Solving integrals using complex analysis
$begingroup$
There are many times when one uses methods from complex analysis to solve integrals that would otherwise be very difficult to solve. My question is:
Can all integrals that are usually solved this way be solved without complex analysis, with the usual integration techniques? Or are there some integrals that can only be solved using complex analysis?
integration complex-analysis residue-calculus
asked Dec 1 '18 at 15:05
J. DionisioJ. Dionisio
10511
$endgroup$
add a comment |
$begingroup$
There are many times when one uses methods from complex analysis to solve integrals that would otherwise be very difficult to solve. My question is:
Can all integrals that are usually solved this way be solved without complex analysis, with the usual integration techniques? Or are there some integrals that can only be solved using complex analysis?
integration complex-analysis residue-calculus
asked Dec 1 '18 at 15:05
J. DionisioJ. Dionisio
10511
$endgroup$
$begingroup$
"So Paul puts up this tremendous damn integral he had obtained by starting out with a complex function that he knew the answer to, taking out the real part of it and leaving only the complex part. He had unwrapped it so it was only possible by contour integration!" A simple example is $$int_0^{2 pi} sec e^{i t} dt = int_0^{2 pi} frac {2 cos(cos t) cosh( sin t)} {cos(2 cos t) + cosh(2 sin t)} dt.$$
$endgroup$
– Maxim
Dec 7 '18 at 10:47
$begingroup$
@Maxim You should rather write ".. so it was as far as he knows only possible by contour integration!". As I explained, one can never be certain that there is no other way to solve it. Anyway, nice example!
$endgroup$
– James
Dec 7 '18 at 13:07
$begingroup$
Related.
$endgroup$
– user26872
Dec 9 '18 at 23:03
add a comment |
$begingroup$
There are many times when one uses methods from complex analysis to solve integrals that would otherwise be very difficult to solve. My question is:
Can all integrals that are usually solved this way be solved without complex analysis, with the usual integration techniques? Or are there some integrals that can only be solved using complex analysis?
integration complex-analysis residue-calculus
asked Dec 1 '18 at 15:05
J. DionisioJ. Dionisio
10511
$endgroup$
There are many times when one uses methods from complex analysis to solve integrals that would otherwise be very difficult to solve. My question is:
Can all integrals that are usually solved this way be solved without complex analysis, with the usual integration techniques? Or are there some integrals that can only be solved using complex analysis?
integration complex-analysis residue-calculus
integration complex-analysis residue-calculus
asked Dec 1 '18 at 15:05
J. DionisioJ. Dionisio
10511
asked Dec 1 '18 at 15:05
J. DionisioJ. Dionisio
10511
asked Dec 1 '18 at 15:05
J. DionisioJ. Dionisio
10511
asked Dec 1 '18 at 15:05
J. DionisioJ. Dionisio
10511
asked Dec 1 '18 at 15:05
J. DionisioJ. Dionisio
10511
10511
$begingroup$
"So Paul puts up this tremendous damn integral he had obtained by starting out with a complex function that he knew the answer to, taking out the real part of it and leaving only the complex part. He had unwrapped it so it was only possible by contour integration!" A simple example is $$int_0^{2 pi} sec e^{i t} dt = int_0^{2 pi} frac {2 cos(cos t) cosh( sin t)} {cos(2 cos t) + cosh(2 sin t)} dt.$$
$endgroup$
– Maxim
Dec 7 '18 at 10:47
$begingroup$
@Maxim You should rather write ".. so it was as far as he knows only possible by contour integration!". As I explained, one can never be certain that there is no other way to solve it. Anyway, nice example!
$endgroup$
– James
Dec 7 '18 at 13:07
$begingroup$
Related.
$endgroup$
– user26872
Dec 9 '18 at 23:03
add a comment |
$begingroup$
"So Paul puts up this tremendous damn integral he had obtained by starting out with a complex function that he knew the answer to, taking out the real part of it and leaving only the complex part. He had unwrapped it so it was only possible by contour integration!" A simple example is $$int_0^{2 pi} sec e^{i t} dt = int_0^{2 pi} frac {2 cos(cos t) cosh( sin t)} {cos(2 cos t) + cosh(2 sin t)} dt.$$
$endgroup$
– Maxim
Dec 7 '18 at 10:47
$begingroup$
@Maxim You should rather write ".. so it was as far as he knows only possible by contour integration!". As I explained, one can never be certain that there is no other way to solve it. Anyway, nice example!
$endgroup$
– James
Dec 7 '18 at 13:07
$begingroup$
Related.
$endgroup$
– user26872
Dec 9 '18 at 23:03
$begingroup$
"So Paul puts up this tremendous damn integral he had obtained by starting out with a complex function that he knew the answer to, taking out the real part of it and leaving only the complex part. He had unwrapped it so it was only possible by contour integration!" A simple example is $$int_0^{2 pi} sec e^{i t} dt = int_0^{2 pi} frac {2 cos(cos t) cosh( sin t)} {cos(2 cos t) + cosh(2 sin t)} dt.$$
$endgroup$
– Maxim
Dec 7 '18 at 10:47
$begingroup$
"So Paul puts up this tremendous damn integral he had obtained by starting out with a complex function that he knew the answer to, taking out the real part of it and leaving only the complex part. He had unwrapped it so it was only possible by contour integration!" A simple example is $$int_0^{2 pi} sec e^{i t} dt = int_0^{2 pi} frac {2 cos(cos t) cosh( sin t)} {cos(2 cos t) + cosh(2 sin t)} dt.$$
$endgroup$
– Maxim
Dec 7 '18 at 10:47
$begingroup$
@Maxim You should rather write ".. so it was as far as he knows only possible by contour integration!". As I explained, one can never be certain that there is no other way to solve it. Anyway, nice example!
$endgroup$
– James
Dec 7 '18 at 13:07
$begingroup$
@Maxim You should rather write ".. so it was as far as he knows only possible by contour integration!". As I explained, one can never be certain that there is no other way to solve it. Anyway, nice example!
$endgroup$
– James
Dec 7 '18 at 13:07
$begingroup$
Related.
$endgroup$
– user26872
Dec 9 '18 at 23:03
$begingroup$
Related.
$endgroup$
– user26872
Dec 9 '18 at 23:03
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
To my best knowledge, elliptic integrals can not be solved without methods from complex analysis.
Also integrals involving the residue theorem to solve them seem to be hard to solve with other methods but some of them can also be solved without using the residue theorem.
But your questions isn't well-posed: Just because we don't know yet how to solve an integral without complex analysis doesn't mean that there cannot be any way to do so.
You should therefore probably ask: Are there integrals for which the only known ways to solve it are methods from complex analysis?
Also note that if there is a nice way to solve an integral using complex analysis, people will most likely not search for a way to solve these integrals by real (and probably very complicated) methods.
answered Dec 6 '18 at 14:40
JamesJames
1,090216
$endgroup$
$begingroup$
The question is if there is an integral that we know for sure that it is impossible to solve with Complex Analysis, if it was proven that it is only possile this way. I'm interested in knowing if such an integral exists
$endgroup$
– J. Dionisio
Dec 6 '18 at 22:02
$begingroup$
What I tried to explain to you is that one just cannot answer this question. How would you know that there is no way to solve it with real methods? No chance because there might be a way you just haven't thought of or a new real method not yet discovered. If you are searching for integrals for which till now no solution with real methods is known yet, you may look at the elliptic integrals.
$endgroup$
– James
Dec 7 '18 at 8:17
add a comment |
$begingroup$
The residue theorem is one of the most powerful means of integration.
At the same time, there are strict conditions for its use, and the integration itself sometimes becomes a very difficult task (task1, task2). The second task allows us to compare the complexity of the application of the theorem on residues with the method of decomposition of a polynomial into elementary ones.
At the same time, there are methods that can compete with the theorem on deductions. First of all, integration by parameter.
answered Dec 7 '18 at 22:47
Yuri NegometyanovYuri Negometyanov
11.7k1729
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To my best knowledge, elliptic integrals can not be solved without methods from complex analysis.
Also integrals involving the residue theorem to solve them seem to be hard to solve with other methods but some of them can also be solved without using the residue theorem.
But your questions isn't well-posed: Just because we don't know yet how to solve an integral without complex analysis doesn't mean that there cannot be any way to do so.
You should therefore probably ask: Are there integrals for which the only known ways to solve it are methods from complex analysis?
Also note that if there is a nice way to solve an integral using complex analysis, people will most likely not search for a way to solve these integrals by real (and probably very complicated) methods.
answered Dec 6 '18 at 14:40
JamesJames
1,090216
$endgroup$
$begingroup$
The question is if there is an integral that we know for sure that it is impossible to solve with Complex Analysis, if it was proven that it is only possile this way. I'm interested in knowing if such an integral exists
$endgroup$
– J. Dionisio
Dec 6 '18 at 22:02
$begingroup$
What I tried to explain to you is that one just cannot answer this question. How would you know that there is no way to solve it with real methods? No chance because there might be a way you just haven't thought of or a new real method not yet discovered. If you are searching for integrals for which till now no solution with real methods is known yet, you may look at the elliptic integrals.
$endgroup$
– James
Dec 7 '18 at 8:17
add a comment |
$begingroup$
To my best knowledge, elliptic integrals can not be solved without methods from complex analysis.
Also integrals involving the residue theorem to solve them seem to be hard to solve with other methods but some of them can also be solved without using the residue theorem.
But your questions isn't well-posed: Just because we don't know yet how to solve an integral without complex analysis doesn't mean that there cannot be any way to do so.
You should therefore probably ask: Are there integrals for which the only known ways to solve it are methods from complex analysis?
Also note that if there is a nice way to solve an integral using complex analysis, people will most likely not search for a way to solve these integrals by real (and probably very complicated) methods.
answered Dec 6 '18 at 14:40
JamesJames
1,090216
$endgroup$
$begingroup$
The question is if there is an integral that we know for sure that it is impossible to solve with Complex Analysis, if it was proven that it is only possile this way. I'm interested in knowing if such an integral exists
$endgroup$
– J. Dionisio
Dec 6 '18 at 22:02
$begingroup$
What I tried to explain to you is that one just cannot answer this question. How would you know that there is no way to solve it with real methods? No chance because there might be a way you just haven't thought of or a new real method not yet discovered. If you are searching for integrals for which till now no solution with real methods is known yet, you may look at the elliptic integrals.
$endgroup$
– James
Dec 7 '18 at 8:17
add a comment |
$begingroup$
To my best knowledge, elliptic integrals can not be solved without methods from complex analysis.
Also integrals involving the residue theorem to solve them seem to be hard to solve with other methods but some of them can also be solved without using the residue theorem.
But your questions isn't well-posed: Just because we don't know yet how to solve an integral without complex analysis doesn't mean that there cannot be any way to do so.
You should therefore probably ask: Are there integrals for which the only known ways to solve it are methods from complex analysis?
Also note that if there is a nice way to solve an integral using complex analysis, people will most likely not search for a way to solve these integrals by real (and probably very complicated) methods.
answered Dec 6 '18 at 14:40
JamesJames
1,090216
$endgroup$
To my best knowledge, elliptic integrals can not be solved without methods from complex analysis.
Also integrals involving the residue theorem to solve them seem to be hard to solve with other methods but some of them can also be solved without using the residue theorem.
But your questions isn't well-posed: Just because we don't know yet how to solve an integral without complex analysis doesn't mean that there cannot be any way to do so.
You should therefore probably ask: Are there integrals for which the only known ways to solve it are methods from complex analysis?
Also note that if there is a nice way to solve an integral using complex analysis, people will most likely not search for a way to solve these integrals by real (and probably very complicated) methods.
answered Dec 6 '18 at 14:40
JamesJames
1,090216
answered Dec 6 '18 at 14:40
JamesJames
1,090216
answered Dec 6 '18 at 14:40
JamesJames
1,090216
answered Dec 6 '18 at 14:40
JamesJames
1,090216
1,090216
$begingroup$
The question is if there is an integral that we know for sure that it is impossible to solve with Complex Analysis, if it was proven that it is only possile this way. I'm interested in knowing if such an integral exists
$endgroup$
– J. Dionisio
Dec 6 '18 at 22:02
$begingroup$
What I tried to explain to you is that one just cannot answer this question. How would you know that there is no way to solve it with real methods? No chance because there might be a way you just haven't thought of or a new real method not yet discovered. If you are searching for integrals for which till now no solution with real methods is known yet, you may look at the elliptic integrals.
$endgroup$
– James
Dec 7 '18 at 8:17
add a comment |
$begingroup$
The question is if there is an integral that we know for sure that it is impossible to solve with Complex Analysis, if it was proven that it is only possile this way. I'm interested in knowing if such an integral exists
$endgroup$
– J. Dionisio
Dec 6 '18 at 22:02
$begingroup$
What I tried to explain to you is that one just cannot answer this question. How would you know that there is no way to solve it with real methods? No chance because there might be a way you just haven't thought of or a new real method not yet discovered. If you are searching for integrals for which till now no solution with real methods is known yet, you may look at the elliptic integrals.
$endgroup$
– James
Dec 7 '18 at 8:17
$begingroup$
The question is if there is an integral that we know for sure that it is impossible to solve with Complex Analysis, if it was proven that it is only possile this way. I'm interested in knowing if such an integral exists
$endgroup$
– J. Dionisio
Dec 6 '18 at 22:02
$begingroup$
The question is if there is an integral that we know for sure that it is impossible to solve with Complex Analysis, if it was proven that it is only possile this way. I'm interested in knowing if such an integral exists
$endgroup$
– J. Dionisio
Dec 6 '18 at 22:02
$begingroup$
What I tried to explain to you is that one just cannot answer this question. How would you know that there is no way to solve it with real methods? No chance because there might be a way you just haven't thought of or a new real method not yet discovered. If you are searching for integrals for which till now no solution with real methods is known yet, you may look at the elliptic integrals.
$endgroup$
– James
Dec 7 '18 at 8:17
$begingroup$
What I tried to explain to you is that one just cannot answer this question. How would you know that there is no way to solve it with real methods? No chance because there might be a way you just haven't thought of or a new real method not yet discovered. If you are searching for integrals for which till now no solution with real methods is known yet, you may look at the elliptic integrals.
$endgroup$
– James
Dec 7 '18 at 8:17
add a comment |
$begingroup$
The residue theorem is one of the most powerful means of integration.
At the same time, there are strict conditions for its use, and the integration itself sometimes becomes a very difficult task (task1, task2). The second task allows us to compare the complexity of the application of the theorem on residues with the method of decomposition of a polynomial into elementary ones.
At the same time, there are methods that can compete with the theorem on deductions. First of all, integration by parameter.
answered Dec 7 '18 at 22:47
Yuri NegometyanovYuri Negometyanov
11.7k1729
$endgroup$
add a comment |
$begingroup$
The residue theorem is one of the most powerful means of integration.
At the same time, there are strict conditions for its use, and the integration itself sometimes becomes a very difficult task (task1, task2). The second task allows us to compare the complexity of the application of the theorem on residues with the method of decomposition of a polynomial into elementary ones.
At the same time, there are methods that can compete with the theorem on deductions. First of all, integration by parameter.
answered Dec 7 '18 at 22:47
Yuri NegometyanovYuri Negometyanov
11.7k1729
$endgroup$
add a comment |
$begingroup$
The residue theorem is one of the most powerful means of integration.
At the same time, there are strict conditions for its use, and the integration itself sometimes becomes a very difficult task (task1, task2). The second task allows us to compare the complexity of the application of the theorem on residues with the method of decomposition of a polynomial into elementary ones.
At the same time, there are methods that can compete with the theorem on deductions. First of all, integration by parameter.
answered Dec 7 '18 at 22:47
Yuri NegometyanovYuri Negometyanov
11.7k1729
$endgroup$
The residue theorem is one of the most powerful means of integration.
At the same time, there are strict conditions for its use, and the integration itself sometimes becomes a very difficult task (task1, task2). The second task allows us to compare the complexity of the application of the theorem on residues with the method of decomposition of a polynomial into elementary ones.
At the same time, there are methods that can compete with the theorem on deductions. First of all, integration by parameter.
answered Dec 7 '18 at 22:47
Yuri NegometyanovYuri Negometyanov
11.7k1729
answered Dec 7 '18 at 22:47
Yuri NegometyanovYuri Negometyanov
11.7k1729
answered Dec 7 '18 at 22:47
Yuri NegometyanovYuri Negometyanov
11.7k1729
answered Dec 7 '18 at 22:47
Yuri NegometyanovYuri Negometyanov
11.7k1729
11.7k1729
add a comment |
add a comment |
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$begingroup$
"So Paul puts up this tremendous damn integral he had obtained by starting out with a complex function that he knew the answer to, taking out the real part of it and leaving only the complex part. He had unwrapped it so it was only possible by contour integration!" A simple example is $$int_0^{2 pi} sec e^{i t} dt = int_0^{2 pi} frac {2 cos(cos t) cosh( sin t)} {cos(2 cos t) + cosh(2 sin t)} dt.$$
$endgroup$
– Maxim
Dec 7 '18 at 10:47
$begingroup$
@Maxim You should rather write ".. so it was as far as he knows only possible by contour integration!". As I explained, one can never be certain that there is no other way to solve it. Anyway, nice example!
$endgroup$
– James
Dec 7 '18 at 13:07
$begingroup$
Related.
$endgroup$
– user26872
Dec 9 '18 at 23:03