MaplePrimes - answers and comments on Question, Float problem for Integral expression

beta(beta) problem (not critical though)

Preben Alsholm — Tue, 12 Feb 2013 21:11:04 Z

In your infinity version there appear factors beta(beta), they should just be beta. See comment at the end, though.
Specifically, you have

1/c1 = 6.274557048*10^8*(Int((32.*sin(0.5000000000e-1*beta)/beta+(960.*(cos(0.5000000000e-1*beta)-40.*sin(0.5000000000e-1*beta)/beta+1600.*sin(0.2500000000e-1*beta)^2/beta^2))/beta^2)^2/((75*(1.50*tanh(beta)*beta(beta)+1))/(75+0.2e-1*tanh(beta)*beta(beta))+2.32*coth(beta)*beta(beta)), beta = 0 .. infinity));
#You see the appearance of beta(beta).
#Let A be the integral (using Int):

A:=Int((32.*sin(0.5000000000e-1*beta)/beta+(960.*(cos(0.5000000000e-1*beta)-40.*sin(0.5000000000e-1*beta)/beta+1600.*sin(0.2500000000e-1*beta)^2/beta^2))/beta^2)^2/((75*(1.50*tanh(beta)*beta(beta)+1))/(75+0.2e-1*tanh(beta)*beta(beta))+2.32*coth(beta)*beta(beta)), beta = 0 .. infinity);
#Now do
subs(beta(beta)=beta,A);
B:=normal(%);
#This integral can be evaluated, but it is not all that easy. The problem is not at beta = 0.
f:=IntegrationTools:-GetIntegrand(B);
asympt(f,beta, 9);
series(f,beta=0, 9);
plot(f,beta=0..100);
evalf(Int(f,beta=0..1000));
evalf(Int(f,beta=0..5000));

More work could be done on this, of course. You could try something like:

evalf[15](Int(f,beta=0..10000,method = _d01akc));
#and
evalf[15](Int(f,beta=0..200000,method = _d01akc,epsilon=1e-10));
#for large values of beta f is bounded from above by
g:=eval(f,{sin=1,cos=1,tanh=1,coth=1,-38400.=38400});
#so you can get an upper bound for the error by computing:
evalf[15](Int(g,beta=200000..infinity));

Comment on beta(beta):

Try the following
restart;
evalf(Int(beta(beta),beta=0..1));
evalf(Int(beta,beta=0..1));
int(beta(beta),beta=0..1);
#The reason the numerical integration works is that in Maple the result of
5(5);
#is 5, as is
5(x);

Function beta can be any value....but that was

mtj4u — Sun, 17 Feb 2013 13:27:45 Z

Function beta can be any value....but that was (0.787*beta)

suggestion for clarity

Axel Vogt — Sun, 17 Feb 2013 16:13:17 Z

A suggestion for clarity: do not use the same name beta for different things.

If you mean a function then name it p, for example, like p(0.787*beta)

Then it makes it clear, that you have to tell us what p stands for,
it does not make much sense to say 'can be any value'

when I put limit range 0-infinity it gives me Float (∞) error

mtj4u — Thu, 21 Feb 2013 10:18:17 Z

Dear Preben Alsholm / Axel Vogt,

Thanks for ur reply and suggestions, Now take a look in attached file For_prime.mw where I have put each in everything in detial, I think this time it you will not create ambiguity

Thanks

Answer to latest version

Preben Alsholm — Thu, 21 Feb 2013 22:07:49 Z

In your uploaded file For_prime.mw you provided an image. It would have spared me some time if you had typed in the integral.
However, I typed it.
I have uploaded a worksheet:

MaplePrimes13-02-16I.mw

Thanks Preben Alsholm! I have tried

mtj4u — Thu, 14 Feb 2013 13:21:21 Z

Thanks Preben Alsholm!

I have tried same as you described and found it very helpful, in first part I calculated for certain range, like 0-200000, then as you mentioned

More work could be done on this, of course. You could try something like:

evalf[15](Int(f,beta=0..10000,method = _d01akc));
#and
evalf[15](Int(f,beta=0..200000,method = _d01akc,epsilon=1e-10));
#for large values of beta f is bounded from above by
g:=eval(f,{sin=1,cos=1,tanh=1,coth=1,-38400.=38400});
#so you can get an upper bound for the error by computing:
evalf[15](Int(g,beta=200000..infinity));

from this, I also calculatee for 200000 to infinity range, but can u plz tell

to get for the whole range i.e 0-infinity, how could I do this, should I add both ranges value?
another thing when I changed some input it give some error so how could I fix it? To_check.mw

Thanks