }}$ is provided, I'll find:




begin{align}
&bbox[10px,#ffd]{int_{large z in rexppars{icbracks{0,pi}}
atop large r > 0}{expo{ic z} over z},dd z}
\[5mm] = &
lim_{epsilon to 0^{large +}}pars{-int_{-r}^{-epsilon}{expo{ic x} over x},dd x - int_{pi}^{0}ic,ddtheta
-int_{epsilon}^{r}{expo{ic x} over x},dd x}
\[5mm] = &
lim_{epsilon to 0^{large +}}pars{int_{epsilon}^{r}{expo{-ic x} over x},dd x + piic
-int_{epsilon}^{r}{expo{ic x} over x},dd x}
\[5mm] = &
piic - 2icint_{0}^{r}{sinpars{x} over x}
,dd x = bbx{piic - 2ic,mrm{Si}pars{r}}
end{align}




$ds{mrm{Si}}$ is the
Sine Integral Function.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    You have a sign wrong on the $-pi i$ term.
    $endgroup$
    – Connor Harris
    Nov 30 '18 at 17:08










  • $begingroup$
    @ConnorHarris Thanks. Fixed.
    $endgroup$
    – Felix Marin
    Dec 3 '18 at 20:29











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

There's no closed form. Take a counterclockwise contour comprising a large semicircle ${r e^{i theta} | 0 leq theta leq pi}$, a small semicircle ${epsilon e^{i theta} | 0 leq theta leq pi}$, and real intervals $[-r, -epsilon]$ and $[epsilon, r]$. This looks like a semicircle with a dimple cut out around the origin. The integrand has no poles in this region, so the integral around the whole contour is zero. If $x$ is real, then $$frac{e^{ix}}{x} = frac{i sin x}{x} = frac{i sin (-x)}{-x} = frac{e^{-ix}}{-x}$$ so the integrals over the real intervals are the same and equal $i int_epsilon^r frac{sin x}{x}, dx$, which has no closed form; the special function $operatorname{Si}(x) = int_0^x frac{sin t}{t}, dt$ is sometimes used.



Finally, as shown in the question you linked, at the $epsilon to 0$ limit, the integral over the small semicircle is $-pi i$ (with the negative sign because the contour traverses this small semicircle in the wrong direction, left-to-right). Thus the integral over the outer semicircle is $$pi i - 2i operatorname{Si}(r).$$



It's possible to show that the $r to infty$ limit of this expression is zero.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But the integral of just the arc! is then $pi i$
    $endgroup$
    – Maria Guthier
    Nov 30 '18 at 21:04
















1












$begingroup$

There's no closed form. Take a counterclockwise contour comprising a large semicircle ${r e^{i theta} | 0 leq theta leq pi}$, a small semicircle ${epsilon e^{i theta} | 0 leq theta leq pi}$, and real intervals $[-r, -epsilon]$ and $[epsilon, r]$. This looks like a semicircle with a dimple cut out around the origin. The integrand has no poles in this region, so the integral around the whole contour is zero. If $x$ is real, then $$frac{e^{ix}}{x} = frac{i sin x}{x} = frac{i sin (-x)}{-x} = frac{e^{-ix}}{-x}$$ so the integrals over the real intervals are the same and equal $i int_epsilon^r frac{sin x}{x}, dx$, which has no closed form; the special function $operatorname{Si}(x) = int_0^x frac{sin t}{t}, dt$ is sometimes used.



Finally, as shown in the question you linked, at the $epsilon to 0$ limit, the integral over the small semicircle is $-pi i$ (with the negative sign because the contour traverses this small semicircle in the wrong direction, left-to-right). Thus the integral over the outer semicircle is $$pi i - 2i operatorname{Si}(r).$$



It's possible to show that the $r to infty$ limit of this expression is zero.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But the integral of just the arc! is then $pi i$
    $endgroup$
    – Maria Guthier
    Nov 30 '18 at 21:04














1












1








1





$begingroup$

There's no closed form. Take a counterclockwise contour comprising a large semicircle ${r e^{i theta} | 0 leq theta leq pi}$, a small semicircle ${epsilon e^{i theta} | 0 leq theta leq pi}$, and real intervals $[-r, -epsilon]$ and $[epsilon, r]$. This looks like a semicircle with a dimple cut out around the origin. The integrand has no poles in this region, so the integral around the whole contour is zero. If $x$ is real, then $$frac{e^{ix}}{x} = frac{i sin x}{x} = frac{i sin (-x)}{-x} = frac{e^{-ix}}{-x}$$ so the integrals over the real intervals are the same and equal $i int_epsilon^r frac{sin x}{x}, dx$, which has no closed form; the special function $operatorname{Si}(x) = int_0^x frac{sin t}{t}, dt$ is sometimes used.



Finally, as shown in the question you linked, at the $epsilon to 0$ limit, the integral over the small semicircle is $-pi i$ (with the negative sign because the contour traverses this small semicircle in the wrong direction, left-to-right). Thus the integral over the outer semicircle is $$pi i - 2i operatorname{Si}(r).$$



It's possible to show that the $r to infty$ limit of this expression is zero.






share|cite|improve this answer











$endgroup$



There's no closed form. Take a counterclockwise contour comprising a large semicircle ${r e^{i theta} | 0 leq theta leq pi}$, a small semicircle ${epsilon e^{i theta} | 0 leq theta leq pi}$, and real intervals $[-r, -epsilon]$ and $[epsilon, r]$. This looks like a semicircle with a dimple cut out around the origin. The integrand has no poles in this region, so the integral around the whole contour is zero. If $x$ is real, then $$frac{e^{ix}}{x} = frac{i sin x}{x} = frac{i sin (-x)}{-x} = frac{e^{-ix}}{-x}$$ so the integrals over the real intervals are the same and equal $i int_epsilon^r frac{sin x}{x}, dx$, which has no closed form; the special function $operatorname{Si}(x) = int_0^x frac{sin t}{t}, dt$ is sometimes used.



Finally, as shown in the question you linked, at the $epsilon to 0$ limit, the integral over the small semicircle is $-pi i$ (with the negative sign because the contour traverses this small semicircle in the wrong direction, left-to-right). Thus the integral over the outer semicircle is $$pi i - 2i operatorname{Si}(r).$$



It's possible to show that the $r to infty$ limit of this expression is zero.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 '18 at 14:22

























answered Nov 30 '18 at 17:08









Connor HarrisConnor Harris

4,350724




4,350724












  • $begingroup$
    But the integral of just the arc! is then $pi i$
    $endgroup$
    – Maria Guthier
    Nov 30 '18 at 21:04


















  • $begingroup$
    But the integral of just the arc! is then $pi i$
    $endgroup$
    – Maria Guthier
    Nov 30 '18 at 21:04
















$begingroup$
But the integral of just the arc! is then $pi i$
$endgroup$
– Maria Guthier
Nov 30 '18 at 21:04




$begingroup$
But the integral of just the arc! is then $pi i$
$endgroup$
– Maria Guthier
Nov 30 '18 at 21:04











1












$begingroup$

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$




By closing the contour along the real axis while an indent $ds{braces{z mid z = epsilonexpo{ictheta}, r > epsilon > 0, theta in bracks{0,pi} }}$ is provided, I'll find:




begin{align}
&bbox[10px,#ffd]{int_{large z in rexppars{icbracks{0,pi}}
atop large r > 0}{expo{ic z} over z},dd z}
\[5mm] = &
lim_{epsilon to 0^{large +}}pars{-int_{-r}^{-epsilon}{expo{ic x} over x},dd x - int_{pi}^{0}ic,ddtheta
-int_{epsilon}^{r}{expo{ic x} over x},dd x}
\[5mm] = &
lim_{epsilon to 0^{large +}}pars{int_{epsilon}^{r}{expo{-ic x} over x},dd x + piic
-int_{epsilon}^{r}{expo{ic x} over x},dd x}
\[5mm] = &
piic - 2icint_{0}^{r}{sinpars{x} over x}
,dd x = bbx{piic - 2ic,mrm{Si}pars{r}}
end{align}




$ds{mrm{Si}}$ is the
Sine Integral Function.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    You have a sign wrong on the $-pi i$ term.
    $endgroup$
    – Connor Harris
    Nov 30 '18 at 17:08










  • $begingroup$
    @ConnorHarris Thanks. Fixed.
    $endgroup$
    – Felix Marin
    Dec 3 '18 at 20:29
















1












$begingroup$

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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newcommand{verts}[1]{leftvert,{#1},rightvert}$




By closing the contour along the real axis while an indent $ds{braces{z mid z = epsilonexpo{ictheta}, r > epsilon > 0, theta in bracks{0,pi} }}$ is provided, I'll find:




begin{align}
&bbox[10px,#ffd]{int_{large z in rexppars{icbracks{0,pi}}
atop large r > 0}{expo{ic z} over z},dd z}
\[5mm] = &
lim_{epsilon to 0^{large +}}pars{-int_{-r}^{-epsilon}{expo{ic x} over x},dd x - int_{pi}^{0}ic,ddtheta
-int_{epsilon}^{r}{expo{ic x} over x},dd x}
\[5mm] = &
lim_{epsilon to 0^{large +}}pars{int_{epsilon}^{r}{expo{-ic x} over x},dd x + piic
-int_{epsilon}^{r}{expo{ic x} over x},dd x}
\[5mm] = &
piic - 2icint_{0}^{r}{sinpars{x} over x}
,dd x = bbx{piic - 2ic,mrm{Si}pars{r}}
end{align}




$ds{mrm{Si}}$ is the
Sine Integral Function.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    You have a sign wrong on the $-pi i$ term.
    $endgroup$
    – Connor Harris
    Nov 30 '18 at 17:08










  • $begingroup$
    @ConnorHarris Thanks. Fixed.
    $endgroup$
    – Felix Marin
    Dec 3 '18 at 20:29














1












1








1





$begingroup$

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
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newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$




By closing the contour along the real axis while an indent $ds{braces{z mid z = epsilonexpo{ictheta}, r > epsilon > 0, theta in bracks{0,pi} }}$ is provided, I'll find:




begin{align}
&bbox[10px,#ffd]{int_{large z in rexppars{icbracks{0,pi}}
atop large r > 0}{expo{ic z} over z},dd z}
\[5mm] = &
lim_{epsilon to 0^{large +}}pars{-int_{-r}^{-epsilon}{expo{ic x} over x},dd x - int_{pi}^{0}ic,ddtheta
-int_{epsilon}^{r}{expo{ic x} over x},dd x}
\[5mm] = &
lim_{epsilon to 0^{large +}}pars{int_{epsilon}^{r}{expo{-ic x} over x},dd x + piic
-int_{epsilon}^{r}{expo{ic x} over x},dd x}
\[5mm] = &
piic - 2icint_{0}^{r}{sinpars{x} over x}
,dd x = bbx{piic - 2ic,mrm{Si}pars{r}}
end{align}




$ds{mrm{Si}}$ is the
Sine Integral Function.







share|cite|improve this answer











$endgroup$



$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
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newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$




By closing the contour along the real axis while an indent $ds{braces{z mid z = epsilonexpo{ictheta}, r > epsilon > 0, theta in bracks{0,pi} }}$ is provided, I'll find:




begin{align}
&bbox[10px,#ffd]{int_{large z in rexppars{icbracks{0,pi}}
atop large r > 0}{expo{ic z} over z},dd z}
\[5mm] = &
lim_{epsilon to 0^{large +}}pars{-int_{-r}^{-epsilon}{expo{ic x} over x},dd x - int_{pi}^{0}ic,ddtheta
-int_{epsilon}^{r}{expo{ic x} over x},dd x}
\[5mm] = &
lim_{epsilon to 0^{large +}}pars{int_{epsilon}^{r}{expo{-ic x} over x},dd x + piic
-int_{epsilon}^{r}{expo{ic x} over x},dd x}
\[5mm] = &
piic - 2icint_{0}^{r}{sinpars{x} over x}
,dd x = bbx{piic - 2ic,mrm{Si}pars{r}}
end{align}




$ds{mrm{Si}}$ is the
Sine Integral Function.








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 '18 at 20:45

























answered Nov 30 '18 at 17:06









Felix MarinFelix Marin

67.3k7107141




67.3k7107141












  • $begingroup$
    You have a sign wrong on the $-pi i$ term.
    $endgroup$
    – Connor Harris
    Nov 30 '18 at 17:08










  • $begingroup$
    @ConnorHarris Thanks. Fixed.
    $endgroup$
    – Felix Marin
    Dec 3 '18 at 20:29


















  • $begingroup$
    You have a sign wrong on the $-pi i$ term.
    $endgroup$
    – Connor Harris
    Nov 30 '18 at 17:08










  • $begingroup$
    @ConnorHarris Thanks. Fixed.
    $endgroup$
    – Felix Marin
    Dec 3 '18 at 20:29
















$begingroup$
You have a sign wrong on the $-pi i$ term.
$endgroup$
– Connor Harris
Nov 30 '18 at 17:08




$begingroup$
You have a sign wrong on the $-pi i$ term.
$endgroup$
– Connor Harris
Nov 30 '18 at 17:08












$begingroup$
@ConnorHarris Thanks. Fixed.
$endgroup$
– Felix Marin
Dec 3 '18 at 20:29




$begingroup$
@ConnorHarris Thanks. Fixed.
$endgroup$
– Felix Marin
Dec 3 '18 at 20:29


















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