Bessel Integration in matlab
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I was trying to do integral in matlab with below equation. Could anyone help me please to do with Riemann sum approach in matlab? Thanks in advance!!
$$int_0^inftye^-k^2/4 J_0(k)kd k$$
integration summation matlab bessel-functions
edited Sep 7 at 14:11
mathreadler
13.9k72058
asked Sep 7 at 3:43
Adilah
132
add a comment |Â
up vote
1
down vote
favorite
I was trying to do integral in matlab with below equation. Could anyone help me please to do with Riemann sum approach in matlab? Thanks in advance!!
$$int_0^inftye^-k^2/4 J_0(k)kd k$$
integration summation matlab bessel-functions
edited Sep 7 at 14:11
mathreadler
13.9k72058
asked Sep 7 at 3:43
Adilah
132
this is not equation. did you miss something?
– Nosrati
Sep 7 at 3:53
Thanks a lot Nosrati! I removed some constant parameters.
– Adilah
Sep 7 at 4:03
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I was trying to do integral in matlab with below equation. Could anyone help me please to do with Riemann sum approach in matlab? Thanks in advance!!
$$int_0^inftye^-k^2/4 J_0(k)kd k$$
integration summation matlab bessel-functions
edited Sep 7 at 14:11
mathreadler
13.9k72058
asked Sep 7 at 3:43
Adilah
132
I was trying to do integral in matlab with below equation. Could anyone help me please to do with Riemann sum approach in matlab? Thanks in advance!!
$$int_0^inftye^-k^2/4 J_0(k)kd k$$
integration summation matlab bessel-functions
integration summation matlab bessel-functions
edited Sep 7 at 14:11
mathreadler
13.9k72058
asked Sep 7 at 3:43
Adilah
132
edited Sep 7 at 14:11
mathreadler
13.9k72058
asked Sep 7 at 3:43
Adilah
132
edited Sep 7 at 14:11
mathreadler
13.9k72058
edited Sep 7 at 14:11
mathreadler
13.9k72058
edited Sep 7 at 14:11
mathreadler
13.9k72058
13.9k72058
asked Sep 7 at 3:43
Adilah
132
asked Sep 7 at 3:43
Adilah
132
asked Sep 7 at 3:43
Adilah
132
132
this is not equation. did you miss something?
– Nosrati
Sep 7 at 3:53
Thanks a lot Nosrati! I removed some constant parameters.
– Adilah
Sep 7 at 4:03
add a comment |Â
this is not equation. did you miss something?
– Nosrati
Sep 7 at 3:53
Thanks a lot Nosrati! I removed some constant parameters.
– Adilah
Sep 7 at 4:03
this is not equation. did you miss something?
– Nosrati
Sep 7 at 3:53
this is not equation. did you miss something?
– Nosrati
Sep 7 at 3:53
Thanks a lot Nosrati! I removed some constant parameters.
– Adilah
Sep 7 at 4:03
Thanks a lot Nosrati! I removed some constant parameters.
– Adilah
Sep 7 at 4:03
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
No need for Riemann sums. By the very definition of $J_0$,
$$ int_0^+infty e^-x^2/4xJ_0(x),dx = sum_ngeq 0frac(-1)^n4^n n!^2underbraceint_0^+infty x^2n+1 e^-x^2/4,dx_2cdot 4^ncdot n!=2sum_ngeq 0frac(-1)^nn!=colorredfrac2e. $$
answered Sep 7 at 13:30
Jack D'Aurizio♦
275k32268640
Wow that was short. Can you please explain or point to reference on how one knows / derives the integral which becomes $2cdot 2^2ncdot n!$ ?
– mathreadler
Sep 7 at 14:12
1
@mathreadler: $int_0^+infty z^n e^-z,dz = n!$.
– Jack D'Aurizio♦
Sep 7 at 14:14
1
Ah, wait it's gamma function isn't it? I didn't see it first.
– mathreadler
Sep 7 at 14:17
add a comment |Â
up vote
0
down vote
There is a closed form (compact form) solution to this integration in the Table of Integrals 7E, Eq 6.631.1 as $$int_0^inftyx^mue^-alpha x^2,J_v(beta,x),dx = fracGammaleft(frac12v+frac12mu+frac12right)beta,alpha^frac12muGammaleft(v+1right)expleft(-fracbeta^28alpharight),M_frac12mu,,frac12vleft(fracbeta^24alpharight)$$
where $Gamma(.)$ is the Gamma function, and $M_mu,v(.)$ is the Whittaker function.
If you still want to do numerically
1- define the function $$f(k)=ke^-k^2/4J_0(k)$$.
2- Define the step size, e.g., dk = 0.01.
3- Initialize a variable total = 0.
4- loop over k from 0 to some upper bound with an increment size dk.
5- for each value of k add to total the value f(k).
6- At the end of the for loop, compare the result with the analytical solution to make sure your work is correct.
answered Sep 7 at 5:10
BlackMath
10018
1
Your initial integral lacks an $x$ term. The actual outcome does not involve $I_0(1/2)$ but just $e^-1$.
– Jack D'Aurizio♦
Sep 7 at 13:31
Right, I treatedkas a constant. I will edit my answer. Thanks
– BlackMath
Sep 7 at 16:33
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
No need for Riemann sums. By the very definition of $J_0$,
$$ int_0^+infty e^-x^2/4xJ_0(x),dx = sum_ngeq 0frac(-1)^n4^n n!^2underbraceint_0^+infty x^2n+1 e^-x^2/4,dx_2cdot 4^ncdot n!=2sum_ngeq 0frac(-1)^nn!=colorredfrac2e. $$
answered Sep 7 at 13:30
Jack D'Aurizio♦
275k32268640
Wow that was short. Can you please explain or point to reference on how one knows / derives the integral which becomes $2cdot 2^2ncdot n!$ ?
– mathreadler
Sep 7 at 14:12
1
@mathreadler: $int_0^+infty z^n e^-z,dz = n!$.
– Jack D'Aurizio♦
Sep 7 at 14:14
1
Ah, wait it's gamma function isn't it? I didn't see it first.
– mathreadler
Sep 7 at 14:17
add a comment |Â
up vote
3
down vote
No need for Riemann sums. By the very definition of $J_0$,
$$ int_0^+infty e^-x^2/4xJ_0(x),dx = sum_ngeq 0frac(-1)^n4^n n!^2underbraceint_0^+infty x^2n+1 e^-x^2/4,dx_2cdot 4^ncdot n!=2sum_ngeq 0frac(-1)^nn!=colorredfrac2e. $$
answered Sep 7 at 13:30
Jack D'Aurizio♦
275k32268640
Wow that was short. Can you please explain or point to reference on how one knows / derives the integral which becomes $2cdot 2^2ncdot n!$ ?
– mathreadler
Sep 7 at 14:12
1
@mathreadler: $int_0^+infty z^n e^-z,dz = n!$.
– Jack D'Aurizio♦
Sep 7 at 14:14
1
Ah, wait it's gamma function isn't it? I didn't see it first.
– mathreadler
Sep 7 at 14:17
add a comment |Â
up vote
3
down vote
up vote
3
down vote
No need for Riemann sums. By the very definition of $J_0$,
$$ int_0^+infty e^-x^2/4xJ_0(x),dx = sum_ngeq 0frac(-1)^n4^n n!^2underbraceint_0^+infty x^2n+1 e^-x^2/4,dx_2cdot 4^ncdot n!=2sum_ngeq 0frac(-1)^nn!=colorredfrac2e. $$
answered Sep 7 at 13:30
Jack D'Aurizio♦
275k32268640
No need for Riemann sums. By the very definition of $J_0$,
$$ int_0^+infty e^-x^2/4xJ_0(x),dx = sum_ngeq 0frac(-1)^n4^n n!^2underbraceint_0^+infty x^2n+1 e^-x^2/4,dx_2cdot 4^ncdot n!=2sum_ngeq 0frac(-1)^nn!=colorredfrac2e. $$
answered Sep 7 at 13:30
Jack D'Aurizio♦
275k32268640
answered Sep 7 at 13:30
Jack D'Aurizio♦
275k32268640
answered Sep 7 at 13:30
Jack D'Aurizio♦
275k32268640
answered Sep 7 at 13:30
Jack D'Aurizio♦
275k32268640
275k32268640
Wow that was short. Can you please explain or point to reference on how one knows / derives the integral which becomes $2cdot 2^2ncdot n!$ ?
– mathreadler
Sep 7 at 14:12
1
@mathreadler: $int_0^+infty z^n e^-z,dz = n!$.
– Jack D'Aurizio♦
Sep 7 at 14:14
1
Ah, wait it's gamma function isn't it? I didn't see it first.
– mathreadler
Sep 7 at 14:17
add a comment |Â
Wow that was short. Can you please explain or point to reference on how one knows / derives the integral which becomes $2cdot 2^2ncdot n!$ ?
– mathreadler
Sep 7 at 14:12
1
@mathreadler: $int_0^+infty z^n e^-z,dz = n!$.
– Jack D'Aurizio♦
Sep 7 at 14:14
1
Ah, wait it's gamma function isn't it? I didn't see it first.
– mathreadler
Sep 7 at 14:17
Wow that was short. Can you please explain or point to reference on how one knows / derives the integral which becomes $2cdot 2^2ncdot n!$ ?
– mathreadler
Sep 7 at 14:12
Wow that was short. Can you please explain or point to reference on how one knows / derives the integral which becomes $2cdot 2^2ncdot n!$ ?
– mathreadler
Sep 7 at 14:12
1
1
@mathreadler: $int_0^+infty z^n e^-z,dz = n!$.
– Jack D'Aurizio♦
Sep 7 at 14:14
@mathreadler: $int_0^+infty z^n e^-z,dz = n!$.
– Jack D'Aurizio♦
Sep 7 at 14:14
1
1
Ah, wait it's gamma function isn't it? I didn't see it first.
– mathreadler
Sep 7 at 14:17
Ah, wait it's gamma function isn't it? I didn't see it first.
– mathreadler
Sep 7 at 14:17
add a comment |Â
up vote
0
down vote
There is a closed form (compact form) solution to this integration in the Table of Integrals 7E, Eq 6.631.1 as $$int_0^inftyx^mue^-alpha x^2,J_v(beta,x),dx = fracGammaleft(frac12v+frac12mu+frac12right)beta,alpha^frac12muGammaleft(v+1right)expleft(-fracbeta^28alpharight),M_frac12mu,,frac12vleft(fracbeta^24alpharight)$$
where $Gamma(.)$ is the Gamma function, and $M_mu,v(.)$ is the Whittaker function.
If you still want to do numerically
1- define the function $$f(k)=ke^-k^2/4J_0(k)$$.
2- Define the step size, e.g., dk = 0.01.
3- Initialize a variable total = 0.
4- loop over k from 0 to some upper bound with an increment size dk.
5- for each value of k add to total the value f(k).
6- At the end of the for loop, compare the result with the analytical solution to make sure your work is correct.
answered Sep 7 at 5:10
BlackMath
10018
1
Your initial integral lacks an $x$ term. The actual outcome does not involve $I_0(1/2)$ but just $e^-1$.
– Jack D'Aurizio♦
Sep 7 at 13:31
Right, I treatedkas a constant. I will edit my answer. Thanks
– BlackMath
Sep 7 at 16:33
add a comment |Â
up vote
0
down vote
There is a closed form (compact form) solution to this integration in the Table of Integrals 7E, Eq 6.631.1 as $$int_0^inftyx^mue^-alpha x^2,J_v(beta,x),dx = fracGammaleft(frac12v+frac12mu+frac12right)beta,alpha^frac12muGammaleft(v+1right)expleft(-fracbeta^28alpharight),M_frac12mu,,frac12vleft(fracbeta^24alpharight)$$
where $Gamma(.)$ is the Gamma function, and $M_mu,v(.)$ is the Whittaker function.
If you still want to do numerically
1- define the function $$f(k)=ke^-k^2/4J_0(k)$$.
2- Define the step size, e.g., dk = 0.01.
3- Initialize a variable total = 0.
4- loop over k from 0 to some upper bound with an increment size dk.
5- for each value of k add to total the value f(k).
6- At the end of the for loop, compare the result with the analytical solution to make sure your work is correct.
answered Sep 7 at 5:10
BlackMath
10018
1
Your initial integral lacks an $x$ term. The actual outcome does not involve $I_0(1/2)$ but just $e^-1$.
– Jack D'Aurizio♦
Sep 7 at 13:31
Right, I treatedkas a constant. I will edit my answer. Thanks
– BlackMath
Sep 7 at 16:33
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There is a closed form (compact form) solution to this integration in the Table of Integrals 7E, Eq 6.631.1 as $$int_0^inftyx^mue^-alpha x^2,J_v(beta,x),dx = fracGammaleft(frac12v+frac12mu+frac12right)beta,alpha^frac12muGammaleft(v+1right)expleft(-fracbeta^28alpharight),M_frac12mu,,frac12vleft(fracbeta^24alpharight)$$
where $Gamma(.)$ is the Gamma function, and $M_mu,v(.)$ is the Whittaker function.
If you still want to do numerically
1- define the function $$f(k)=ke^-k^2/4J_0(k)$$.
2- Define the step size, e.g., dk = 0.01.
3- Initialize a variable total = 0.
4- loop over k from 0 to some upper bound with an increment size dk.
5- for each value of k add to total the value f(k).
6- At the end of the for loop, compare the result with the analytical solution to make sure your work is correct.
answered Sep 7 at 5:10
BlackMath
10018
There is a closed form (compact form) solution to this integration in the Table of Integrals 7E, Eq 6.631.1 as $$int_0^inftyx^mue^-alpha x^2,J_v(beta,x),dx = fracGammaleft(frac12v+frac12mu+frac12right)beta,alpha^frac12muGammaleft(v+1right)expleft(-fracbeta^28alpharight),M_frac12mu,,frac12vleft(fracbeta^24alpharight)$$
where $Gamma(.)$ is the Gamma function, and $M_mu,v(.)$ is the Whittaker function.
If you still want to do numerically
1- define the function $$f(k)=ke^-k^2/4J_0(k)$$.
2- Define the step size, e.g., dk = 0.01.
3- Initialize a variable total = 0.
4- loop over k from 0 to some upper bound with an increment size dk.
5- for each value of k add to total the value f(k).
6- At the end of the for loop, compare the result with the analytical solution to make sure your work is correct.
answered Sep 7 at 5:10
BlackMath
10018
edited Sep 7 at 16:45
answered Sep 7 at 5:10
BlackMath
10018
answered Sep 7 at 5:10
BlackMath
10018
answered Sep 7 at 5:10
BlackMath
10018
10018
1
Your initial integral lacks an $x$ term. The actual outcome does not involve $I_0(1/2)$ but just $e^-1$.
– Jack D'Aurizio♦
Sep 7 at 13:31
Right, I treatedkas a constant. I will edit my answer. Thanks
– BlackMath
Sep 7 at 16:33
add a comment |Â
1
Your initial integral lacks an $x$ term. The actual outcome does not involve $I_0(1/2)$ but just $e^-1$.
– Jack D'Aurizio♦
Sep 7 at 13:31
Right, I treatedkas a constant. I will edit my answer. Thanks
– BlackMath
Sep 7 at 16:33
1
1
Your initial integral lacks an $x$ term. The actual outcome does not involve $I_0(1/2)$ but just $e^-1$.
– Jack D'Aurizio♦
Sep 7 at 13:31
Your initial integral lacks an $x$ term. The actual outcome does not involve $I_0(1/2)$ but just $e^-1$.
– Jack D'Aurizio♦
Sep 7 at 13:31
Right, I treated
k as a constant. I will edit my answer. Thanks– BlackMath
Sep 7 at 16:33
Right, I treated
k as a constant. I will edit my answer. Thanks– BlackMath
Sep 7 at 16:33
add a comment |Â
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this is not equation. did you miss something?
– Nosrati
Sep 7 at 3:53
Thanks a lot Nosrati! I removed some constant parameters.
– Adilah
Sep 7 at 4:03