Integral with DiracDelta. Can Mathematica be made to solve this?How to solve this integralIntegral of DiracDelta giving an unusual answerHow can I solve this integral analytically?DiracDelta behaves incorrectly on multidimensional integralPartial derivative with Dirac Delta as initial conditionDensity of states from numerical integration and then differentiationIntegral Representation of DiracDeltaCan this integral be calculated using Mathematica?Symbolic IntegralMathematica won't evaluate this integral
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Integral with DiracDelta. Can Mathematica be made to solve this?
How to solve this integralIntegral of DiracDelta giving an unusual answerHow can I solve this integral analytically?DiracDelta behaves incorrectly on multidimensional integralPartial derivative with Dirac Delta as initial conditionDensity of states from numerical integration and then differentiationIntegral Representation of DiracDeltaCan this integral be calculated using Mathematica?Symbolic IntegralMathematica won't evaluate this integral
$begingroup$
This was an exam question.
$$
int_0^2 pi delta(sin^2(theta) -x ) ,dtheta
$$
Direct use of Integrate on it does not give the solution. Is there a trick or workaround? Here is the code I used
ClearAll[theta,x]
integrand = DiracDelta[Sin[theta]^2 - x]
Integrate[ integrand, theta, 0, 2 Pi]
Here is the key solution analytical solution
The solution uses this known relation (half way down the Wikipedia page)
Where the sum above is over all zeros of $g(x)$ in the integration interval. Mathematica does not seem to know this relation?
ps. Maple can't do it either.
calculus-and-analysis
asked 3 hours ago
NasserNasser
59.2k491209
$endgroup$
add a comment |
$begingroup$
This was an exam question.
$$
int_0^2 pi delta(sin^2(theta) -x ) ,dtheta
$$
Direct use of Integrate on it does not give the solution. Is there a trick or workaround? Here is the code I used
ClearAll[theta,x]
integrand = DiracDelta[Sin[theta]^2 - x]
Integrate[ integrand, theta, 0, 2 Pi]
Here is the key solution analytical solution
The solution uses this known relation (half way down the Wikipedia page)
Where the sum above is over all zeros of $g(x)$ in the integration interval. Mathematica does not seem to know this relation?
ps. Maple can't do it either.
calculus-and-analysis
asked 3 hours ago
NasserNasser
59.2k491209
$endgroup$
add a comment |
$begingroup$
This was an exam question.
$$
int_0^2 pi delta(sin^2(theta) -x ) ,dtheta
$$
Direct use of Integrate on it does not give the solution. Is there a trick or workaround? Here is the code I used
ClearAll[theta,x]
integrand = DiracDelta[Sin[theta]^2 - x]
Integrate[ integrand, theta, 0, 2 Pi]
Here is the key solution analytical solution
The solution uses this known relation (half way down the Wikipedia page)
Where the sum above is over all zeros of $g(x)$ in the integration interval. Mathematica does not seem to know this relation?
ps. Maple can't do it either.
calculus-and-analysis
asked 3 hours ago
NasserNasser
59.2k491209
$endgroup$
This was an exam question.
$$
int_0^2 pi delta(sin^2(theta) -x ) ,dtheta
$$
Direct use of Integrate on it does not give the solution. Is there a trick or workaround? Here is the code I used
ClearAll[theta,x]
integrand = DiracDelta[Sin[theta]^2 - x]
Integrate[ integrand, theta, 0, 2 Pi]
Here is the key solution analytical solution
The solution uses this known relation (half way down the Wikipedia page)
Where the sum above is over all zeros of $g(x)$ in the integration interval. Mathematica does not seem to know this relation?
ps. Maple can't do it either.
calculus-and-analysis
calculus-and-analysis
asked 3 hours ago
NasserNasser
59.2k491209
asked 3 hours ago
NasserNasser
59.2k491209
edited 2 hours ago
Nasser
asked 3 hours ago
NasserNasser
59.2k491209
asked 3 hours ago
NasserNasser
59.2k491209
asked 3 hours ago
NasserNasser
59.2k491209
59.2k491209
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You need to give Integrate assumptions:
Integrate[DiracDelta[Sin[θ]^2-x], θ, 0, 2 π, Assumptions -> 0<x<1]
2/Sqrt[-(-1 + x) x]
Unfortunately, Integrate is not quite smart enough to use the assumption x ∈ Reals:
Integrate[DiracDelta[Sin[θ]^2-x], θ, 0, 2 π, Assumptions -> x ∈ Reals]
Integrate[DiracDelta[-x + Sin[θ]^2], θ, 0, 2 π,
Assumptions -> x ∈ Reals]
answered 2 hours ago
Carl WollCarl Woll
77.7k3102204
$endgroup$
$begingroup$
Nice! For some reason, this does not work for exponents other than2(e.g.,Sin[θ]^3 - x). My code does not work either, presumably for the same reason (Solveis unable to determine the roots). Any idea how to work around this?
$endgroup$
– AccidentalFourierTransform
2 hours ago
2
$begingroup$
@AccidentalFourierTransform One hack is to replacexwithEulerGamma, that will produce a result forSin[θ]^3 - x, which you can then extrapolate into a generic answer.
$endgroup$
– Carl Woll
2 hours ago
add a comment |
$begingroup$
f[θ_, x_] := Sin[θ]^2 - x
Derivative[1, 0][f][θ, x] /. Solve[f[θ, x] == 0, 0 < θ < 2 π, θ, Reals]
Integrate[DiracDelta[x - θ]/Abs[%], θ, 0, 2 π] // Total
This code should also work for other functions $f$, presumably.
answered 2 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,18311042
$endgroup$
$begingroup$
You have an extra[θ, x], and yourSolvewould work better if you added the domainReals. Probably more robust than just relying onIntegrate.
$endgroup$
– Carl Woll
2 hours ago
$begingroup$
@CarlWoll Ah, yes, thank you!
$endgroup$
– AccidentalFourierTransform
2 hours ago
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
$begingroup$
You need to give Integrate assumptions:
Integrate[DiracDelta[Sin[θ]^2-x], θ, 0, 2 π, Assumptions -> 0<x<1]
2/Sqrt[-(-1 + x) x]
Unfortunately, Integrate is not quite smart enough to use the assumption x ∈ Reals:
Integrate[DiracDelta[Sin[θ]^2-x], θ, 0, 2 π, Assumptions -> x ∈ Reals]
Integrate[DiracDelta[-x + Sin[θ]^2], θ, 0, 2 π,
Assumptions -> x ∈ Reals]
answered 2 hours ago
Carl WollCarl Woll
77.7k3102204
$endgroup$
$begingroup$
Nice! For some reason, this does not work for exponents other than2(e.g.,Sin[θ]^3 - x). My code does not work either, presumably for the same reason (Solveis unable to determine the roots). Any idea how to work around this?
$endgroup$
– AccidentalFourierTransform
2 hours ago
2
$begingroup$
@AccidentalFourierTransform One hack is to replacexwithEulerGamma, that will produce a result forSin[θ]^3 - x, which you can then extrapolate into a generic answer.
$endgroup$
– Carl Woll
2 hours ago
add a comment |
$begingroup$
You need to give Integrate assumptions:
Integrate[DiracDelta[Sin[θ]^2-x], θ, 0, 2 π, Assumptions -> 0<x<1]
2/Sqrt[-(-1 + x) x]
Unfortunately, Integrate is not quite smart enough to use the assumption x ∈ Reals:
Integrate[DiracDelta[Sin[θ]^2-x], θ, 0, 2 π, Assumptions -> x ∈ Reals]
Integrate[DiracDelta[-x + Sin[θ]^2], θ, 0, 2 π,
Assumptions -> x ∈ Reals]
answered 2 hours ago
Carl WollCarl Woll
77.7k3102204
$endgroup$
$begingroup$
Nice! For some reason, this does not work for exponents other than2(e.g.,Sin[θ]^3 - x). My code does not work either, presumably for the same reason (Solveis unable to determine the roots). Any idea how to work around this?
$endgroup$
– AccidentalFourierTransform
2 hours ago
2
$begingroup$
@AccidentalFourierTransform One hack is to replacexwithEulerGamma, that will produce a result forSin[θ]^3 - x, which you can then extrapolate into a generic answer.
$endgroup$
– Carl Woll
2 hours ago
add a comment |
$begingroup$
You need to give Integrate assumptions:
Integrate[DiracDelta[Sin[θ]^2-x], θ, 0, 2 π, Assumptions -> 0<x<1]
2/Sqrt[-(-1 + x) x]
Unfortunately, Integrate is not quite smart enough to use the assumption x ∈ Reals:
Integrate[DiracDelta[Sin[θ]^2-x], θ, 0, 2 π, Assumptions -> x ∈ Reals]
Integrate[DiracDelta[-x + Sin[θ]^2], θ, 0, 2 π,
Assumptions -> x ∈ Reals]
answered 2 hours ago
Carl WollCarl Woll
77.7k3102204
$endgroup$
You need to give Integrate assumptions:
Integrate[DiracDelta[Sin[θ]^2-x], θ, 0, 2 π, Assumptions -> 0<x<1]
2/Sqrt[-(-1 + x) x]
Unfortunately, Integrate is not quite smart enough to use the assumption x ∈ Reals:
Integrate[DiracDelta[Sin[θ]^2-x], θ, 0, 2 π, Assumptions -> x ∈ Reals]
Integrate[DiracDelta[-x + Sin[θ]^2], θ, 0, 2 π,
Assumptions -> x ∈ Reals]
answered 2 hours ago
Carl WollCarl Woll
77.7k3102204
answered 2 hours ago
Carl WollCarl Woll
77.7k3102204
answered 2 hours ago
Carl WollCarl Woll
77.7k3102204
answered 2 hours ago
Carl WollCarl Woll
77.7k3102204
77.7k3102204
$begingroup$
Nice! For some reason, this does not work for exponents other than2(e.g.,Sin[θ]^3 - x). My code does not work either, presumably for the same reason (Solveis unable to determine the roots). Any idea how to work around this?
$endgroup$
– AccidentalFourierTransform
2 hours ago
2
$begingroup$
@AccidentalFourierTransform One hack is to replacexwithEulerGamma, that will produce a result forSin[θ]^3 - x, which you can then extrapolate into a generic answer.
$endgroup$
– Carl Woll
2 hours ago
add a comment |
$begingroup$
Nice! For some reason, this does not work for exponents other than2(e.g.,Sin[θ]^3 - x). My code does not work either, presumably for the same reason (Solveis unable to determine the roots). Any idea how to work around this?
$endgroup$
– AccidentalFourierTransform
2 hours ago
2
$begingroup$
@AccidentalFourierTransform One hack is to replacexwithEulerGamma, that will produce a result forSin[θ]^3 - x, which you can then extrapolate into a generic answer.
$endgroup$
– Carl Woll
2 hours ago
$begingroup$
Nice! For some reason, this does not work for exponents other than
2 (e.g., Sin[θ]^3 - x). My code does not work either, presumably for the same reason (Solve is unable to determine the roots). Any idea how to work around this?$endgroup$
– AccidentalFourierTransform
2 hours ago
$begingroup$
Nice! For some reason, this does not work for exponents other than
2 (e.g., Sin[θ]^3 - x). My code does not work either, presumably for the same reason (Solve is unable to determine the roots). Any idea how to work around this?$endgroup$
– AccidentalFourierTransform
2 hours ago
2
2
$begingroup$
@AccidentalFourierTransform One hack is to replace
x with EulerGamma, that will produce a result for Sin[θ]^3 - x, which you can then extrapolate into a generic answer.$endgroup$
– Carl Woll
2 hours ago
$begingroup$
@AccidentalFourierTransform One hack is to replace
x with EulerGamma, that will produce a result for Sin[θ]^3 - x, which you can then extrapolate into a generic answer.$endgroup$
– Carl Woll
2 hours ago
add a comment |
$begingroup$
f[θ_, x_] := Sin[θ]^2 - x
Derivative[1, 0][f][θ, x] /. Solve[f[θ, x] == 0, 0 < θ < 2 π, θ, Reals]
Integrate[DiracDelta[x - θ]/Abs[%], θ, 0, 2 π] // Total
This code should also work for other functions $f$, presumably.
answered 2 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,18311042
$endgroup$
$begingroup$
You have an extra[θ, x], and yourSolvewould work better if you added the domainReals. Probably more robust than just relying onIntegrate.
$endgroup$
– Carl Woll
2 hours ago
$begingroup$
@CarlWoll Ah, yes, thank you!
$endgroup$
– AccidentalFourierTransform
2 hours ago
add a comment |
$begingroup$
f[θ_, x_] := Sin[θ]^2 - x
Derivative[1, 0][f][θ, x] /. Solve[f[θ, x] == 0, 0 < θ < 2 π, θ, Reals]
Integrate[DiracDelta[x - θ]/Abs[%], θ, 0, 2 π] // Total
This code should also work for other functions $f$, presumably.
answered 2 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,18311042
$endgroup$
$begingroup$
You have an extra[θ, x], and yourSolvewould work better if you added the domainReals. Probably more robust than just relying onIntegrate.
$endgroup$
– Carl Woll
2 hours ago
$begingroup$
@CarlWoll Ah, yes, thank you!
$endgroup$
– AccidentalFourierTransform
2 hours ago
add a comment |
$begingroup$
f[θ_, x_] := Sin[θ]^2 - x
Derivative[1, 0][f][θ, x] /. Solve[f[θ, x] == 0, 0 < θ < 2 π, θ, Reals]
Integrate[DiracDelta[x - θ]/Abs[%], θ, 0, 2 π] // Total
This code should also work for other functions $f$, presumably.
answered 2 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,18311042
$endgroup$
f[θ_, x_] := Sin[θ]^2 - x
Derivative[1, 0][f][θ, x] /. Solve[f[θ, x] == 0, 0 < θ < 2 π, θ, Reals]
Integrate[DiracDelta[x - θ]/Abs[%], θ, 0, 2 π] // Total
This code should also work for other functions $f$, presumably.
answered 2 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,18311042
edited 2 hours ago
answered 2 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,18311042
answered 2 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,18311042
answered 2 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,18311042
5,18311042
$begingroup$
You have an extra[θ, x], and yourSolvewould work better if you added the domainReals. Probably more robust than just relying onIntegrate.
$endgroup$
– Carl Woll
2 hours ago
$begingroup$
@CarlWoll Ah, yes, thank you!
$endgroup$
– AccidentalFourierTransform
2 hours ago
add a comment |
$begingroup$
You have an extra[θ, x], and yourSolvewould work better if you added the domainReals. Probably more robust than just relying onIntegrate.
$endgroup$
– Carl Woll
2 hours ago
$begingroup$
@CarlWoll Ah, yes, thank you!
$endgroup$
– AccidentalFourierTransform
2 hours ago
$begingroup$
You have an extra
[θ, x], and your Solve would work better if you added the domain Reals. Probably more robust than just relying on Integrate.$endgroup$
– Carl Woll
2 hours ago
$begingroup$
You have an extra
[θ, x], and your Solve would work better if you added the domain Reals. Probably more robust than just relying on Integrate.$endgroup$
– Carl Woll
2 hours ago
$begingroup$
@CarlWoll Ah, yes, thank you!
$endgroup$
– AccidentalFourierTransform
2 hours ago
$begingroup$
@CarlWoll Ah, yes, thank you!
$endgroup$
– AccidentalFourierTransform
2 hours ago
add a comment |
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