Mixing
Calculating $int_-infty^inftyleft( fraccosleft (x right )x^4 + 1 right)dx$ via the Residue Theorem?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
In the text, "Function Theory of One Complex Variable" by Robert E. Greene and Steven G. Krantz. I'm inquiring if my proof of $(1)$ is valid ?
$textProposition , , , (1) $
$$int_-infty^inftyleft( fraccosleft (x right )x^4 + 1 right)dx=fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right) $$
The first step on our quest to prove $(1)$, is one must consider a Semicircular Contour $gamma_R$, assuming $R > 1$ define
$$gamma_R^1(t) = t + i0 , , textif , , -R leq t leq R $$
$$gamma_R^2(t) = Re^it , , textif , , , , , , 0leq t leq pi.$$
One can call these two curves taken together $gamma_R$ or $gamma$, after picking our $gamma_R$ one can consider $$oint_gamma _Rfrace^izz^4+1dz.$$
It's trivial that,
$$oint_gamma_Rg(z)~ dz = oint_gamma_R^1 g(z) dz+ oint_gamma_R^2g(z) ~dz.$$
It's imperative that
$$displaystyle oint_gamma_R^1 g(z) dz rightarrow lim_R rightarrow infty int_-R^R frace^ix1+x^4 , , textas, , R rightarrow infty $$
Using the Estimation Lemma one be relived to see
$$bigg |oint_gamma_R^2 g(z) dz bigg | leq bigtextlength(gamma_R^2) big cdot sup_gamma_R^2|g(z)|leq pi cdot frac1R^4 - 1. $$
Now it's safe to say that
$$lim_R rightarrow inftybigg | oint_gamma_R^2frac11+z^4 dz bigg| rightarrow 0. $$
It's easy to note after all our struggle that
$$ int_-infty^infty fraccos(x)1+x^4 = Re int_-infty^infty frace^ix1+x^4 = Rebigg( fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right)bigg) = fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right) $$
The finial leg of our conquest is to consider that
$$oint_gamma _Rfrace^izz^4+1dz = 2 pi i sum_j=1,2,3,4 Ind_gamma cdot operatornameRes_f(P_j)$$
It's easy to calculate that
$$operatornameRes_z = sqrt[4]-1 Big(frace^izz^4+1 Big) = frac14sqrt[4]-1 , , -e^(-1)^3/4$$
$$operatornameRes_z = sqrt[4]-1 Big(frace^izz^4+1 Big) = frac14sqrt[4]-1 , , e^(-1)^3/4$$
$$operatornameRes_z = -(-1)^3/4 Big(frace^izz^4+1 Big) = frac14sqrt[4]-1 , , e^sqrt[4]-1$$
$$operatornameRes_z = (-1)^3/4 Big(frace^izz^4+1 Big) = frac14sqrt[4]-1 , , e^sqrt[4]-1$$
Putting the pieces together we have that
$$oint_gamma _Rfrace^izz^4+1dz = 2 pi i + frac14sqrt[4]-1 , , -e^(-1)^3/4 + frac14sqrt[4]-1 , , e^(-1)^3/4 + frac14sqrt[4]-1 , , e^sqrt[4]-1 + frac14sqrt[4]-1 , , e^sqrt[4]-1= fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right).$$
complex-analysis proof-verification contour-integration residue-calculus
asked Aug 5 at 23:24
Zophikel
429518
add a comment |Â
up vote
1
down vote
favorite
In the text, "Function Theory of One Complex Variable" by Robert E. Greene and Steven G. Krantz. I'm inquiring if my proof of $(1)$ is valid ?
$textProposition , , , (1) $
$$int_-infty^inftyleft( fraccosleft (x right )x^4 + 1 right)dx=fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right) $$
The first step on our quest to prove $(1)$, is one must consider a Semicircular Contour $gamma_R$, assuming $R > 1$ define
$$gamma_R^1(t) = t + i0 , , textif , , -R leq t leq R $$
$$gamma_R^2(t) = Re^it , , textif , , , , , , 0leq t leq pi.$$
One can call these two curves taken together $gamma_R$ or $gamma$, after picking our $gamma_R$ one can consider $$oint_gamma _Rfrace^izz^4+1dz.$$
It's trivial that,
$$oint_gamma_Rg(z)~ dz = oint_gamma_R^1 g(z) dz+ oint_gamma_R^2g(z) ~dz.$$
It's imperative that
$$displaystyle oint_gamma_R^1 g(z) dz rightarrow lim_R rightarrow infty int_-R^R frace^ix1+x^4 , , textas, , R rightarrow infty $$
Using the Estimation Lemma one be relived to see
$$bigg |oint_gamma_R^2 g(z) dz bigg | leq bigtextlength(gamma_R^2) big cdot sup_gamma_R^2|g(z)|leq pi cdot frac1R^4 - 1. $$
Now it's safe to say that
$$lim_R rightarrow inftybigg | oint_gamma_R^2frac11+z^4 dz bigg| rightarrow 0. $$
It's easy to note after all our struggle that
$$ int_-infty^infty fraccos(x)1+x^4 = Re int_-infty^infty frace^ix1+x^4 = Rebigg( fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right)bigg) = fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right) $$
The finial leg of our conquest is to consider that
$$oint_gamma _Rfrace^izz^4+1dz = 2 pi i sum_j=1,2,3,4 Ind_gamma cdot operatornameRes_f(P_j)$$
It's easy to calculate that
$$operatornameRes_z = sqrt[4]-1 Big(frace^izz^4+1 Big) = frac14sqrt[4]-1 , , -e^(-1)^3/4$$
$$operatornameRes_z = sqrt[4]-1 Big(frace^izz^4+1 Big) = frac14sqrt[4]-1 , , e^(-1)^3/4$$
$$operatornameRes_z = -(-1)^3/4 Big(frace^izz^4+1 Big) = frac14sqrt[4]-1 , , e^sqrt[4]-1$$
$$operatornameRes_z = (-1)^3/4 Big(frace^izz^4+1 Big) = frac14sqrt[4]-1 , , e^sqrt[4]-1$$
Putting the pieces together we have that
$$oint_gamma _Rfrace^izz^4+1dz = 2 pi i + frac14sqrt[4]-1 , , -e^(-1)^3/4 + frac14sqrt[4]-1 , , e^(-1)^3/4 + frac14sqrt[4]-1 , , e^sqrt[4]-1 + frac14sqrt[4]-1 , , e^sqrt[4]-1= fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right).$$
complex-analysis proof-verification contour-integration residue-calculus
asked Aug 5 at 23:24
Zophikel
429518
Note that $sin(x)+cos(x) = sqrt2sin(x+pi/4)$.
– marty cohen
Aug 5 at 23:37
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In the text, "Function Theory of One Complex Variable" by Robert E. Greene and Steven G. Krantz. I'm inquiring if my proof of $(1)$ is valid ?
$textProposition , , , (1) $
$$int_-infty^inftyleft( fraccosleft (x right )x^4 + 1 right)dx=fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right) $$
The first step on our quest to prove $(1)$, is one must consider a Semicircular Contour $gamma_R$, assuming $R > 1$ define
$$gamma_R^1(t) = t + i0 , , textif , , -R leq t leq R $$
$$gamma_R^2(t) = Re^it , , textif , , , , , , 0leq t leq pi.$$
One can call these two curves taken together $gamma_R$ or $gamma$, after picking our $gamma_R$ one can consider $$oint_gamma _Rfrace^izz^4+1dz.$$
It's trivial that,
$$oint_gamma_Rg(z)~ dz = oint_gamma_R^1 g(z) dz+ oint_gamma_R^2g(z) ~dz.$$
It's imperative that
$$displaystyle oint_gamma_R^1 g(z) dz rightarrow lim_R rightarrow infty int_-R^R frace^ix1+x^4 , , textas, , R rightarrow infty $$
Using the Estimation Lemma one be relived to see
$$bigg |oint_gamma_R^2 g(z) dz bigg | leq bigtextlength(gamma_R^2) big cdot sup_gamma_R^2|g(z)|leq pi cdot frac1R^4 - 1. $$
Now it's safe to say that
$$lim_R rightarrow inftybigg | oint_gamma_R^2frac11+z^4 dz bigg| rightarrow 0. $$
It's easy to note after all our struggle that
$$ int_-infty^infty fraccos(x)1+x^4 = Re int_-infty^infty frace^ix1+x^4 = Rebigg( fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right)bigg) = fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right) $$
The finial leg of our conquest is to consider that
$$oint_gamma _Rfrace^izz^4+1dz = 2 pi i sum_j=1,2,3,4 Ind_gamma cdot operatornameRes_f(P_j)$$
It's easy to calculate that
$$operatornameRes_z = sqrt[4]-1 Big(frace^izz^4+1 Big) = frac14sqrt[4]-1 , , -e^(-1)^3/4$$
$$operatornameRes_z = sqrt[4]-1 Big(frace^izz^4+1 Big) = frac14sqrt[4]-1 , , e^(-1)^3/4$$
$$operatornameRes_z = -(-1)^3/4 Big(frace^izz^4+1 Big) = frac14sqrt[4]-1 , , e^sqrt[4]-1$$
$$operatornameRes_z = (-1)^3/4 Big(frace^izz^4+1 Big) = frac14sqrt[4]-1 , , e^sqrt[4]-1$$
Putting the pieces together we have that
$$oint_gamma _Rfrace^izz^4+1dz = 2 pi i + frac14sqrt[4]-1 , , -e^(-1)^3/4 + frac14sqrt[4]-1 , , e^(-1)^3/4 + frac14sqrt[4]-1 , , e^sqrt[4]-1 + frac14sqrt[4]-1 , , e^sqrt[4]-1= fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right).$$
complex-analysis proof-verification contour-integration residue-calculus
asked Aug 5 at 23:24
Zophikel
429518
In the text, "Function Theory of One Complex Variable" by Robert E. Greene and Steven G. Krantz. I'm inquiring if my proof of $(1)$ is valid ?
$textProposition , , , (1) $
$$int_-infty^inftyleft( fraccosleft (x right )x^4 + 1 right)dx=fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right) $$
The first step on our quest to prove $(1)$, is one must consider a Semicircular Contour $gamma_R$, assuming $R > 1$ define
$$gamma_R^1(t) = t + i0 , , textif , , -R leq t leq R $$
$$gamma_R^2(t) = Re^it , , textif , , , , , , 0leq t leq pi.$$
One can call these two curves taken together $gamma_R$ or $gamma$, after picking our $gamma_R$ one can consider $$oint_gamma _Rfrace^izz^4+1dz.$$
It's trivial that,
$$oint_gamma_Rg(z)~ dz = oint_gamma_R^1 g(z) dz+ oint_gamma_R^2g(z) ~dz.$$
It's imperative that
$$displaystyle oint_gamma_R^1 g(z) dz rightarrow lim_R rightarrow infty int_-R^R frace^ix1+x^4 , , textas, , R rightarrow infty $$
Using the Estimation Lemma one be relived to see
$$bigg |oint_gamma_R^2 g(z) dz bigg | leq bigtextlength(gamma_R^2) big cdot sup_gamma_R^2|g(z)|leq pi cdot frac1R^4 - 1. $$
Now it's safe to say that
$$lim_R rightarrow inftybigg | oint_gamma_R^2frac11+z^4 dz bigg| rightarrow 0. $$
It's easy to note after all our struggle that
$$ int_-infty^infty fraccos(x)1+x^4 = Re int_-infty^infty frace^ix1+x^4 = Rebigg( fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right)bigg) = fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right) $$
The finial leg of our conquest is to consider that
$$oint_gamma _Rfrace^izz^4+1dz = 2 pi i sum_j=1,2,3,4 Ind_gamma cdot operatornameRes_f(P_j)$$
It's easy to calculate that
$$operatornameRes_z = sqrt[4]-1 Big(frace^izz^4+1 Big) = frac14sqrt[4]-1 , , -e^(-1)^3/4$$
$$operatornameRes_z = sqrt[4]-1 Big(frace^izz^4+1 Big) = frac14sqrt[4]-1 , , e^(-1)^3/4$$
$$operatornameRes_z = -(-1)^3/4 Big(frace^izz^4+1 Big) = frac14sqrt[4]-1 , , e^sqrt[4]-1$$
$$operatornameRes_z = (-1)^3/4 Big(frace^izz^4+1 Big) = frac14sqrt[4]-1 , , e^sqrt[4]-1$$
Putting the pieces together we have that
$$oint_gamma _Rfrace^izz^4+1dz = 2 pi i + frac14sqrt[4]-1 , , -e^(-1)^3/4 + frac14sqrt[4]-1 , , e^(-1)^3/4 + frac14sqrt[4]-1 , , e^sqrt[4]-1 + frac14sqrt[4]-1 , , e^sqrt[4]-1= fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right).$$
complex-analysis proof-verification contour-integration residue-calculus
asked Aug 5 at 23:24
Zophikel
429518
edited Aug 6 at 22:32
asked Aug 5 at 23:24
Zophikel
429518
asked Aug 5 at 23:24
Zophikel
429518
asked Aug 5 at 23:24
Zophikel
429518
429518
Note that $sin(x)+cos(x) = sqrt2sin(x+pi/4)$.
– marty cohen
Aug 5 at 23:37
add a comment |Â
Note that $sin(x)+cos(x) = sqrt2sin(x+pi/4)$.
– marty cohen
Aug 5 at 23:37
Note that $sin(x)+cos(x) = sqrt2sin(x+pi/4)$.
– marty cohen
Aug 5 at 23:37
Note that $sin(x)+cos(x) = sqrt2sin(x+pi/4)$.
– marty cohen
Aug 5 at 23:37
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
Most of it is correct, but there's a problem near the end, when you apply the residue theorem. The polynomial $x^4+1$ has four roots: $expleft(fracpi i4right)$, $expleft(frac3pi i4right)$, $expleft(frac5pi i4right)$, and $expleft(frac7pi i4right)$. But only the first two matter, since they're the only ones which are in the semicircle bounded by the image of $gamma_R$.
answered Aug 5 at 23:33
José Carlos Santos
115k1698177
Thanks for pointing out the issue looking back at my textbook it seems one should only sum the poles that lie inside our contour, which is $frac14sqrt[4]-1 , , -e^(-1)^3/4$ and $frac14sqrt[4]-1 , , e^(-1)^3/4$. I'll have answer my question with the correct final steps where I apply the Residue Theorem
– Zophikel
Aug 5 at 23:42
add a comment |Â
up vote
0
down vote
Just for the fun of it,
I put this into Wolfy
and got
$int_0^∞ cos(x)/(x^4 + 1) dx
= -frac14 (-1)^3/4 e^-(-1)^1/4 (i + e^i sqrt2) À
≈0.772138
$.
answered Aug 5 at 23:35
marty cohen
69.3k446122
Dude that Is not a really helpful answer and doesn't even address my question I know what the integral approximation is and what I want to know is whether my proof is correct or not.
– Zophikel
Aug 8 at 1:12
add a comment |Â
up vote
0
down vote
accepted
From the previous attempt the last line,
$$oint_gamma _Rfrace^izz^4+1dz = 2 pi i + frac14sqrt[4]-1 , , -e^(-1)^3/4 + frac14sqrt[4]-1 , , e^(-1)^3/4 + frac14sqrt[4]-1 , , e^sqrt[4]-1 + frac14sqrt[4]-1 , , e^sqrt[4]-1= fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right).$$
Is incorrect and one should sum the poles inside $gamma_R$ hence,
$$oint_gamma _Rfrace^izz^4+1dz = 2 pi i +operatornameReslimitslimits_ z=omegafrace^izz^4+1 = fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right). $$
answered Aug 6 at 17:14
Zophikel
429518
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Most of it is correct, but there's a problem near the end, when you apply the residue theorem. The polynomial $x^4+1$ has four roots: $expleft(fracpi i4right)$, $expleft(frac3pi i4right)$, $expleft(frac5pi i4right)$, and $expleft(frac7pi i4right)$. But only the first two matter, since they're the only ones which are in the semicircle bounded by the image of $gamma_R$.
answered Aug 5 at 23:33
José Carlos Santos
115k1698177
Thanks for pointing out the issue looking back at my textbook it seems one should only sum the poles that lie inside our contour, which is $frac14sqrt[4]-1 , , -e^(-1)^3/4$ and $frac14sqrt[4]-1 , , e^(-1)^3/4$. I'll have answer my question with the correct final steps where I apply the Residue Theorem
– Zophikel
Aug 5 at 23:42
add a comment |Â
up vote
1
down vote
Most of it is correct, but there's a problem near the end, when you apply the residue theorem. The polynomial $x^4+1$ has four roots: $expleft(fracpi i4right)$, $expleft(frac3pi i4right)$, $expleft(frac5pi i4right)$, and $expleft(frac7pi i4right)$. But only the first two matter, since they're the only ones which are in the semicircle bounded by the image of $gamma_R$.
answered Aug 5 at 23:33
José Carlos Santos
115k1698177
Thanks for pointing out the issue looking back at my textbook it seems one should only sum the poles that lie inside our contour, which is $frac14sqrt[4]-1 , , -e^(-1)^3/4$ and $frac14sqrt[4]-1 , , e^(-1)^3/4$. I'll have answer my question with the correct final steps where I apply the Residue Theorem
– Zophikel
Aug 5 at 23:42
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Most of it is correct, but there's a problem near the end, when you apply the residue theorem. The polynomial $x^4+1$ has four roots: $expleft(fracpi i4right)$, $expleft(frac3pi i4right)$, $expleft(frac5pi i4right)$, and $expleft(frac7pi i4right)$. But only the first two matter, since they're the only ones which are in the semicircle bounded by the image of $gamma_R$.
answered Aug 5 at 23:33
José Carlos Santos
115k1698177
Most of it is correct, but there's a problem near the end, when you apply the residue theorem. The polynomial $x^4+1$ has four roots: $expleft(fracpi i4right)$, $expleft(frac3pi i4right)$, $expleft(frac5pi i4right)$, and $expleft(frac7pi i4right)$. But only the first two matter, since they're the only ones which are in the semicircle bounded by the image of $gamma_R$.
answered Aug 5 at 23:33
José Carlos Santos
115k1698177
answered Aug 5 at 23:33
José Carlos Santos
115k1698177
answered Aug 5 at 23:33
José Carlos Santos
115k1698177
answered Aug 5 at 23:33
José Carlos Santos
115k1698177
115k1698177
Thanks for pointing out the issue looking back at my textbook it seems one should only sum the poles that lie inside our contour, which is $frac14sqrt[4]-1 , , -e^(-1)^3/4$ and $frac14sqrt[4]-1 , , e^(-1)^3/4$. I'll have answer my question with the correct final steps where I apply the Residue Theorem
– Zophikel
Aug 5 at 23:42
add a comment |Â
Thanks for pointing out the issue looking back at my textbook it seems one should only sum the poles that lie inside our contour, which is $frac14sqrt[4]-1 , , -e^(-1)^3/4$ and $frac14sqrt[4]-1 , , e^(-1)^3/4$. I'll have answer my question with the correct final steps where I apply the Residue Theorem
– Zophikel
Aug 5 at 23:42
Thanks for pointing out the issue looking back at my textbook it seems one should only sum the poles that lie inside our contour, which is $frac14sqrt[4]-1 , , -e^(-1)^3/4$ and $frac14sqrt[4]-1 , , e^(-1)^3/4$. I'll have answer my question with the correct final steps where I apply the Residue Theorem
– Zophikel
Aug 5 at 23:42
Thanks for pointing out the issue looking back at my textbook it seems one should only sum the poles that lie inside our contour, which is $frac14sqrt[4]-1 , , -e^(-1)^3/4$ and $frac14sqrt[4]-1 , , e^(-1)^3/4$. I'll have answer my question with the correct final steps where I apply the Residue Theorem
– Zophikel
Aug 5 at 23:42
add a comment |Â
up vote
0
down vote
Just for the fun of it,
I put this into Wolfy
and got
$int_0^∞ cos(x)/(x^4 + 1) dx
= -frac14 (-1)^3/4 e^-(-1)^1/4 (i + e^i sqrt2) À
≈0.772138
$.
answered Aug 5 at 23:35
marty cohen
69.3k446122
Dude that Is not a really helpful answer and doesn't even address my question I know what the integral approximation is and what I want to know is whether my proof is correct or not.
– Zophikel
Aug 8 at 1:12
add a comment |Â
up vote
0
down vote
Just for the fun of it,
I put this into Wolfy
and got
$int_0^∞ cos(x)/(x^4 + 1) dx
= -frac14 (-1)^3/4 e^-(-1)^1/4 (i + e^i sqrt2) À
≈0.772138
$.
answered Aug 5 at 23:35
marty cohen
69.3k446122
Dude that Is not a really helpful answer and doesn't even address my question I know what the integral approximation is and what I want to know is whether my proof is correct or not.
– Zophikel
Aug 8 at 1:12
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Just for the fun of it,
I put this into Wolfy
and got
$int_0^∞ cos(x)/(x^4 + 1) dx
= -frac14 (-1)^3/4 e^-(-1)^1/4 (i + e^i sqrt2) À
≈0.772138
$.
answered Aug 5 at 23:35
marty cohen
69.3k446122
Just for the fun of it,
I put this into Wolfy
and got
$int_0^∞ cos(x)/(x^4 + 1) dx
= -frac14 (-1)^3/4 e^-(-1)^1/4 (i + e^i sqrt2) À
≈0.772138
$.
answered Aug 5 at 23:35
marty cohen
69.3k446122
answered Aug 5 at 23:35
marty cohen
69.3k446122
answered Aug 5 at 23:35
marty cohen
69.3k446122
answered Aug 5 at 23:35
marty cohen
69.3k446122
69.3k446122
Dude that Is not a really helpful answer and doesn't even address my question I know what the integral approximation is and what I want to know is whether my proof is correct or not.
– Zophikel
Aug 8 at 1:12
add a comment |Â
Dude that Is not a really helpful answer and doesn't even address my question I know what the integral approximation is and what I want to know is whether my proof is correct or not.
– Zophikel
Aug 8 at 1:12
Dude that Is not a really helpful answer and doesn't even address my question I know what the integral approximation is and what I want to know is whether my proof is correct or not.
– Zophikel
Aug 8 at 1:12
Dude that Is not a really helpful answer and doesn't even address my question I know what the integral approximation is and what I want to know is whether my proof is correct or not.
– Zophikel
Aug 8 at 1:12
add a comment |Â
up vote
0
down vote
accepted
From the previous attempt the last line,
$$oint_gamma _Rfrace^izz^4+1dz = 2 pi i + frac14sqrt[4]-1 , , -e^(-1)^3/4 + frac14sqrt[4]-1 , , e^(-1)^3/4 + frac14sqrt[4]-1 , , e^sqrt[4]-1 + frac14sqrt[4]-1 , , e^sqrt[4]-1= fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right).$$
Is incorrect and one should sum the poles inside $gamma_R$ hence,
$$oint_gamma _Rfrace^izz^4+1dz = 2 pi i +operatornameReslimitslimits_ z=omegafrace^izz^4+1 = fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right). $$
answered Aug 6 at 17:14
Zophikel
429518
add a comment |Â
up vote
0
down vote
accepted
From the previous attempt the last line,
$$oint_gamma _Rfrace^izz^4+1dz = 2 pi i + frac14sqrt[4]-1 , , -e^(-1)^3/4 + frac14sqrt[4]-1 , , e^(-1)^3/4 + frac14sqrt[4]-1 , , e^sqrt[4]-1 + frac14sqrt[4]-1 , , e^sqrt[4]-1= fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right).$$
Is incorrect and one should sum the poles inside $gamma_R$ hence,
$$oint_gamma _Rfrace^izz^4+1dz = 2 pi i +operatornameReslimitslimits_ z=omegafrace^izz^4+1 = fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right). $$
answered Aug 6 at 17:14
Zophikel
429518
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
From the previous attempt the last line,
$$oint_gamma _Rfrace^izz^4+1dz = 2 pi i + frac14sqrt[4]-1 , , -e^(-1)^3/4 + frac14sqrt[4]-1 , , e^(-1)^3/4 + frac14sqrt[4]-1 , , e^sqrt[4]-1 + frac14sqrt[4]-1 , , e^sqrt[4]-1= fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right).$$
Is incorrect and one should sum the poles inside $gamma_R$ hence,
$$oint_gamma _Rfrace^izz^4+1dz = 2 pi i +operatornameReslimitslimits_ z=omegafrace^izz^4+1 = fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right). $$
answered Aug 6 at 17:14
Zophikel
429518
From the previous attempt the last line,
$$oint_gamma _Rfrace^izz^4+1dz = 2 pi i + frac14sqrt[4]-1 , , -e^(-1)^3/4 + frac14sqrt[4]-1 , , e^(-1)^3/4 + frac14sqrt[4]-1 , , e^sqrt[4]-1 + frac14sqrt[4]-1 , , e^sqrt[4]-1= fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right).$$
Is incorrect and one should sum the poles inside $gamma_R$ hence,
$$oint_gamma _Rfrace^izz^4+1dz = 2 pi i +operatornameReslimitslimits_ z=omegafrace^izz^4+1 = fracsqrt22 e^fracsqrt22 left(pi sinleft (fracsqrt22 right ) + pi cosleft (fracsqrt22 right )right). $$
answered Aug 6 at 17:14
Zophikel
429518
edited Aug 6 at 17:24
answered Aug 6 at 17:14
Zophikel
429518
answered Aug 6 at 17:14
Zophikel
429518
answered Aug 6 at 17:14
Zophikel
429518
429518
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873429%2fcalculating-int-infty-infty-left-frac-cos-left-x-right-x4%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Note that $sin(x)+cos(x) = sqrt2sin(x+pi/4)$.
– marty cohen
Aug 5 at 23:37