A problem when integrate Cos[n*x]*Cos[k*x]
$begingroup$
When integrate the indefinite integral Cos[nx]Cos[kx] about x, where both k and n are positive integer, the result is Pi when n equals to k and 0 when n is unequal to k. However, the code
sol = Integrate[Cos[n*x]*Cos[k*x], {x, -Pi, Pi},
Assumptions -> n ∈ Integers && k ∈ Integers && n > 0 && k > 0]
gives the result (k Sin[π k + π n] - n Sin[π k + π n] + .
k Sin[π k - π n] + n Sin[π k - π n])/(k^2 - n^2)
And then use the Simplify function,
Simplify[sol, Assumptions -> n ∈ Integers && k ∈ Integers && n > 0 && k > 0]
gives the result 0. Shouldn't that Integrate returns a Piecewise function like Piecewise[{{Pi, n == k}, {0, n != k}}] instead?
calculus-and-analysis
edited 2 hours ago
Mr.Wizard♦
231k294751042
asked 4 hours ago
shelure21shelure21
184
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shelure21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
When integrate the indefinite integral Cos[nx]Cos[kx] about x, where both k and n are positive integer, the result is Pi when n equals to k and 0 when n is unequal to k. However, the code
sol = Integrate[Cos[n*x]*Cos[k*x], {x, -Pi, Pi},
Assumptions -> n ∈ Integers && k ∈ Integers && n > 0 && k > 0]
gives the result (k Sin[π k + π n] - n Sin[π k + π n] + .
k Sin[π k - π n] + n Sin[π k - π n])/(k^2 - n^2)
And then use the Simplify function,
Simplify[sol, Assumptions -> n ∈ Integers && k ∈ Integers && n > 0 && k > 0]
gives the result 0. Shouldn't that Integrate returns a Piecewise function like Piecewise[{{Pi, n == k}, {0, n != k}}] instead?
calculus-and-analysis
edited 2 hours ago
Mr.Wizard♦
231k294751042
asked 4 hours ago
shelure21shelure21
184
New contributor
shelure21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
When integrate the indefinite integral Cos[nx]Cos[kx] about x, where both k and n are positive integer, the result is Pi when n equals to k and 0 when n is unequal to k. However, the code
sol = Integrate[Cos[n*x]*Cos[k*x], {x, -Pi, Pi},
Assumptions -> n ∈ Integers && k ∈ Integers && n > 0 && k > 0]
gives the result (k Sin[π k + π n] - n Sin[π k + π n] + .
k Sin[π k - π n] + n Sin[π k - π n])/(k^2 - n^2)
And then use the Simplify function,
Simplify[sol, Assumptions -> n ∈ Integers && k ∈ Integers && n > 0 && k > 0]
gives the result 0. Shouldn't that Integrate returns a Piecewise function like Piecewise[{{Pi, n == k}, {0, n != k}}] instead?
calculus-and-analysis
edited 2 hours ago
Mr.Wizard♦
231k294751042
asked 4 hours ago
shelure21shelure21
184
New contributor
shelure21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
When integrate the indefinite integral Cos[nx]Cos[kx] about x, where both k and n are positive integer, the result is Pi when n equals to k and 0 when n is unequal to k. However, the code
sol = Integrate[Cos[n*x]*Cos[k*x], {x, -Pi, Pi},
Assumptions -> n ∈ Integers && k ∈ Integers && n > 0 && k > 0]
gives the result (k Sin[π k + π n] - n Sin[π k + π n] + .
k Sin[π k - π n] + n Sin[π k - π n])/(k^2 - n^2)
And then use the Simplify function,
Simplify[sol, Assumptions -> n ∈ Integers && k ∈ Integers && n > 0 && k > 0]
gives the result 0. Shouldn't that Integrate returns a Piecewise function like Piecewise[{{Pi, n == k}, {0, n != k}}] instead?
calculus-and-analysis
calculus-and-analysis
edited 2 hours ago
Mr.Wizard♦
231k294751042
asked 4 hours ago
shelure21shelure21
184
New contributor
shelure21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Mr.Wizard♦
231k294751042
asked 4 hours ago
shelure21shelure21
184
New contributor
shelure21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Mr.Wizard♦
231k294751042
edited 2 hours ago
Mr.Wizard♦
231k294751042
edited 2 hours ago
Mr.Wizard♦
231k294751042
231k294751042
asked 4 hours ago
shelure21shelure21
184
New contributor
shelure21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
shelure21shelure21
184
asked 4 hours ago
shelure21shelure21
184
184
New contributor
shelure21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
shelure21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is well know issue. One way to handle it is
Simplify[ sol,
Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k != n]
(* 0 *)
And
Simplify[ Limit[sol, k -> n],
Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k == n ]
(* Pi *)
See
should-integrate-detect-orthogonality-of-functions-in-the-integrand
And
What assumptions to use to check for orthogonality
And
should-integrate-have-given-zero-for-this-integral
And
proper-way-to-simplify-integral-result-in-mathematica-given-integer-constraints
And
usage-of-assuming-for-integration
answered 3 hours ago
NasserNasser
57.7k488205
$endgroup$
$begingroup$
You can shorten theLimittoLimit[sol, k -> n, Assumptions -> Element[n, Integers]]
$endgroup$
– Bob Hanlon
32 mins ago
$begingroup$
@BobHanlon thanks. I am sure you are right. I was only copying what the OP had in there. But good point.
$endgroup$
– Nasser
17 mins ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is well know issue. One way to handle it is
Simplify[ sol,
Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k != n]
(* 0 *)
And
Simplify[ Limit[sol, k -> n],
Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k == n ]
(* Pi *)
See
should-integrate-detect-orthogonality-of-functions-in-the-integrand
And
What assumptions to use to check for orthogonality
And
should-integrate-have-given-zero-for-this-integral
And
proper-way-to-simplify-integral-result-in-mathematica-given-integer-constraints
And
usage-of-assuming-for-integration
answered 3 hours ago
NasserNasser
57.7k488205
$endgroup$
$begingroup$
You can shorten theLimittoLimit[sol, k -> n, Assumptions -> Element[n, Integers]]
$endgroup$
– Bob Hanlon
32 mins ago
$begingroup$
@BobHanlon thanks. I am sure you are right. I was only copying what the OP had in there. But good point.
$endgroup$
– Nasser
17 mins ago
add a comment |
$begingroup$
This is well know issue. One way to handle it is
Simplify[ sol,
Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k != n]
(* 0 *)
And
Simplify[ Limit[sol, k -> n],
Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k == n ]
(* Pi *)
See
should-integrate-detect-orthogonality-of-functions-in-the-integrand
And
What assumptions to use to check for orthogonality
And
should-integrate-have-given-zero-for-this-integral
And
proper-way-to-simplify-integral-result-in-mathematica-given-integer-constraints
And
usage-of-assuming-for-integration
answered 3 hours ago
NasserNasser
57.7k488205
$endgroup$
$begingroup$
You can shorten theLimittoLimit[sol, k -> n, Assumptions -> Element[n, Integers]]
$endgroup$
– Bob Hanlon
32 mins ago
$begingroup$
@BobHanlon thanks. I am sure you are right. I was only copying what the OP had in there. But good point.
$endgroup$
– Nasser
17 mins ago
add a comment |
$begingroup$
This is well know issue. One way to handle it is
Simplify[ sol,
Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k != n]
(* 0 *)
And
Simplify[ Limit[sol, k -> n],
Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k == n ]
(* Pi *)
See
should-integrate-detect-orthogonality-of-functions-in-the-integrand
And
What assumptions to use to check for orthogonality
And
should-integrate-have-given-zero-for-this-integral
And
proper-way-to-simplify-integral-result-in-mathematica-given-integer-constraints
And
usage-of-assuming-for-integration
answered 3 hours ago
NasserNasser
57.7k488205
$endgroup$
This is well know issue. One way to handle it is
Simplify[ sol,
Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k != n]
(* 0 *)
And
Simplify[ Limit[sol, k -> n],
Assumptions -> Element[n, Integers] && Element[k, Integers] && n > 0 && k > 0 && k == n ]
(* Pi *)
See
should-integrate-detect-orthogonality-of-functions-in-the-integrand
And
What assumptions to use to check for orthogonality
And
should-integrate-have-given-zero-for-this-integral
And
proper-way-to-simplify-integral-result-in-mathematica-given-integer-constraints
And
usage-of-assuming-for-integration
answered 3 hours ago
NasserNasser
57.7k488205
answered 3 hours ago
NasserNasser
57.7k488205
answered 3 hours ago
NasserNasser
57.7k488205
answered 3 hours ago
NasserNasser
57.7k488205
57.7k488205
$begingroup$
You can shorten theLimittoLimit[sol, k -> n, Assumptions -> Element[n, Integers]]
$endgroup$
– Bob Hanlon
32 mins ago
$begingroup$
@BobHanlon thanks. I am sure you are right. I was only copying what the OP had in there. But good point.
$endgroup$
– Nasser
17 mins ago
add a comment |
$begingroup$
You can shorten theLimittoLimit[sol, k -> n, Assumptions -> Element[n, Integers]]
$endgroup$
– Bob Hanlon
32 mins ago
$begingroup$
@BobHanlon thanks. I am sure you are right. I was only copying what the OP had in there. But good point.
$endgroup$
– Nasser
17 mins ago
$begingroup$
You can shorten the
Limit to Limit[sol, k -> n, Assumptions -> Element[n, Integers]]$endgroup$
– Bob Hanlon
32 mins ago
$begingroup$
You can shorten the
Limit to Limit[sol, k -> n, Assumptions -> Element[n, Integers]]$endgroup$
– Bob Hanlon
32 mins ago
$begingroup$
@BobHanlon thanks. I am sure you are right. I was only copying what the OP had in there. But good point.
$endgroup$
– Nasser
17 mins ago
$begingroup$
@BobHanlon thanks. I am sure you are right. I was only copying what the OP had in there. But good point.
$endgroup$
– Nasser
17 mins ago
add a comment |
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