Mixing
How to integrate this Prandtl-Meyer function
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
This is the Prandtl-Meyer function that can be found in any aerodynamics textbook. However, none of those books gave detailed steps of obtaining this integration. Could you please give me any hints on how to attack this integration?
calculus integration
asked Aug 3 at 7:58
8cold8hot
32
add a comment |Â
up vote
0
down vote
favorite
This is the Prandtl-Meyer function that can be found in any aerodynamics textbook. However, none of those books gave detailed steps of obtaining this integration. Could you please give me any hints on how to attack this integration?
calculus integration
asked Aug 3 at 7:58
8cold8hot
32
Try integrate (x^2-1)^(1/2)/x/(1+k^2 x^2) . WA gives something nicer than integrate (x^2-1)^(1/2)/x/(1+k x^2) . Recombine the logarithms. Still hidden complex numbers but better.
– Claude Leibovici
Aug 3 at 16:50
An interesting problem would be to compute $M$ for a given value of $I(M)$. Do you know if this has been addressed anywhere ?
– Claude Leibovici
Aug 4 at 5:59
I did not find the application case of computing M from a given value of I(M). It might be due to the reason that usually the data type are in the form of "I(M2)-I(M1)".
– 8cold8hot
Aug 4 at 8:44
I just asked because, from a numerical point of view, the problem is interesting. May I confess that I cannot resist an equation ? Cheers.
– Claude Leibovici
Aug 4 at 8:51
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This is the Prandtl-Meyer function that can be found in any aerodynamics textbook. However, none of those books gave detailed steps of obtaining this integration. Could you please give me any hints on how to attack this integration?
calculus integration
asked Aug 3 at 7:58
8cold8hot
32
This is the Prandtl-Meyer function that can be found in any aerodynamics textbook. However, none of those books gave detailed steps of obtaining this integration. Could you please give me any hints on how to attack this integration?
calculus integration
asked Aug 3 at 7:58
8cold8hot
32
edited Aug 4 at 8:36
asked Aug 3 at 7:58
8cold8hot
32
asked Aug 3 at 7:58
8cold8hot
32
asked Aug 3 at 7:58
8cold8hot
32
32
Try integrate (x^2-1)^(1/2)/x/(1+k^2 x^2) . WA gives something nicer than integrate (x^2-1)^(1/2)/x/(1+k x^2) . Recombine the logarithms. Still hidden complex numbers but better.
– Claude Leibovici
Aug 3 at 16:50
An interesting problem would be to compute $M$ for a given value of $I(M)$. Do you know if this has been addressed anywhere ?
– Claude Leibovici
Aug 4 at 5:59
I did not find the application case of computing M from a given value of I(M). It might be due to the reason that usually the data type are in the form of "I(M2)-I(M1)".
– 8cold8hot
Aug 4 at 8:44
I just asked because, from a numerical point of view, the problem is interesting. May I confess that I cannot resist an equation ? Cheers.
– Claude Leibovici
Aug 4 at 8:51
add a comment |Â
Try integrate (x^2-1)^(1/2)/x/(1+k^2 x^2) . WA gives something nicer than integrate (x^2-1)^(1/2)/x/(1+k x^2) . Recombine the logarithms. Still hidden complex numbers but better.
– Claude Leibovici
Aug 3 at 16:50
An interesting problem would be to compute $M$ for a given value of $I(M)$. Do you know if this has been addressed anywhere ?
– Claude Leibovici
Aug 4 at 5:59
I did not find the application case of computing M from a given value of I(M). It might be due to the reason that usually the data type are in the form of "I(M2)-I(M1)".
– 8cold8hot
Aug 4 at 8:44
I just asked because, from a numerical point of view, the problem is interesting. May I confess that I cannot resist an equation ? Cheers.
– Claude Leibovici
Aug 4 at 8:51
Try integrate (x^2-1)^(1/2)/x/(1+k^2 x^2) . WA gives something nicer than integrate (x^2-1)^(1/2)/x/(1+k x^2) . Recombine the logarithms. Still hidden complex numbers but better.
– Claude Leibovici
Aug 3 at 16:50
Try integrate (x^2-1)^(1/2)/x/(1+k^2 x^2) . WA gives something nicer than integrate (x^2-1)^(1/2)/x/(1+k x^2) . Recombine the logarithms. Still hidden complex numbers but better.
– Claude Leibovici
Aug 3 at 16:50
An interesting problem would be to compute $M$ for a given value of $I(M)$. Do you know if this has been addressed anywhere ?
– Claude Leibovici
Aug 4 at 5:59
An interesting problem would be to compute $M$ for a given value of $I(M)$. Do you know if this has been addressed anywhere ?
– Claude Leibovici
Aug 4 at 5:59
I did not find the application case of computing M from a given value of I(M). It might be due to the reason that usually the data type are in the form of "I(M2)-I(M1)".
– 8cold8hot
Aug 4 at 8:44
I did not find the application case of computing M from a given value of I(M). It might be due to the reason that usually the data type are in the form of "I(M2)-I(M1)".
– 8cold8hot
Aug 4 at 8:44
I just asked because, from a numerical point of view, the problem is interesting. May I confess that I cannot resist an equation ? Cheers.
– Claude Leibovici
Aug 4 at 8:51
I just asked because, from a numerical point of view, the problem is interesting. May I confess that I cannot resist an equation ? Cheers.
– Claude Leibovici
Aug 4 at 8:51
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
If you consider the problem of the antiderivative, that is to say, compute
$$I(M)=intfracsqrtM^2-1M left(1+k M^2right),dM$$ I am almost sure that Mathematica will give a result for it (may be messy, but a result). Try it and, please, report in the post the result you obtain.
Now, with regard to the integral, what I suspect is that could exist a problem for the evaluation at the lower bound. What I suggest is that you study the limit of $I(M)$ when $M to 1$. This could explain why you do not get the result.
Trying to compute it, let
$$u=sqrtM^2-1 implies M=sqrtu^2+1implies dM=fracusqrtu^2+1,du$$ which makes
$$I=int fracu^2left(u^2+1right) left(k u^2+(k+1)right),du$$ Now, partial fraction decomposition
$$fracu^2left(u^2+1right) left(k u^2+(k+1)right)=frack+1k u^2+(k+1)-frac1u^2+1$$ and things are becoming simple.
$$I=fracsqrtk+1sqrtk tan ^-1left(fracsqrtk
sqrtk+1uright)-tan ^-1(u)$$
Back to $M$ and simplifying the radicals,
$$I(M)=sqrtfrack+1ktan ^-1left(sqrtfrackk+1(M^2-1) right)-tan ^-1left(sqrtM^2-1right)$$ which is the final result for the definite integral since $I(1)=0$.
Make $k=frac 12 (gamma -1)$ to get the result given in the Wikipedia page.
answered Aug 3 at 9:48
Claude Leibovici
111k1054126
Thank you so much. The answer is very clear. The antiderivative is updated. I cannot figure out where the complex part comes from or eliminate it.
– 8cold8hot
Aug 3 at 13:45
@8cold8hot. You are very welcome ! As I suspected, using the antiderivative you give, $I(1)$ is indeterminate. Could try to compute it and see if I am correct ?
– Claude Leibovici
Aug 3 at 14:17
@8cold8hot. You can recombine the two logarithms making the product of their arguments; this will remove the $i$'s but you will stay with hidden complex numbers.
– Claude Leibovici
Aug 3 at 14:32
I updated the question to include the figure of I(1). If I use the FullSimply function in the Mathematica, it can give a determinate result.
– 8cold8hot
Aug 4 at 8:40
@8cold8hot. Good to know but the problem is that this limit is again a complex number.
– Claude Leibovici
Aug 4 at 8:57
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If you consider the problem of the antiderivative, that is to say, compute
$$I(M)=intfracsqrtM^2-1M left(1+k M^2right),dM$$ I am almost sure that Mathematica will give a result for it (may be messy, but a result). Try it and, please, report in the post the result you obtain.
Now, with regard to the integral, what I suspect is that could exist a problem for the evaluation at the lower bound. What I suggest is that you study the limit of $I(M)$ when $M to 1$. This could explain why you do not get the result.
Trying to compute it, let
$$u=sqrtM^2-1 implies M=sqrtu^2+1implies dM=fracusqrtu^2+1,du$$ which makes
$$I=int fracu^2left(u^2+1right) left(k u^2+(k+1)right),du$$ Now, partial fraction decomposition
$$fracu^2left(u^2+1right) left(k u^2+(k+1)right)=frack+1k u^2+(k+1)-frac1u^2+1$$ and things are becoming simple.
$$I=fracsqrtk+1sqrtk tan ^-1left(fracsqrtk
sqrtk+1uright)-tan ^-1(u)$$
Back to $M$ and simplifying the radicals,
$$I(M)=sqrtfrack+1ktan ^-1left(sqrtfrackk+1(M^2-1) right)-tan ^-1left(sqrtM^2-1right)$$ which is the final result for the definite integral since $I(1)=0$.
Make $k=frac 12 (gamma -1)$ to get the result given in the Wikipedia page.
answered Aug 3 at 9:48
Claude Leibovici
111k1054126
Thank you so much. The answer is very clear. The antiderivative is updated. I cannot figure out where the complex part comes from or eliminate it.
– 8cold8hot
Aug 3 at 13:45
@8cold8hot. You are very welcome ! As I suspected, using the antiderivative you give, $I(1)$ is indeterminate. Could try to compute it and see if I am correct ?
– Claude Leibovici
Aug 3 at 14:17
@8cold8hot. You can recombine the two logarithms making the product of their arguments; this will remove the $i$'s but you will stay with hidden complex numbers.
– Claude Leibovici
Aug 3 at 14:32
I updated the question to include the figure of I(1). If I use the FullSimply function in the Mathematica, it can give a determinate result.
– 8cold8hot
Aug 4 at 8:40
@8cold8hot. Good to know but the problem is that this limit is again a complex number.
– Claude Leibovici
Aug 4 at 8:57
add a comment |Â
up vote
2
down vote
accepted
If you consider the problem of the antiderivative, that is to say, compute
$$I(M)=intfracsqrtM^2-1M left(1+k M^2right),dM$$ I am almost sure that Mathematica will give a result for it (may be messy, but a result). Try it and, please, report in the post the result you obtain.
Now, with regard to the integral, what I suspect is that could exist a problem for the evaluation at the lower bound. What I suggest is that you study the limit of $I(M)$ when $M to 1$. This could explain why you do not get the result.
Trying to compute it, let
$$u=sqrtM^2-1 implies M=sqrtu^2+1implies dM=fracusqrtu^2+1,du$$ which makes
$$I=int fracu^2left(u^2+1right) left(k u^2+(k+1)right),du$$ Now, partial fraction decomposition
$$fracu^2left(u^2+1right) left(k u^2+(k+1)right)=frack+1k u^2+(k+1)-frac1u^2+1$$ and things are becoming simple.
$$I=fracsqrtk+1sqrtk tan ^-1left(fracsqrtk
sqrtk+1uright)-tan ^-1(u)$$
Back to $M$ and simplifying the radicals,
$$I(M)=sqrtfrack+1ktan ^-1left(sqrtfrackk+1(M^2-1) right)-tan ^-1left(sqrtM^2-1right)$$ which is the final result for the definite integral since $I(1)=0$.
Make $k=frac 12 (gamma -1)$ to get the result given in the Wikipedia page.
answered Aug 3 at 9:48
Claude Leibovici
111k1054126
Thank you so much. The answer is very clear. The antiderivative is updated. I cannot figure out where the complex part comes from or eliminate it.
– 8cold8hot
Aug 3 at 13:45
@8cold8hot. You are very welcome ! As I suspected, using the antiderivative you give, $I(1)$ is indeterminate. Could try to compute it and see if I am correct ?
– Claude Leibovici
Aug 3 at 14:17
@8cold8hot. You can recombine the two logarithms making the product of their arguments; this will remove the $i$'s but you will stay with hidden complex numbers.
– Claude Leibovici
Aug 3 at 14:32
I updated the question to include the figure of I(1). If I use the FullSimply function in the Mathematica, it can give a determinate result.
– 8cold8hot
Aug 4 at 8:40
@8cold8hot. Good to know but the problem is that this limit is again a complex number.
– Claude Leibovici
Aug 4 at 8:57
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If you consider the problem of the antiderivative, that is to say, compute
$$I(M)=intfracsqrtM^2-1M left(1+k M^2right),dM$$ I am almost sure that Mathematica will give a result for it (may be messy, but a result). Try it and, please, report in the post the result you obtain.
Now, with regard to the integral, what I suspect is that could exist a problem for the evaluation at the lower bound. What I suggest is that you study the limit of $I(M)$ when $M to 1$. This could explain why you do not get the result.
Trying to compute it, let
$$u=sqrtM^2-1 implies M=sqrtu^2+1implies dM=fracusqrtu^2+1,du$$ which makes
$$I=int fracu^2left(u^2+1right) left(k u^2+(k+1)right),du$$ Now, partial fraction decomposition
$$fracu^2left(u^2+1right) left(k u^2+(k+1)right)=frack+1k u^2+(k+1)-frac1u^2+1$$ and things are becoming simple.
$$I=fracsqrtk+1sqrtk tan ^-1left(fracsqrtk
sqrtk+1uright)-tan ^-1(u)$$
Back to $M$ and simplifying the radicals,
$$I(M)=sqrtfrack+1ktan ^-1left(sqrtfrackk+1(M^2-1) right)-tan ^-1left(sqrtM^2-1right)$$ which is the final result for the definite integral since $I(1)=0$.
Make $k=frac 12 (gamma -1)$ to get the result given in the Wikipedia page.
answered Aug 3 at 9:48
Claude Leibovici
111k1054126
If you consider the problem of the antiderivative, that is to say, compute
$$I(M)=intfracsqrtM^2-1M left(1+k M^2right),dM$$ I am almost sure that Mathematica will give a result for it (may be messy, but a result). Try it and, please, report in the post the result you obtain.
Now, with regard to the integral, what I suspect is that could exist a problem for the evaluation at the lower bound. What I suggest is that you study the limit of $I(M)$ when $M to 1$. This could explain why you do not get the result.
Trying to compute it, let
$$u=sqrtM^2-1 implies M=sqrtu^2+1implies dM=fracusqrtu^2+1,du$$ which makes
$$I=int fracu^2left(u^2+1right) left(k u^2+(k+1)right),du$$ Now, partial fraction decomposition
$$fracu^2left(u^2+1right) left(k u^2+(k+1)right)=frack+1k u^2+(k+1)-frac1u^2+1$$ and things are becoming simple.
$$I=fracsqrtk+1sqrtk tan ^-1left(fracsqrtk
sqrtk+1uright)-tan ^-1(u)$$
Back to $M$ and simplifying the radicals,
$$I(M)=sqrtfrack+1ktan ^-1left(sqrtfrackk+1(M^2-1) right)-tan ^-1left(sqrtM^2-1right)$$ which is the final result for the definite integral since $I(1)=0$.
Make $k=frac 12 (gamma -1)$ to get the result given in the Wikipedia page.
answered Aug 3 at 9:48
Claude Leibovici
111k1054126
edited Aug 3 at 12:50
answered Aug 3 at 9:48
Claude Leibovici
111k1054126
answered Aug 3 at 9:48
Claude Leibovici
111k1054126
answered Aug 3 at 9:48
Claude Leibovici
111k1054126
111k1054126
Thank you so much. The answer is very clear. The antiderivative is updated. I cannot figure out where the complex part comes from or eliminate it.
– 8cold8hot
Aug 3 at 13:45
@8cold8hot. You are very welcome ! As I suspected, using the antiderivative you give, $I(1)$ is indeterminate. Could try to compute it and see if I am correct ?
– Claude Leibovici
Aug 3 at 14:17
@8cold8hot. You can recombine the two logarithms making the product of their arguments; this will remove the $i$'s but you will stay with hidden complex numbers.
– Claude Leibovici
Aug 3 at 14:32
I updated the question to include the figure of I(1). If I use the FullSimply function in the Mathematica, it can give a determinate result.
– 8cold8hot
Aug 4 at 8:40
@8cold8hot. Good to know but the problem is that this limit is again a complex number.
– Claude Leibovici
Aug 4 at 8:57
add a comment |Â
Thank you so much. The answer is very clear. The antiderivative is updated. I cannot figure out where the complex part comes from or eliminate it.
– 8cold8hot
Aug 3 at 13:45
@8cold8hot. You are very welcome ! As I suspected, using the antiderivative you give, $I(1)$ is indeterminate. Could try to compute it and see if I am correct ?
– Claude Leibovici
Aug 3 at 14:17
@8cold8hot. You can recombine the two logarithms making the product of their arguments; this will remove the $i$'s but you will stay with hidden complex numbers.
– Claude Leibovici
Aug 3 at 14:32
I updated the question to include the figure of I(1). If I use the FullSimply function in the Mathematica, it can give a determinate result.
– 8cold8hot
Aug 4 at 8:40
@8cold8hot. Good to know but the problem is that this limit is again a complex number.
– Claude Leibovici
Aug 4 at 8:57
Thank you so much. The answer is very clear. The antiderivative is updated. I cannot figure out where the complex part comes from or eliminate it.
– 8cold8hot
Aug 3 at 13:45
Thank you so much. The answer is very clear. The antiderivative is updated. I cannot figure out where the complex part comes from or eliminate it.
– 8cold8hot
Aug 3 at 13:45
@8cold8hot. You are very welcome ! As I suspected, using the antiderivative you give, $I(1)$ is indeterminate. Could try to compute it and see if I am correct ?
– Claude Leibovici
Aug 3 at 14:17
@8cold8hot. You are very welcome ! As I suspected, using the antiderivative you give, $I(1)$ is indeterminate. Could try to compute it and see if I am correct ?
– Claude Leibovici
Aug 3 at 14:17
@8cold8hot. You can recombine the two logarithms making the product of their arguments; this will remove the $i$'s but you will stay with hidden complex numbers.
– Claude Leibovici
Aug 3 at 14:32
@8cold8hot. You can recombine the two logarithms making the product of their arguments; this will remove the $i$'s but you will stay with hidden complex numbers.
– Claude Leibovici
Aug 3 at 14:32
I updated the question to include the figure of I(1). If I use the FullSimply function in the Mathematica, it can give a determinate result.
– 8cold8hot
Aug 4 at 8:40
I updated the question to include the figure of I(1). If I use the FullSimply function in the Mathematica, it can give a determinate result.
– 8cold8hot
Aug 4 at 8:40
@8cold8hot. Good to know but the problem is that this limit is again a complex number.
– Claude Leibovici
Aug 4 at 8:57
@8cold8hot. Good to know but the problem is that this limit is again a complex number.
– Claude Leibovici
Aug 4 at 8:57
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2870850%2fhow-to-integrate-this-prandtl-meyer-function%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Try integrate (x^2-1)^(1/2)/x/(1+k^2 x^2) . WA gives something nicer than integrate (x^2-1)^(1/2)/x/(1+k x^2) . Recombine the logarithms. Still hidden complex numbers but better.
– Claude Leibovici
Aug 3 at 16:50
An interesting problem would be to compute $M$ for a given value of $I(M)$. Do you know if this has been addressed anywhere ?
– Claude Leibovici
Aug 4 at 5:59
I did not find the application case of computing M from a given value of I(M). It might be due to the reason that usually the data type are in the form of "I(M2)-I(M1)".
– 8cold8hot
Aug 4 at 8:44
I just asked because, from a numerical point of view, the problem is interesting. May I confess that I cannot resist an equation ? Cheers.
– Claude Leibovici
Aug 4 at 8:51