Showing that $int_0^infty fraccos(x)x^2+3dx = frace^-sqrt3 pi2 sqrt3$ via Contour Integration?
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In the text "Function Theory of One Complex Variable" Third Edition, I'm inquiring if my proof of $textProposition (1)$ is sound ?
$textProposition (1)$
$$int_0^infty fraccos(x)x^2+3dx = frace^-sqrt3 pi2 sqrt3$$
$textProof$
Assume that $R>1$ define $gamma_R$ such that,
$$gamma_R^1(t) = t + i0 , , textif , , -R leq t leq R$$
$$gamma_R^2(t) = Re^it , textif , , , , , , 0leq t leq pi.$$
$, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , $ 
Consider our choice $f$ and that,
$$oint_gamma_R frace^izz^2+3 , dz.$$
Clearly it's obvious that
$$oint_gamma_R frace^izz^2+3dz = sum_psi = 1,2 oint_gamma_R^psifrace^izz^2+3dz.$$
It's trivial that,
$$oint_gamma_R^1e^iz/(z^2+3)dx rightarrow int_0^infty frace^ixx^2+3operatornamedx$$
It's natural to claim that,
$$Bigg| lim_R rightarrow inftyoint_ gamma_R^2 frace^izz^2+3 dz Bigg| rightarrow 0. $$
Using the Estimation Lemma one can be relived that,
$$bigg |oint_gamma_R^2 frace^izz^2+3 dz bigg | leq bigtextlength(gamma_R^2) big cdot sup_gamma_R^2|frace^izz^2+3|leq pi R cdot frac1R^2 - 3 rightarrow 0 , textas , R rightarrow infty$$
Thus,
$$ operatornameReint_0^infty fraccos(x)x^2+3dx = operatornameRe frace^-sqrt3 pi2 sqrt3 = frace^-sqrt3 pi2 sqrt3.$$
However we need to consider that
$$oint_gamma_R frace^izz^2+3 , dz = 2 pi i sum_j=1,2 operatornameInd_gamma cdot operatornameRes_f(P_j)$$
It's easy to observe that,
$$oint_gamma_R frace^izz^2+3 , dz = big(2pi i cdot fracie^sqrt32 sqrt3) = frace^-sqrt3 pi2 sqrt3$$
complex-analysis proof-verification contour-integration
edited Aug 8 at 7:44
Batominovski
23.4k22779
asked Aug 7 at 18:28
Zophikel
429518
 |Â
show 6 more comments
up vote
2
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In the text "Function Theory of One Complex Variable" Third Edition, I'm inquiring if my proof of $textProposition (1)$ is sound ?
$textProposition (1)$
$$int_0^infty fraccos(x)x^2+3dx = frace^-sqrt3 pi2 sqrt3$$
$textProof$
Assume that $R>1$ define $gamma_R$ such that,
$$gamma_R^1(t) = t + i0 , , textif , , -R leq t leq R$$
$$gamma_R^2(t) = Re^it , textif , , , , , , 0leq t leq pi.$$
$, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , $
Consider our choice $f$ and that,
$$oint_gamma_R frace^izz^2+3 , dz.$$
Clearly it's obvious that
$$oint_gamma_R frace^izz^2+3dz = sum_psi = 1,2 oint_gamma_R^psifrace^izz^2+3dz.$$
It's trivial that,
$$oint_gamma_R^1e^iz/(z^2+3)dx rightarrow int_0^infty frace^ixx^2+3operatornamedx$$
It's natural to claim that,
$$Bigg| lim_R rightarrow inftyoint_ gamma_R^2 frace^izz^2+3 dz Bigg| rightarrow 0. $$
Using the Estimation Lemma one can be relived that,
$$bigg |oint_gamma_R^2 frace^izz^2+3 dz bigg | leq bigtextlength(gamma_R^2) big cdot sup_gamma_R^2|frace^izz^2+3|leq pi R cdot frac1R^2 - 3 rightarrow 0 , textas , R rightarrow infty$$
Thus,
$$ operatornameReint_0^infty fraccos(x)x^2+3dx = operatornameRe frace^-sqrt3 pi2 sqrt3 = frace^-sqrt3 pi2 sqrt3.$$
However we need to consider that
$$oint_gamma_R frace^izz^2+3 , dz = 2 pi i sum_j=1,2 operatornameInd_gamma cdot operatornameRes_f(P_j)$$
It's easy to observe that,
$$oint_gamma_R frace^izz^2+3 , dz = big(2pi i cdot fracie^sqrt32 sqrt3) = frace^-sqrt3 pi2 sqrt3$$
complex-analysis proof-verification contour-integration
edited Aug 8 at 7:44
Batominovski
23.4k22779
asked Aug 7 at 18:28
Zophikel
429518
1
Where does $frac1R^2-1$ come from?
– Lord Shark the Unknown
Aug 7 at 18:41
1
Thanks for pointing out the typo it's been fixed.
– Zophikel
Aug 7 at 18:47
1
The integral is over $Bbb R$, or only over $Bbb R_ge 0$? (The contour goes from $-R$ to $R$. I was looking for a factor $2$ or $1/2$ on the one or the other side...)
– dan_fulea
Aug 7 at 19:14
1
More generally, $$int_0^infty,fraccos(kx)x^2+1,textdx=fracpi2,expbig(-|k|big)text for all kinmathbbR,.$$ In your case, $$int_0^infty,fraccos(x)x^2+3,textdx=frac1sqrt3,int_0^infty,fraccos(sqrt3t)t^2+1,textdt=fracpi2sqrt3,exp(-sqrt3),,$$ where $t:=sqrt3x$. See, for example, the last two hidden boxes in the question of math.stackexchange.com/questions/2870410/….
– Batominovski
Aug 7 at 19:37
1
My comment is that there are two errors. Note that $$lim_Rtoinfty,int_gamma_R^1,fracexp(textiz)z^2+3,textdz=2,int_0^infty,fraccos(x)x^2+3,textdx$$ as you probably forgot that $2,cos(x)=exp(+textix)+exp(-textix)$, and that $$2pitexticdotleft(fracexp(-sqrt3)2sqrt3textiright)=fracpi,exp(-sqrt3)sqrt3$$ is what you should get after applying the Residue Theorem. Dividing this result by $2$, you will get the answer you are looking for.
– Batominovski
Aug 8 at 7:50
 |Â
show 6 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In the text "Function Theory of One Complex Variable" Third Edition, I'm inquiring if my proof of $textProposition (1)$ is sound ?
$textProposition (1)$
$$int_0^infty fraccos(x)x^2+3dx = frace^-sqrt3 pi2 sqrt3$$
$textProof$
Assume that $R>1$ define $gamma_R$ such that,
$$gamma_R^1(t) = t + i0 , , textif , , -R leq t leq R$$
$$gamma_R^2(t) = Re^it , textif , , , , , , 0leq t leq pi.$$
$, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , $
Consider our choice $f$ and that,
$$oint_gamma_R frace^izz^2+3 , dz.$$
Clearly it's obvious that
$$oint_gamma_R frace^izz^2+3dz = sum_psi = 1,2 oint_gamma_R^psifrace^izz^2+3dz.$$
It's trivial that,
$$oint_gamma_R^1e^iz/(z^2+3)dx rightarrow int_0^infty frace^ixx^2+3operatornamedx$$
It's natural to claim that,
$$Bigg| lim_R rightarrow inftyoint_ gamma_R^2 frace^izz^2+3 dz Bigg| rightarrow 0. $$
Using the Estimation Lemma one can be relived that,
$$bigg |oint_gamma_R^2 frace^izz^2+3 dz bigg | leq bigtextlength(gamma_R^2) big cdot sup_gamma_R^2|frace^izz^2+3|leq pi R cdot frac1R^2 - 3 rightarrow 0 , textas , R rightarrow infty$$
Thus,
$$ operatornameReint_0^infty fraccos(x)x^2+3dx = operatornameRe frace^-sqrt3 pi2 sqrt3 = frace^-sqrt3 pi2 sqrt3.$$
However we need to consider that
$$oint_gamma_R frace^izz^2+3 , dz = 2 pi i sum_j=1,2 operatornameInd_gamma cdot operatornameRes_f(P_j)$$
It's easy to observe that,
$$oint_gamma_R frace^izz^2+3 , dz = big(2pi i cdot fracie^sqrt32 sqrt3) = frace^-sqrt3 pi2 sqrt3$$
complex-analysis proof-verification contour-integration
edited Aug 8 at 7:44
Batominovski
23.4k22779
asked Aug 7 at 18:28
Zophikel
429518
In the text "Function Theory of One Complex Variable" Third Edition, I'm inquiring if my proof of $textProposition (1)$ is sound ?
$textProposition (1)$
$$int_0^infty fraccos(x)x^2+3dx = frace^-sqrt3 pi2 sqrt3$$
$textProof$
Assume that $R>1$ define $gamma_R$ such that,
$$gamma_R^1(t) = t + i0 , , textif , , -R leq t leq R$$
$$gamma_R^2(t) = Re^it , textif , , , , , , 0leq t leq pi.$$
$, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , $
Consider our choice $f$ and that,
$$oint_gamma_R frace^izz^2+3 , dz.$$
Clearly it's obvious that
$$oint_gamma_R frace^izz^2+3dz = sum_psi = 1,2 oint_gamma_R^psifrace^izz^2+3dz.$$
It's trivial that,
$$oint_gamma_R^1e^iz/(z^2+3)dx rightarrow int_0^infty frace^ixx^2+3operatornamedx$$
It's natural to claim that,
$$Bigg| lim_R rightarrow inftyoint_ gamma_R^2 frace^izz^2+3 dz Bigg| rightarrow 0. $$
Using the Estimation Lemma one can be relived that,
$$bigg |oint_gamma_R^2 frace^izz^2+3 dz bigg | leq bigtextlength(gamma_R^2) big cdot sup_gamma_R^2|frace^izz^2+3|leq pi R cdot frac1R^2 - 3 rightarrow 0 , textas , R rightarrow infty$$
Thus,
$$ operatornameReint_0^infty fraccos(x)x^2+3dx = operatornameRe frace^-sqrt3 pi2 sqrt3 = frace^-sqrt3 pi2 sqrt3.$$
However we need to consider that
$$oint_gamma_R frace^izz^2+3 , dz = 2 pi i sum_j=1,2 operatornameInd_gamma cdot operatornameRes_f(P_j)$$
It's easy to observe that,
$$oint_gamma_R frace^izz^2+3 , dz = big(2pi i cdot fracie^sqrt32 sqrt3) = frace^-sqrt3 pi2 sqrt3$$
complex-analysis proof-verification contour-integration
edited Aug 8 at 7:44
Batominovski
23.4k22779
asked Aug 7 at 18:28
Zophikel
429518
edited Aug 8 at 7:44
Batominovski
23.4k22779
edited Aug 8 at 7:44
Batominovski
23.4k22779
edited Aug 8 at 7:44
Batominovski
23.4k22779
23.4k22779
asked Aug 7 at 18:28
Zophikel
429518
asked Aug 7 at 18:28
Zophikel
429518
asked Aug 7 at 18:28
Zophikel
429518
429518
1
Where does $frac1R^2-1$ come from?
– Lord Shark the Unknown
Aug 7 at 18:41
1
Thanks for pointing out the typo it's been fixed.
– Zophikel
Aug 7 at 18:47
1
The integral is over $Bbb R$, or only over $Bbb R_ge 0$? (The contour goes from $-R$ to $R$. I was looking for a factor $2$ or $1/2$ on the one or the other side...)
– dan_fulea
Aug 7 at 19:14
1
More generally, $$int_0^infty,fraccos(kx)x^2+1,textdx=fracpi2,expbig(-|k|big)text for all kinmathbbR,.$$ In your case, $$int_0^infty,fraccos(x)x^2+3,textdx=frac1sqrt3,int_0^infty,fraccos(sqrt3t)t^2+1,textdt=fracpi2sqrt3,exp(-sqrt3),,$$ where $t:=sqrt3x$. See, for example, the last two hidden boxes in the question of math.stackexchange.com/questions/2870410/….
– Batominovski
Aug 7 at 19:37
1
My comment is that there are two errors. Note that $$lim_Rtoinfty,int_gamma_R^1,fracexp(textiz)z^2+3,textdz=2,int_0^infty,fraccos(x)x^2+3,textdx$$ as you probably forgot that $2,cos(x)=exp(+textix)+exp(-textix)$, and that $$2pitexticdotleft(fracexp(-sqrt3)2sqrt3textiright)=fracpi,exp(-sqrt3)sqrt3$$ is what you should get after applying the Residue Theorem. Dividing this result by $2$, you will get the answer you are looking for.
– Batominovski
Aug 8 at 7:50
 |Â
show 6 more comments
1
Where does $frac1R^2-1$ come from?
– Lord Shark the Unknown
Aug 7 at 18:41
1
Thanks for pointing out the typo it's been fixed.
– Zophikel
Aug 7 at 18:47
1
The integral is over $Bbb R$, or only over $Bbb R_ge 0$? (The contour goes from $-R$ to $R$. I was looking for a factor $2$ or $1/2$ on the one or the other side...)
– dan_fulea
Aug 7 at 19:14
1
More generally, $$int_0^infty,fraccos(kx)x^2+1,textdx=fracpi2,expbig(-|k|big)text for all kinmathbbR,.$$ In your case, $$int_0^infty,fraccos(x)x^2+3,textdx=frac1sqrt3,int_0^infty,fraccos(sqrt3t)t^2+1,textdt=fracpi2sqrt3,exp(-sqrt3),,$$ where $t:=sqrt3x$. See, for example, the last two hidden boxes in the question of math.stackexchange.com/questions/2870410/….
– Batominovski
Aug 7 at 19:37
1
My comment is that there are two errors. Note that $$lim_Rtoinfty,int_gamma_R^1,fracexp(textiz)z^2+3,textdz=2,int_0^infty,fraccos(x)x^2+3,textdx$$ as you probably forgot that $2,cos(x)=exp(+textix)+exp(-textix)$, and that $$2pitexticdotleft(fracexp(-sqrt3)2sqrt3textiright)=fracpi,exp(-sqrt3)sqrt3$$ is what you should get after applying the Residue Theorem. Dividing this result by $2$, you will get the answer you are looking for.
– Batominovski
Aug 8 at 7:50
1
1
Where does $frac1R^2-1$ come from?
– Lord Shark the Unknown
Aug 7 at 18:41
Where does $frac1R^2-1$ come from?
– Lord Shark the Unknown
Aug 7 at 18:41
1
1
Thanks for pointing out the typo it's been fixed.
– Zophikel
Aug 7 at 18:47
Thanks for pointing out the typo it's been fixed.
– Zophikel
Aug 7 at 18:47
1
1
The integral is over $Bbb R$, or only over $Bbb R_ge 0$? (The contour goes from $-R$ to $R$. I was looking for a factor $2$ or $1/2$ on the one or the other side...)
– dan_fulea
Aug 7 at 19:14
The integral is over $Bbb R$, or only over $Bbb R_ge 0$? (The contour goes from $-R$ to $R$. I was looking for a factor $2$ or $1/2$ on the one or the other side...)
– dan_fulea
Aug 7 at 19:14
1
1
More generally, $$int_0^infty,fraccos(kx)x^2+1,textdx=fracpi2,expbig(-|k|big)text for all kinmathbbR,.$$ In your case, $$int_0^infty,fraccos(x)x^2+3,textdx=frac1sqrt3,int_0^infty,fraccos(sqrt3t)t^2+1,textdt=fracpi2sqrt3,exp(-sqrt3),,$$ where $t:=sqrt3x$. See, for example, the last two hidden boxes in the question of math.stackexchange.com/questions/2870410/….
– Batominovski
Aug 7 at 19:37
More generally, $$int_0^infty,fraccos(kx)x^2+1,textdx=fracpi2,expbig(-|k|big)text for all kinmathbbR,.$$ In your case, $$int_0^infty,fraccos(x)x^2+3,textdx=frac1sqrt3,int_0^infty,fraccos(sqrt3t)t^2+1,textdt=fracpi2sqrt3,exp(-sqrt3),,$$ where $t:=sqrt3x$. See, for example, the last two hidden boxes in the question of math.stackexchange.com/questions/2870410/….
– Batominovski
Aug 7 at 19:37
1
1
My comment is that there are two errors. Note that $$lim_Rtoinfty,int_gamma_R^1,fracexp(textiz)z^2+3,textdz=2,int_0^infty,fraccos(x)x^2+3,textdx$$ as you probably forgot that $2,cos(x)=exp(+textix)+exp(-textix)$, and that $$2pitexticdotleft(fracexp(-sqrt3)2sqrt3textiright)=fracpi,exp(-sqrt3)sqrt3$$ is what you should get after applying the Residue Theorem. Dividing this result by $2$, you will get the answer you are looking for.
– Batominovski
Aug 8 at 7:50
My comment is that there are two errors. Note that $$lim_Rtoinfty,int_gamma_R^1,fracexp(textiz)z^2+3,textdz=2,int_0^infty,fraccos(x)x^2+3,textdx$$ as you probably forgot that $2,cos(x)=exp(+textix)+exp(-textix)$, and that $$2pitexticdotleft(fracexp(-sqrt3)2sqrt3textiright)=fracpi,exp(-sqrt3)sqrt3$$ is what you should get after applying the Residue Theorem. Dividing this result by $2$, you will get the answer you are looking for.
– Batominovski
Aug 8 at 7:50
 |Â
show 6 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
Thanks to @Batominovski's comments and insights I was able to note that there was errors in the original proof and a new one can be found below.
$textProof$
To proceed one must consider that,
$$ int_0^infty,fraccos(x)x^2+3,textdx=frac1sqrt3,int_0^infty,fraccos(sqrt3t)t^2+1,textdt=fracpi2sqrt3,e^(-sqrt3)$$
$t:=sqrt3x$
Assume that $R>1$ define $gamma_R$ such that,
$$gamma_R^1(t) = t + i0 , , textif , , -R leq t leq R$$
$$gamma_R^2(t) = Re^it , textif , , , , , , 0leq t leq pi.$$
Consider our choice $f$ and that,
$$frac1sqrt3,oint_gamma_R,frace^(sqrt3t)z^2+1,textdz $$
Clearly it's simple that
$$frac1sqrt3 oint_gamma_Rfrace^(sqrt3z)z^2+1dz = sum_psi = 1,2 frac1sqrt3 oint_gamma_R^psifrace^(sqrt3z)z^2+1dz.$$
Now it's imperative to claim that
$$lim_Rtoinfty,oint_gamma_R^1,fracexp(sqrt3z)z^2+3,textdz= frac1sqrt3 ,int_0^infty, , frace^(sqrt3t)z^2+1,textdz = fracpi2sqrt3,e^(-sqrt3).$$
Now at this leg of our journey important to conjecture that,
$$Bigg| lim_R rightarrow inftyoint_ gamma_R^2frace^(sqrt3z)z^2+1dz Bigg| rightarrow 0. $$
Thanks to Estimation Lemma we can be relived that,
$$bigg |oint_gamma_R^2 frace^(sqrt3z)z^2+3dz bigg | leq bigtextlength(gamma_R^2) big cdot sup_gamma_R^2|frace^(sqrt3z)z^2+3|leq pi R cdot frac1R^2 - 3 rightarrow 0 , textas , R rightarrow infty.$$
Thus, taking
$$operatornameRe bigg( frac1sqrt3,int_0^infty,fraccos(sqrt3t)t^2+1textdt bigg) , = fracpi2sqrt3,e^(-sqrt3)$$
However we need to consider that
$$frac1sqrt3 oint_gamma_R frace^(sqrt3z)z^2+1, dz = 2 pi i sum_j=1,2 operatornameInd_gamma cdot operatornameRes_f(P_j) $$
After a trivial calculation we have that,
$$frac1sqrt3 oint_gamma_R frace^(sqrt3z)z^2+1, dz = 2 pi i + operatornameResfrace^(sqrt3z)z^2+1dz = fracpi2sqrt3,e^(-sqrt3).$$
answered Aug 8 at 15:23
Zophikel
429518
@Batominovski I managed to get a correct solution using your suggestions in the comments section, is their an alternate approach without using the Residue Theorem ?
– Zophikel
Aug 8 at 15:38
1
Rather than posting your (additional) work as an answer, better to edit it into the original post.
– Doug M
Aug 17 at 0:58
Actually this isn't (additional) work this is a proof addressing the errors that were spotted by @Batominovski
– Zophikel
Aug 17 at 1:44
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Thanks to @Batominovski's comments and insights I was able to note that there was errors in the original proof and a new one can be found below.
$textProof$
To proceed one must consider that,
$$ int_0^infty,fraccos(x)x^2+3,textdx=frac1sqrt3,int_0^infty,fraccos(sqrt3t)t^2+1,textdt=fracpi2sqrt3,e^(-sqrt3)$$
$t:=sqrt3x$
Assume that $R>1$ define $gamma_R$ such that,
$$gamma_R^1(t) = t + i0 , , textif , , -R leq t leq R$$
$$gamma_R^2(t) = Re^it , textif , , , , , , 0leq t leq pi.$$
Consider our choice $f$ and that,
$$frac1sqrt3,oint_gamma_R,frace^(sqrt3t)z^2+1,textdz $$
Clearly it's simple that
$$frac1sqrt3 oint_gamma_Rfrace^(sqrt3z)z^2+1dz = sum_psi = 1,2 frac1sqrt3 oint_gamma_R^psifrace^(sqrt3z)z^2+1dz.$$
Now it's imperative to claim that
$$lim_Rtoinfty,oint_gamma_R^1,fracexp(sqrt3z)z^2+3,textdz= frac1sqrt3 ,int_0^infty, , frace^(sqrt3t)z^2+1,textdz = fracpi2sqrt3,e^(-sqrt3).$$
Now at this leg of our journey important to conjecture that,
$$Bigg| lim_R rightarrow inftyoint_ gamma_R^2frace^(sqrt3z)z^2+1dz Bigg| rightarrow 0. $$
Thanks to Estimation Lemma we can be relived that,
$$bigg |oint_gamma_R^2 frace^(sqrt3z)z^2+3dz bigg | leq bigtextlength(gamma_R^2) big cdot sup_gamma_R^2|frace^(sqrt3z)z^2+3|leq pi R cdot frac1R^2 - 3 rightarrow 0 , textas , R rightarrow infty.$$
Thus, taking
$$operatornameRe bigg( frac1sqrt3,int_0^infty,fraccos(sqrt3t)t^2+1textdt bigg) , = fracpi2sqrt3,e^(-sqrt3)$$
However we need to consider that
$$frac1sqrt3 oint_gamma_R frace^(sqrt3z)z^2+1, dz = 2 pi i sum_j=1,2 operatornameInd_gamma cdot operatornameRes_f(P_j) $$
After a trivial calculation we have that,
$$frac1sqrt3 oint_gamma_R frace^(sqrt3z)z^2+1, dz = 2 pi i + operatornameResfrace^(sqrt3z)z^2+1dz = fracpi2sqrt3,e^(-sqrt3).$$
answered Aug 8 at 15:23
Zophikel
429518
@Batominovski I managed to get a correct solution using your suggestions in the comments section, is their an alternate approach without using the Residue Theorem ?
– Zophikel
Aug 8 at 15:38
1
Rather than posting your (additional) work as an answer, better to edit it into the original post.
– Doug M
Aug 17 at 0:58
Actually this isn't (additional) work this is a proof addressing the errors that were spotted by @Batominovski
– Zophikel
Aug 17 at 1:44
add a comment |Â
up vote
0
down vote
accepted
Thanks to @Batominovski's comments and insights I was able to note that there was errors in the original proof and a new one can be found below.
$textProof$
To proceed one must consider that,
$$ int_0^infty,fraccos(x)x^2+3,textdx=frac1sqrt3,int_0^infty,fraccos(sqrt3t)t^2+1,textdt=fracpi2sqrt3,e^(-sqrt3)$$
$t:=sqrt3x$
Assume that $R>1$ define $gamma_R$ such that,
$$gamma_R^1(t) = t + i0 , , textif , , -R leq t leq R$$
$$gamma_R^2(t) = Re^it , textif , , , , , , 0leq t leq pi.$$
Consider our choice $f$ and that,
$$frac1sqrt3,oint_gamma_R,frace^(sqrt3t)z^2+1,textdz $$
Clearly it's simple that
$$frac1sqrt3 oint_gamma_Rfrace^(sqrt3z)z^2+1dz = sum_psi = 1,2 frac1sqrt3 oint_gamma_R^psifrace^(sqrt3z)z^2+1dz.$$
Now it's imperative to claim that
$$lim_Rtoinfty,oint_gamma_R^1,fracexp(sqrt3z)z^2+3,textdz= frac1sqrt3 ,int_0^infty, , frace^(sqrt3t)z^2+1,textdz = fracpi2sqrt3,e^(-sqrt3).$$
Now at this leg of our journey important to conjecture that,
$$Bigg| lim_R rightarrow inftyoint_ gamma_R^2frace^(sqrt3z)z^2+1dz Bigg| rightarrow 0. $$
Thanks to Estimation Lemma we can be relived that,
$$bigg |oint_gamma_R^2 frace^(sqrt3z)z^2+3dz bigg | leq bigtextlength(gamma_R^2) big cdot sup_gamma_R^2|frace^(sqrt3z)z^2+3|leq pi R cdot frac1R^2 - 3 rightarrow 0 , textas , R rightarrow infty.$$
Thus, taking
$$operatornameRe bigg( frac1sqrt3,int_0^infty,fraccos(sqrt3t)t^2+1textdt bigg) , = fracpi2sqrt3,e^(-sqrt3)$$
However we need to consider that
$$frac1sqrt3 oint_gamma_R frace^(sqrt3z)z^2+1, dz = 2 pi i sum_j=1,2 operatornameInd_gamma cdot operatornameRes_f(P_j) $$
After a trivial calculation we have that,
$$frac1sqrt3 oint_gamma_R frace^(sqrt3z)z^2+1, dz = 2 pi i + operatornameResfrace^(sqrt3z)z^2+1dz = fracpi2sqrt3,e^(-sqrt3).$$
answered Aug 8 at 15:23
Zophikel
429518
@Batominovski I managed to get a correct solution using your suggestions in the comments section, is their an alternate approach without using the Residue Theorem ?
– Zophikel
Aug 8 at 15:38
1
Rather than posting your (additional) work as an answer, better to edit it into the original post.
– Doug M
Aug 17 at 0:58
Actually this isn't (additional) work this is a proof addressing the errors that were spotted by @Batominovski
– Zophikel
Aug 17 at 1:44
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Thanks to @Batominovski's comments and insights I was able to note that there was errors in the original proof and a new one can be found below.
$textProof$
To proceed one must consider that,
$$ int_0^infty,fraccos(x)x^2+3,textdx=frac1sqrt3,int_0^infty,fraccos(sqrt3t)t^2+1,textdt=fracpi2sqrt3,e^(-sqrt3)$$
$t:=sqrt3x$
Assume that $R>1$ define $gamma_R$ such that,
$$gamma_R^1(t) = t + i0 , , textif , , -R leq t leq R$$
$$gamma_R^2(t) = Re^it , textif , , , , , , 0leq t leq pi.$$
Consider our choice $f$ and that,
$$frac1sqrt3,oint_gamma_R,frace^(sqrt3t)z^2+1,textdz $$
Clearly it's simple that
$$frac1sqrt3 oint_gamma_Rfrace^(sqrt3z)z^2+1dz = sum_psi = 1,2 frac1sqrt3 oint_gamma_R^psifrace^(sqrt3z)z^2+1dz.$$
Now it's imperative to claim that
$$lim_Rtoinfty,oint_gamma_R^1,fracexp(sqrt3z)z^2+3,textdz= frac1sqrt3 ,int_0^infty, , frace^(sqrt3t)z^2+1,textdz = fracpi2sqrt3,e^(-sqrt3).$$
Now at this leg of our journey important to conjecture that,
$$Bigg| lim_R rightarrow inftyoint_ gamma_R^2frace^(sqrt3z)z^2+1dz Bigg| rightarrow 0. $$
Thanks to Estimation Lemma we can be relived that,
$$bigg |oint_gamma_R^2 frace^(sqrt3z)z^2+3dz bigg | leq bigtextlength(gamma_R^2) big cdot sup_gamma_R^2|frace^(sqrt3z)z^2+3|leq pi R cdot frac1R^2 - 3 rightarrow 0 , textas , R rightarrow infty.$$
Thus, taking
$$operatornameRe bigg( frac1sqrt3,int_0^infty,fraccos(sqrt3t)t^2+1textdt bigg) , = fracpi2sqrt3,e^(-sqrt3)$$
However we need to consider that
$$frac1sqrt3 oint_gamma_R frace^(sqrt3z)z^2+1, dz = 2 pi i sum_j=1,2 operatornameInd_gamma cdot operatornameRes_f(P_j) $$
After a trivial calculation we have that,
$$frac1sqrt3 oint_gamma_R frace^(sqrt3z)z^2+1, dz = 2 pi i + operatornameResfrace^(sqrt3z)z^2+1dz = fracpi2sqrt3,e^(-sqrt3).$$
answered Aug 8 at 15:23
Zophikel
429518
Thanks to @Batominovski's comments and insights I was able to note that there was errors in the original proof and a new one can be found below.
$textProof$
To proceed one must consider that,
$$ int_0^infty,fraccos(x)x^2+3,textdx=frac1sqrt3,int_0^infty,fraccos(sqrt3t)t^2+1,textdt=fracpi2sqrt3,e^(-sqrt3)$$
$t:=sqrt3x$
Assume that $R>1$ define $gamma_R$ such that,
$$gamma_R^1(t) = t + i0 , , textif , , -R leq t leq R$$
$$gamma_R^2(t) = Re^it , textif , , , , , , 0leq t leq pi.$$
Consider our choice $f$ and that,
$$frac1sqrt3,oint_gamma_R,frace^(sqrt3t)z^2+1,textdz $$
Clearly it's simple that
$$frac1sqrt3 oint_gamma_Rfrace^(sqrt3z)z^2+1dz = sum_psi = 1,2 frac1sqrt3 oint_gamma_R^psifrace^(sqrt3z)z^2+1dz.$$
Now it's imperative to claim that
$$lim_Rtoinfty,oint_gamma_R^1,fracexp(sqrt3z)z^2+3,textdz= frac1sqrt3 ,int_0^infty, , frace^(sqrt3t)z^2+1,textdz = fracpi2sqrt3,e^(-sqrt3).$$
Now at this leg of our journey important to conjecture that,
$$Bigg| lim_R rightarrow inftyoint_ gamma_R^2frace^(sqrt3z)z^2+1dz Bigg| rightarrow 0. $$
Thanks to Estimation Lemma we can be relived that,
$$bigg |oint_gamma_R^2 frace^(sqrt3z)z^2+3dz bigg | leq bigtextlength(gamma_R^2) big cdot sup_gamma_R^2|frace^(sqrt3z)z^2+3|leq pi R cdot frac1R^2 - 3 rightarrow 0 , textas , R rightarrow infty.$$
Thus, taking
$$operatornameRe bigg( frac1sqrt3,int_0^infty,fraccos(sqrt3t)t^2+1textdt bigg) , = fracpi2sqrt3,e^(-sqrt3)$$
However we need to consider that
$$frac1sqrt3 oint_gamma_R frace^(sqrt3z)z^2+1, dz = 2 pi i sum_j=1,2 operatornameInd_gamma cdot operatornameRes_f(P_j) $$
After a trivial calculation we have that,
$$frac1sqrt3 oint_gamma_R frace^(sqrt3z)z^2+1, dz = 2 pi i + operatornameResfrace^(sqrt3z)z^2+1dz = fracpi2sqrt3,e^(-sqrt3).$$
answered Aug 8 at 15:23
Zophikel
429518
edited Aug 17 at 0:41
answered Aug 8 at 15:23
Zophikel
429518
answered Aug 8 at 15:23
Zophikel
429518
answered Aug 8 at 15:23
Zophikel
429518
429518
@Batominovski I managed to get a correct solution using your suggestions in the comments section, is their an alternate approach without using the Residue Theorem ?
– Zophikel
Aug 8 at 15:38
1
Rather than posting your (additional) work as an answer, better to edit it into the original post.
– Doug M
Aug 17 at 0:58
Actually this isn't (additional) work this is a proof addressing the errors that were spotted by @Batominovski
– Zophikel
Aug 17 at 1:44
add a comment |Â
@Batominovski I managed to get a correct solution using your suggestions in the comments section, is their an alternate approach without using the Residue Theorem ?
– Zophikel
Aug 8 at 15:38
1
Rather than posting your (additional) work as an answer, better to edit it into the original post.
– Doug M
Aug 17 at 0:58
Actually this isn't (additional) work this is a proof addressing the errors that were spotted by @Batominovski
– Zophikel
Aug 17 at 1:44
@Batominovski I managed to get a correct solution using your suggestions in the comments section, is their an alternate approach without using the Residue Theorem ?
– Zophikel
Aug 8 at 15:38
@Batominovski I managed to get a correct solution using your suggestions in the comments section, is their an alternate approach without using the Residue Theorem ?
– Zophikel
Aug 8 at 15:38
1
1
Rather than posting your (additional) work as an answer, better to edit it into the original post.
– Doug M
Aug 17 at 0:58
Rather than posting your (additional) work as an answer, better to edit it into the original post.
– Doug M
Aug 17 at 0:58
Actually this isn't (additional) work this is a proof addressing the errors that were spotted by @Batominovski
– Zophikel
Aug 17 at 1:44
Actually this isn't (additional) work this is a proof addressing the errors that were spotted by @Batominovski
– Zophikel
Aug 17 at 1:44
add a comment |Â
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1
Where does $frac1R^2-1$ come from?
– Lord Shark the Unknown
Aug 7 at 18:41
1
Thanks for pointing out the typo it's been fixed.
– Zophikel
Aug 7 at 18:47
1
The integral is over $Bbb R$, or only over $Bbb R_ge 0$? (The contour goes from $-R$ to $R$. I was looking for a factor $2$ or $1/2$ on the one or the other side...)
– dan_fulea
Aug 7 at 19:14
1
More generally, $$int_0^infty,fraccos(kx)x^2+1,textdx=fracpi2,expbig(-|k|big)text for all kinmathbbR,.$$ In your case, $$int_0^infty,fraccos(x)x^2+3,textdx=frac1sqrt3,int_0^infty,fraccos(sqrt3t)t^2+1,textdt=fracpi2sqrt3,exp(-sqrt3),,$$ where $t:=sqrt3x$. See, for example, the last two hidden boxes in the question of math.stackexchange.com/questions/2870410/….
– Batominovski
Aug 7 at 19:37
1
My comment is that there are two errors. Note that $$lim_Rtoinfty,int_gamma_R^1,fracexp(textiz)z^2+3,textdz=2,int_0^infty,fraccos(x)x^2+3,textdx$$ as you probably forgot that $2,cos(x)=exp(+textix)+exp(-textix)$, and that $$2pitexticdotleft(fracexp(-sqrt3)2sqrt3textiright)=fracpi,exp(-sqrt3)sqrt3$$ is what you should get after applying the Residue Theorem. Dividing this result by $2$, you will get the answer you are looking for.
– Batominovski
Aug 8 at 7:50