Integration of Bessel function multiplied with an algebraic and trigonometric functions.
Clash Royale CLAN TAG#URR8PPP
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I tried my best to solve the following definite integration.
Struggling a lot, ended with no luck. I know similar but a bit little simplified identity.
The question is how to change this identity for my case which include two extra constants or how to solve this integral in first place.
integral-equations
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This question came from our site for users of Wolfram Mathematica.
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up vote
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down vote
favorite
I tried my best to solve the following definite integration.
Struggling a lot, ended with no luck. I know similar but a bit little simplified identity.
The question is how to change this identity for my case which include two extra constants or how to solve this integral in first place.
integral-equations
migrated from mathematica.stackexchange.com yesterday
This question came from our site for users of Wolfram Mathematica.
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I tried my best to solve the following definite integration.
Struggling a lot, ended with no luck. I know similar but a bit little simplified identity.
The question is how to change this identity for my case which include two extra constants or how to solve this integral in first place.
integral-equations
I tried my best to solve the following definite integration.
Struggling a lot, ended with no luck. I know similar but a bit little simplified identity.
The question is how to change this identity for my case which include two extra constants or how to solve this integral in first place.
integral-equations
asked yesterday
Ubaid Ullah
31
31
migrated from mathematica.stackexchange.com yesterday
This question came from our site for users of Wolfram Mathematica.
migrated from mathematica.stackexchange.com yesterday
This question came from our site for users of Wolfram Mathematica.
add a comment |Â
add a comment |Â
1 Answer
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You want to compute
$$I=int_0^X J_1( alpha t),e^-b t^2,dt$$
As you did, let $alpha t=x$ to make
$$I=frac 1 alphaint_0^ alpha X J_1( x),e^-beta x^2,dx qquad textwithqquad beta=fracbalpha^2 $$
Now, use the expansion
$$ J_1( x)=sum_m=0^infty frac(-1)^m m! ,2^2 m+1 , Gamma (m+2)x^2 m+1$$ which makes that you face now the problem of
$$K_m=int x^2 m+1,e^-beta x^2,dx=-frac Gamma left(m+1, beta x^2right) 2beta ^(m+1) ,$$ and
$$L_m=int_0^ alpha X x^2 m+1,e^-beta x^2,dx=fracGamma (m+1)-Gamma left(m+1, beta alpha ^2X^2right) 2 beta ^m+1 $$
Thanks Claude Leibovici, I will try to solve like this but if in case I found some issues I will get in touch to you.
â Ubaid Ullah
yesterday
@UbaidUllah. You are welcome ! If you think it is good, may be, you could accept my answer in order other people know that is a possible solution. Cheers.
â Claude Leibovici
yesterday
Great Claude Leibovici, Your answer solve my problem. There is a typo "The upper limit of integral is $$ alpha x$$" and the answer confuses x with X.
â Ubaid Ullah
yesterday
@UbaidUllah. Sorry for the typo !
â Claude Leibovici
yesterday
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You want to compute
$$I=int_0^X J_1( alpha t),e^-b t^2,dt$$
As you did, let $alpha t=x$ to make
$$I=frac 1 alphaint_0^ alpha X J_1( x),e^-beta x^2,dx qquad textwithqquad beta=fracbalpha^2 $$
Now, use the expansion
$$ J_1( x)=sum_m=0^infty frac(-1)^m m! ,2^2 m+1 , Gamma (m+2)x^2 m+1$$ which makes that you face now the problem of
$$K_m=int x^2 m+1,e^-beta x^2,dx=-frac Gamma left(m+1, beta x^2right) 2beta ^(m+1) ,$$ and
$$L_m=int_0^ alpha X x^2 m+1,e^-beta x^2,dx=fracGamma (m+1)-Gamma left(m+1, beta alpha ^2X^2right) 2 beta ^m+1 $$
Thanks Claude Leibovici, I will try to solve like this but if in case I found some issues I will get in touch to you.
â Ubaid Ullah
yesterday
@UbaidUllah. You are welcome ! If you think it is good, may be, you could accept my answer in order other people know that is a possible solution. Cheers.
â Claude Leibovici
yesterday
Great Claude Leibovici, Your answer solve my problem. There is a typo "The upper limit of integral is $$ alpha x$$" and the answer confuses x with X.
â Ubaid Ullah
yesterday
@UbaidUllah. Sorry for the typo !
â Claude Leibovici
yesterday
add a comment |Â
up vote
1
down vote
accepted
You want to compute
$$I=int_0^X J_1( alpha t),e^-b t^2,dt$$
As you did, let $alpha t=x$ to make
$$I=frac 1 alphaint_0^ alpha X J_1( x),e^-beta x^2,dx qquad textwithqquad beta=fracbalpha^2 $$
Now, use the expansion
$$ J_1( x)=sum_m=0^infty frac(-1)^m m! ,2^2 m+1 , Gamma (m+2)x^2 m+1$$ which makes that you face now the problem of
$$K_m=int x^2 m+1,e^-beta x^2,dx=-frac Gamma left(m+1, beta x^2right) 2beta ^(m+1) ,$$ and
$$L_m=int_0^ alpha X x^2 m+1,e^-beta x^2,dx=fracGamma (m+1)-Gamma left(m+1, beta alpha ^2X^2right) 2 beta ^m+1 $$
Thanks Claude Leibovici, I will try to solve like this but if in case I found some issues I will get in touch to you.
â Ubaid Ullah
yesterday
@UbaidUllah. You are welcome ! If you think it is good, may be, you could accept my answer in order other people know that is a possible solution. Cheers.
â Claude Leibovici
yesterday
Great Claude Leibovici, Your answer solve my problem. There is a typo "The upper limit of integral is $$ alpha x$$" and the answer confuses x with X.
â Ubaid Ullah
yesterday
@UbaidUllah. Sorry for the typo !
â Claude Leibovici
yesterday
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You want to compute
$$I=int_0^X J_1( alpha t),e^-b t^2,dt$$
As you did, let $alpha t=x$ to make
$$I=frac 1 alphaint_0^ alpha X J_1( x),e^-beta x^2,dx qquad textwithqquad beta=fracbalpha^2 $$
Now, use the expansion
$$ J_1( x)=sum_m=0^infty frac(-1)^m m! ,2^2 m+1 , Gamma (m+2)x^2 m+1$$ which makes that you face now the problem of
$$K_m=int x^2 m+1,e^-beta x^2,dx=-frac Gamma left(m+1, beta x^2right) 2beta ^(m+1) ,$$ and
$$L_m=int_0^ alpha X x^2 m+1,e^-beta x^2,dx=fracGamma (m+1)-Gamma left(m+1, beta alpha ^2X^2right) 2 beta ^m+1 $$
You want to compute
$$I=int_0^X J_1( alpha t),e^-b t^2,dt$$
As you did, let $alpha t=x$ to make
$$I=frac 1 alphaint_0^ alpha X J_1( x),e^-beta x^2,dx qquad textwithqquad beta=fracbalpha^2 $$
Now, use the expansion
$$ J_1( x)=sum_m=0^infty frac(-1)^m m! ,2^2 m+1 , Gamma (m+2)x^2 m+1$$ which makes that you face now the problem of
$$K_m=int x^2 m+1,e^-beta x^2,dx=-frac Gamma left(m+1, beta x^2right) 2beta ^(m+1) ,$$ and
$$L_m=int_0^ alpha X x^2 m+1,e^-beta x^2,dx=fracGamma (m+1)-Gamma left(m+1, beta alpha ^2X^2right) 2 beta ^m+1 $$
edited yesterday
answered yesterday
Claude Leibovici
111k1054126
111k1054126
Thanks Claude Leibovici, I will try to solve like this but if in case I found some issues I will get in touch to you.
â Ubaid Ullah
yesterday
@UbaidUllah. You are welcome ! If you think it is good, may be, you could accept my answer in order other people know that is a possible solution. Cheers.
â Claude Leibovici
yesterday
Great Claude Leibovici, Your answer solve my problem. There is a typo "The upper limit of integral is $$ alpha x$$" and the answer confuses x with X.
â Ubaid Ullah
yesterday
@UbaidUllah. Sorry for the typo !
â Claude Leibovici
yesterday
add a comment |Â
Thanks Claude Leibovici, I will try to solve like this but if in case I found some issues I will get in touch to you.
â Ubaid Ullah
yesterday
@UbaidUllah. You are welcome ! If you think it is good, may be, you could accept my answer in order other people know that is a possible solution. Cheers.
â Claude Leibovici
yesterday
Great Claude Leibovici, Your answer solve my problem. There is a typo "The upper limit of integral is $$ alpha x$$" and the answer confuses x with X.
â Ubaid Ullah
yesterday
@UbaidUllah. Sorry for the typo !
â Claude Leibovici
yesterday
Thanks Claude Leibovici, I will try to solve like this but if in case I found some issues I will get in touch to you.
â Ubaid Ullah
yesterday
Thanks Claude Leibovici, I will try to solve like this but if in case I found some issues I will get in touch to you.
â Ubaid Ullah
yesterday
@UbaidUllah. You are welcome ! If you think it is good, may be, you could accept my answer in order other people know that is a possible solution. Cheers.
â Claude Leibovici
yesterday
@UbaidUllah. You are welcome ! If you think it is good, may be, you could accept my answer in order other people know that is a possible solution. Cheers.
â Claude Leibovici
yesterday
Great Claude Leibovici, Your answer solve my problem. There is a typo "The upper limit of integral is $$ alpha x$$" and the answer confuses x with X.
â Ubaid Ullah
yesterday
Great Claude Leibovici, Your answer solve my problem. There is a typo "The upper limit of integral is $$ alpha x$$" and the answer confuses x with X.
â Ubaid Ullah
yesterday
@UbaidUllah. Sorry for the typo !
â Claude Leibovici
yesterday
@UbaidUllah. Sorry for the typo !
â Claude Leibovici
yesterday
add a comment |Â
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