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Contour integral of $int_-infty^infty frace^ax1+e^xdx $ with non rectangular contour
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Is there a way to solve the integral of $$int_-infty^infty frace^ax1+e^xdx $$ for $$ain (0,1)$$
without using the rectangular region like in this post but still using a contour integral?
Perhaps using a semicircular region, circular region, or freshnel contour perhaps? I just don't have a lot of experience with the rectangular region problems.
Thanks.
integration complex-analysis contour-integration
edited Aug 4 at 6:50
Manthanein
6,0071436
asked Jul 28 at 2:01
MathIsHard
1,122415
add a comment |Â
up vote
2
down vote
favorite
Is there a way to solve the integral of $$int_-infty^infty frace^ax1+e^xdx $$ for $$ain (0,1)$$
without using the rectangular region like in this post but still using a contour integral?
Perhaps using a semicircular region, circular region, or freshnel contour perhaps? I just don't have a lot of experience with the rectangular region problems.
Thanks.
integration complex-analysis contour-integration
edited Aug 4 at 6:50
Manthanein
6,0071436
asked Jul 28 at 2:01
MathIsHard
1,122415
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Is there a way to solve the integral of $$int_-infty^infty frace^ax1+e^xdx $$ for $$ain (0,1)$$
without using the rectangular region like in this post but still using a contour integral?
Perhaps using a semicircular region, circular region, or freshnel contour perhaps? I just don't have a lot of experience with the rectangular region problems.
Thanks.
integration complex-analysis contour-integration
edited Aug 4 at 6:50
Manthanein
6,0071436
asked Jul 28 at 2:01
MathIsHard
1,122415
Is there a way to solve the integral of $$int_-infty^infty frace^ax1+e^xdx $$ for $$ain (0,1)$$
without using the rectangular region like in this post but still using a contour integral?
Perhaps using a semicircular region, circular region, or freshnel contour perhaps? I just don't have a lot of experience with the rectangular region problems.
Thanks.
integration complex-analysis contour-integration
edited Aug 4 at 6:50
Manthanein
6,0071436
asked Jul 28 at 2:01
MathIsHard
1,122415
edited Aug 4 at 6:50
Manthanein
6,0071436
edited Aug 4 at 6:50
Manthanein
6,0071436
edited Aug 4 at 6:50
Manthanein
6,0071436
6,0071436
asked Jul 28 at 2:01
MathIsHard
1,122415
asked Jul 28 at 2:01
MathIsHard
1,122415
asked Jul 28 at 2:01
MathIsHard
1,122415
1,122415
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
After substituting $xmapstolog(x)$, we use the keyhole contour from this answer with $n=a-1$ and $m=1$:
$$
beginalign
int_-infty^inftyfrace^ax1+e^x,mathrmdx
&=int_0^inftyfracx^a-11+x,mathrmdx\
&=picsc(pi a)
endalign
$$
answered Jul 28 at 6:01
robjohn♦
258k25296612
add a comment |Â
up vote
2
down vote
(Not using contour integration ; Sorry )
$$I=int_-infty^infty frac e^ax1+e^x dx=int_-infty^infty frac e^xcdot e^axe^x+e^2x dx$$
Use the substitution $e^x=t$
$$I=int_0^infty frac t^a-11+tdt =B(a,1-a)=Gamma(a)Gamma(1-a)=frac pisin (pi a)$$
answered Jul 28 at 2:27
Manthanein
6,0071436
add a comment |Â
up vote
0
down vote
Sure!
Take contour $C$ as an infinitely large semicircle on the upper half plane, centered at the origin.
Let $f(z)=frace^az1+e^z$.
Note that $$oint_C f(z)dz=int^infty_-inftyf(z)dz+int_textarcf(z)dz$$ and the arc integral vanishes.
The poles enclosed are at $z=(2n+1)pi i$ where $nge 0$.
It is easy to show that $$textRes_z=(2n+1)pi if(z)=-e^(2n+1)api i$$
Therefore, the sum of residues is a geometric series:
$$2pi isumtextRes=-2pi icdot e^api icdotfrac11-e^2api i= fracpisin(api)$$
As a result, by Residue theorem,
$$colorredint^infty_-inftyfrace^ax1+e^x= fracpisin(api)$$
As mentioned in the comments, the geometric series does not converge. However, if we use residue method to evaluate
$$lim_kto aint^infty_-inftyfrace^kx1+e^xdx$$, we would always obtain a convergent geometric sum, no mattet in which direction $a$ is approached, except horizontally. (I can attach a proof on request) Unfortunately, the integral would evaluate to a divergent series if $k=a$. I don’t know an explanation for this phenomenon.
answered Jul 28 at 2:51
Szeto
3,8781421
The geometric series in the sum of residues is not convergent.
– i707107
Jul 28 at 3:30
Thank you for your answer @ Szeto. What does that mean if the sum of the residues isn't convergent though. Could you expand on what you mean @i707107?
– MathIsHard
Jul 28 at 5:26
When $Sigma mathrmRes$ is done, the series is geometric with the ratio $r=e^2api i$. For the geometric series to converge, one must have $|r|<1$, but this one has $|r|=|e^2api i|=1$.
– i707107
Jul 28 at 5:33
@i707107 I will think about this for some more time. If I still cannot solve this issue, I may temporarily delete this answer and ask a question about this.
– Szeto
Jul 28 at 5:38
Neither the sum of the residues NOR the integral over the "arc" converges as the radius of the arc approaches infinity. This analysis is flawed.
– Mark Viola
Jul 28 at 12:43
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
After substituting $xmapstolog(x)$, we use the keyhole contour from this answer with $n=a-1$ and $m=1$:
$$
beginalign
int_-infty^inftyfrace^ax1+e^x,mathrmdx
&=int_0^inftyfracx^a-11+x,mathrmdx\
&=picsc(pi a)
endalign
$$
answered Jul 28 at 6:01
robjohn♦
258k25296612
add a comment |Â
up vote
2
down vote
accepted
After substituting $xmapstolog(x)$, we use the keyhole contour from this answer with $n=a-1$ and $m=1$:
$$
beginalign
int_-infty^inftyfrace^ax1+e^x,mathrmdx
&=int_0^inftyfracx^a-11+x,mathrmdx\
&=picsc(pi a)
endalign
$$
answered Jul 28 at 6:01
robjohn♦
258k25296612
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
After substituting $xmapstolog(x)$, we use the keyhole contour from this answer with $n=a-1$ and $m=1$:
$$
beginalign
int_-infty^inftyfrace^ax1+e^x,mathrmdx
&=int_0^inftyfracx^a-11+x,mathrmdx\
&=picsc(pi a)
endalign
$$
answered Jul 28 at 6:01
robjohn♦
258k25296612
After substituting $xmapstolog(x)$, we use the keyhole contour from this answer with $n=a-1$ and $m=1$:
$$
beginalign
int_-infty^inftyfrace^ax1+e^x,mathrmdx
&=int_0^inftyfracx^a-11+x,mathrmdx\
&=picsc(pi a)
endalign
$$
answered Jul 28 at 6:01
robjohn♦
258k25296612
answered Jul 28 at 6:01
robjohn♦
258k25296612
answered Jul 28 at 6:01
robjohn♦
258k25296612
answered Jul 28 at 6:01
robjohn♦
258k25296612
258k25296612
add a comment |Â
add a comment |Â
up vote
2
down vote
(Not using contour integration ; Sorry )
$$I=int_-infty^infty frac e^ax1+e^x dx=int_-infty^infty frac e^xcdot e^axe^x+e^2x dx$$
Use the substitution $e^x=t$
$$I=int_0^infty frac t^a-11+tdt =B(a,1-a)=Gamma(a)Gamma(1-a)=frac pisin (pi a)$$
answered Jul 28 at 2:27
Manthanein
6,0071436
add a comment |Â
up vote
2
down vote
(Not using contour integration ; Sorry )
$$I=int_-infty^infty frac e^ax1+e^x dx=int_-infty^infty frac e^xcdot e^axe^x+e^2x dx$$
Use the substitution $e^x=t$
$$I=int_0^infty frac t^a-11+tdt =B(a,1-a)=Gamma(a)Gamma(1-a)=frac pisin (pi a)$$
answered Jul 28 at 2:27
Manthanein
6,0071436
add a comment |Â
up vote
2
down vote
up vote
2
down vote
(Not using contour integration ; Sorry )
$$I=int_-infty^infty frac e^ax1+e^x dx=int_-infty^infty frac e^xcdot e^axe^x+e^2x dx$$
Use the substitution $e^x=t$
$$I=int_0^infty frac t^a-11+tdt =B(a,1-a)=Gamma(a)Gamma(1-a)=frac pisin (pi a)$$
answered Jul 28 at 2:27
Manthanein
6,0071436
(Not using contour integration ; Sorry )
$$I=int_-infty^infty frac e^ax1+e^x dx=int_-infty^infty frac e^xcdot e^axe^x+e^2x dx$$
Use the substitution $e^x=t$
$$I=int_0^infty frac t^a-11+tdt =B(a,1-a)=Gamma(a)Gamma(1-a)=frac pisin (pi a)$$
answered Jul 28 at 2:27
Manthanein
6,0071436
answered Jul 28 at 2:27
Manthanein
6,0071436
answered Jul 28 at 2:27
Manthanein
6,0071436
answered Jul 28 at 2:27
Manthanein
6,0071436
6,0071436
add a comment |Â
add a comment |Â
up vote
0
down vote
Sure!
Take contour $C$ as an infinitely large semicircle on the upper half plane, centered at the origin.
Let $f(z)=frace^az1+e^z$.
Note that $$oint_C f(z)dz=int^infty_-inftyf(z)dz+int_textarcf(z)dz$$ and the arc integral vanishes.
The poles enclosed are at $z=(2n+1)pi i$ where $nge 0$.
It is easy to show that $$textRes_z=(2n+1)pi if(z)=-e^(2n+1)api i$$
Therefore, the sum of residues is a geometric series:
$$2pi isumtextRes=-2pi icdot e^api icdotfrac11-e^2api i= fracpisin(api)$$
As a result, by Residue theorem,
$$colorredint^infty_-inftyfrace^ax1+e^x= fracpisin(api)$$
As mentioned in the comments, the geometric series does not converge. However, if we use residue method to evaluate
$$lim_kto aint^infty_-inftyfrace^kx1+e^xdx$$, we would always obtain a convergent geometric sum, no mattet in which direction $a$ is approached, except horizontally. (I can attach a proof on request) Unfortunately, the integral would evaluate to a divergent series if $k=a$. I don’t know an explanation for this phenomenon.
answered Jul 28 at 2:51
Szeto
3,8781421
The geometric series in the sum of residues is not convergent.
– i707107
Jul 28 at 3:30
Thank you for your answer @ Szeto. What does that mean if the sum of the residues isn't convergent though. Could you expand on what you mean @i707107?
– MathIsHard
Jul 28 at 5:26
When $Sigma mathrmRes$ is done, the series is geometric with the ratio $r=e^2api i$. For the geometric series to converge, one must have $|r|<1$, but this one has $|r|=|e^2api i|=1$.
– i707107
Jul 28 at 5:33
@i707107 I will think about this for some more time. If I still cannot solve this issue, I may temporarily delete this answer and ask a question about this.
– Szeto
Jul 28 at 5:38
Neither the sum of the residues NOR the integral over the "arc" converges as the radius of the arc approaches infinity. This analysis is flawed.
– Mark Viola
Jul 28 at 12:43
add a comment |Â
up vote
0
down vote
Sure!
Take contour $C$ as an infinitely large semicircle on the upper half plane, centered at the origin.
Let $f(z)=frace^az1+e^z$.
Note that $$oint_C f(z)dz=int^infty_-inftyf(z)dz+int_textarcf(z)dz$$ and the arc integral vanishes.
The poles enclosed are at $z=(2n+1)pi i$ where $nge 0$.
It is easy to show that $$textRes_z=(2n+1)pi if(z)=-e^(2n+1)api i$$
Therefore, the sum of residues is a geometric series:
$$2pi isumtextRes=-2pi icdot e^api icdotfrac11-e^2api i= fracpisin(api)$$
As a result, by Residue theorem,
$$colorredint^infty_-inftyfrace^ax1+e^x= fracpisin(api)$$
As mentioned in the comments, the geometric series does not converge. However, if we use residue method to evaluate
$$lim_kto aint^infty_-inftyfrace^kx1+e^xdx$$, we would always obtain a convergent geometric sum, no mattet in which direction $a$ is approached, except horizontally. (I can attach a proof on request) Unfortunately, the integral would evaluate to a divergent series if $k=a$. I don’t know an explanation for this phenomenon.
answered Jul 28 at 2:51
Szeto
3,8781421
The geometric series in the sum of residues is not convergent.
– i707107
Jul 28 at 3:30
Thank you for your answer @ Szeto. What does that mean if the sum of the residues isn't convergent though. Could you expand on what you mean @i707107?
– MathIsHard
Jul 28 at 5:26
When $Sigma mathrmRes$ is done, the series is geometric with the ratio $r=e^2api i$. For the geometric series to converge, one must have $|r|<1$, but this one has $|r|=|e^2api i|=1$.
– i707107
Jul 28 at 5:33
@i707107 I will think about this for some more time. If I still cannot solve this issue, I may temporarily delete this answer and ask a question about this.
– Szeto
Jul 28 at 5:38
Neither the sum of the residues NOR the integral over the "arc" converges as the radius of the arc approaches infinity. This analysis is flawed.
– Mark Viola
Jul 28 at 12:43
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Sure!
Take contour $C$ as an infinitely large semicircle on the upper half plane, centered at the origin.
Let $f(z)=frace^az1+e^z$.
Note that $$oint_C f(z)dz=int^infty_-inftyf(z)dz+int_textarcf(z)dz$$ and the arc integral vanishes.
The poles enclosed are at $z=(2n+1)pi i$ where $nge 0$.
It is easy to show that $$textRes_z=(2n+1)pi if(z)=-e^(2n+1)api i$$
Therefore, the sum of residues is a geometric series:
$$2pi isumtextRes=-2pi icdot e^api icdotfrac11-e^2api i= fracpisin(api)$$
As a result, by Residue theorem,
$$colorredint^infty_-inftyfrace^ax1+e^x= fracpisin(api)$$
As mentioned in the comments, the geometric series does not converge. However, if we use residue method to evaluate
$$lim_kto aint^infty_-inftyfrace^kx1+e^xdx$$, we would always obtain a convergent geometric sum, no mattet in which direction $a$ is approached, except horizontally. (I can attach a proof on request) Unfortunately, the integral would evaluate to a divergent series if $k=a$. I don’t know an explanation for this phenomenon.
answered Jul 28 at 2:51
Szeto
3,8781421
Sure!
Take contour $C$ as an infinitely large semicircle on the upper half plane, centered at the origin.
Let $f(z)=frace^az1+e^z$.
Note that $$oint_C f(z)dz=int^infty_-inftyf(z)dz+int_textarcf(z)dz$$ and the arc integral vanishes.
The poles enclosed are at $z=(2n+1)pi i$ where $nge 0$.
It is easy to show that $$textRes_z=(2n+1)pi if(z)=-e^(2n+1)api i$$
Therefore, the sum of residues is a geometric series:
$$2pi isumtextRes=-2pi icdot e^api icdotfrac11-e^2api i= fracpisin(api)$$
As a result, by Residue theorem,
$$colorredint^infty_-inftyfrace^ax1+e^x= fracpisin(api)$$
As mentioned in the comments, the geometric series does not converge. However, if we use residue method to evaluate
$$lim_kto aint^infty_-inftyfrace^kx1+e^xdx$$, we would always obtain a convergent geometric sum, no mattet in which direction $a$ is approached, except horizontally. (I can attach a proof on request) Unfortunately, the integral would evaluate to a divergent series if $k=a$. I don’t know an explanation for this phenomenon.
answered Jul 28 at 2:51
Szeto
3,8781421
edited Jul 28 at 12:31
answered Jul 28 at 2:51
Szeto
3,8781421
answered Jul 28 at 2:51
Szeto
3,8781421
answered Jul 28 at 2:51
Szeto
3,8781421
3,8781421
The geometric series in the sum of residues is not convergent.
– i707107
Jul 28 at 3:30
Thank you for your answer @ Szeto. What does that mean if the sum of the residues isn't convergent though. Could you expand on what you mean @i707107?
– MathIsHard
Jul 28 at 5:26
When $Sigma mathrmRes$ is done, the series is geometric with the ratio $r=e^2api i$. For the geometric series to converge, one must have $|r|<1$, but this one has $|r|=|e^2api i|=1$.
– i707107
Jul 28 at 5:33
@i707107 I will think about this for some more time. If I still cannot solve this issue, I may temporarily delete this answer and ask a question about this.
– Szeto
Jul 28 at 5:38
Neither the sum of the residues NOR the integral over the "arc" converges as the radius of the arc approaches infinity. This analysis is flawed.
– Mark Viola
Jul 28 at 12:43
add a comment |Â
The geometric series in the sum of residues is not convergent.
– i707107
Jul 28 at 3:30
Thank you for your answer @ Szeto. What does that mean if the sum of the residues isn't convergent though. Could you expand on what you mean @i707107?
– MathIsHard
Jul 28 at 5:26
When $Sigma mathrmRes$ is done, the series is geometric with the ratio $r=e^2api i$. For the geometric series to converge, one must have $|r|<1$, but this one has $|r|=|e^2api i|=1$.
– i707107
Jul 28 at 5:33
@i707107 I will think about this for some more time. If I still cannot solve this issue, I may temporarily delete this answer and ask a question about this.
– Szeto
Jul 28 at 5:38
Neither the sum of the residues NOR the integral over the "arc" converges as the radius of the arc approaches infinity. This analysis is flawed.
– Mark Viola
Jul 28 at 12:43
The geometric series in the sum of residues is not convergent.
– i707107
Jul 28 at 3:30
The geometric series in the sum of residues is not convergent.
– i707107
Jul 28 at 3:30
Thank you for your answer @ Szeto. What does that mean if the sum of the residues isn't convergent though. Could you expand on what you mean @i707107?
– MathIsHard
Jul 28 at 5:26
Thank you for your answer @ Szeto. What does that mean if the sum of the residues isn't convergent though. Could you expand on what you mean @i707107?
– MathIsHard
Jul 28 at 5:26
When $Sigma mathrmRes$ is done, the series is geometric with the ratio $r=e^2api i$. For the geometric series to converge, one must have $|r|<1$, but this one has $|r|=|e^2api i|=1$.
– i707107
Jul 28 at 5:33
When $Sigma mathrmRes$ is done, the series is geometric with the ratio $r=e^2api i$. For the geometric series to converge, one must have $|r|<1$, but this one has $|r|=|e^2api i|=1$.
– i707107
Jul 28 at 5:33
@i707107 I will think about this for some more time. If I still cannot solve this issue, I may temporarily delete this answer and ask a question about this.
– Szeto
Jul 28 at 5:38
@i707107 I will think about this for some more time. If I still cannot solve this issue, I may temporarily delete this answer and ask a question about this.
– Szeto
Jul 28 at 5:38
Neither the sum of the residues NOR the integral over the "arc" converges as the radius of the arc approaches infinity. This analysis is flawed.
– Mark Viola
Jul 28 at 12:43
Neither the sum of the residues NOR the integral over the "arc" converges as the radius of the arc approaches infinity. This analysis is flawed.
– Mark Viola
Jul 28 at 12:43
add a comment |Â
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