is there a matlab code for gaussion integration!!!

 syms x;
 syms E;
  syms r;
  syms D;
  syms c;
syms m;

 g(x)=exp((-2./h).*(sqrt(2*m*(E-c))).*(sqrt(x*(x+r))-r*log((sqrt(r)+sqrt(x+r))-D)))

Answer 1

[John D'Errico]

Did you look on the file exchange? (Clearly not.)

https://www.mathworks.com/matlabcentral/fileexchange/8679-gaussquad

https://www.mathworks.com/matlabcentral/fileexchange/1130-quadg

They both look decent, though guassquad is purely a gauss-legendre code, gaussg a more general code for standard weight functions, though it is a guass-kronrod scheme.

In fact, even my sympoly toolbox provides gaussian quadrature rules, thus the weights and nodes for any standard weight function class. So look at the guassquadrule function.

https://www.mathworks.com/matlabcentral/fileexchange/9577-symbolic-polynomial-manipulation

So using guassquadrule from sympoly, the nodes and weights for a 4 point Gauss-Hermite rule would be:

[nodes,weights] = gaussquadrule(4,'hermite')
nodes =
         -1.65068012388579         -0.52464762327529          0.52464762327529          1.65068012388579
weights =
         0.081312835447245         0.804914090005513         0.804914090005513        0.0813128354472448

To compute the integral of (x^5 - 3*x^2 - 2)*exp(-x.^2), from -inf to inf, I might do:

fun = @(x) (x.^5 -3*x.^2 - 2);

dot(weights,fun(nodes))
ans =
          -6.2035884781693

Syms confirms that it was correct:

syms x
int((x.^5 -3*x.^2 - 2)*exp(-x^2),[-inf,inf])
ans =
-(7*pi^(1/2))/2

vpa(ans)
ans =
-6.203588478169306095543586191694

However, will you find a tool that performs gaussian integration for a function that contains symbolic parameters? No, you won't find anything for a good reason. Gaussian integration is a numerical integration procedure, not really designed to do symbolic integration.

Can you use gaussian quadrature for a symbolic function? Well, yes.

So, here a Gauss-Laguerre integral of (a*x*2 + b*x + c)*exp(-x), over the domain [0,inf].

syms a b c
fun = @(x) a*x.^2 + b*x + c;
[nodes,weights] = gaussquadrule(5,'laguerre');
vpa(dot(weights,fun(nodes)),13)
ans =
2.0*a + 1.0*b + 1.0*c

What Gaussian quadrature rule you expect to apply to this function:

g(x)=exp((-2./h).*(sqrt(2*m*(E-c))).*(sqrt(x*(x+r))-r*log((sqrt(r)+sqrt(x+r))-D)))

I have no idea, since I don't see any domain provided, nor do I see any standard weight function in there that can be extracted. So I have a funny feeling that you may be confused as to the purpose and utility of classical Gaussian quadratures.

If you look at Gaussian quadrature rules, they presume a weight function from among several standard forms, AND a domain of integration. Gauss-Legendre assumes a unit weight function, so is applicable to integration of a general function, over the interval [-1,1]. For example, suppose you wanted to compute the integral of cos(exp(x)), over the interval [0,1].

syms x
int(cos(exp(x)),[-1,1])
ans =
cosint(exp(1)) - cosint(exp(-1))

vpa(ans)
ans =
0.67038594208938451613294763492665

Luckily, the symbolic TB is smart enough to do the work, because I'm way too sleepy to think.

Gauss-Legendre presumes an interval of [-1,1], although the nodes and weights can be transformed for other intervals.

fun = @(x) cos(exp(x));
[nodes,weights] = gaussquadrule(15,'legendre');
dot(weights,fun(nodes))
ans =
         0.670385942089469

[Aalaa Abu alrob]

thanks for your attention i give the other variable value and i want to integrate from -inf to inf but i cant get any solution m=.5; E=20; r=2; c=1; h=6.6*10^-34; D=sqrt(r+1)-r*log(1+sqrt(r+1)); g(x)=exp((-2./h).*(sqrt(2*m*(E-c))).*(sqrt(x*(x+r))-r*log((sqrt(r)+sqrt(x+r))-D))) G(x)=int(g(x),[-inf,inf])

answer G(x) =

int(exp(26417569354791965812949041543118848*log((x + 2)^(1/2) + 2^(1/2) + 626122553190977/2251799813685248) - 13208784677395982906474520771559424*(x*(x + 2))^(1/2)), x, -Inf, Inf)

[John D'Errico]

This question, as I thought, has relatively little to do with Gaussian integration, except that sometimes Gaussian integration sometimes allows you to do an integral over an infinite interval.

m=.5;
E=20;
r=2;
c=1;
h=6.6e-34;
D=sqrt(r+1)-r*log(1+sqrt(r+1));
syms x

g = exp((-2./h).*(sqrt(2*m*(E-c))).*(sqrt(x*(x+r))-r*log((sqrt(r)+sqrt(x+r))-D)));

pretty(vpa(g,5))
             34                                        34
exp(2.6418 10   log(sqrt(x + 2.0) + 1.6923) - 1.3209 10   sqrt(x (x + 2.0)))

Your question is how to compute the integral of g, with limits of [-inf,inf].

Again, Gaussian integration has nothing to do with it. Basic mathematics does. The question for you should really be, does that integral exist?

If we look inside the exponential, effectively h(x) = log(g(x)), we would see that as x-->inf

limit(h,inf) approaches -inf

However, the limit of h(x) as x approaches -inf? It goes to:

limit(26417569354791965812949041543118848*log((x + 2)^(1/2) + 2^(1/2) + 626122553190977/2251799813685248) - 13208784677395982906474520771559424*(x*(x + 2))^(1/2),-inf)
ans =
- Inf + pi*13208784677395982906474520771559424i

Your integral is divergent. It has no solution.

[Aalaa Abu alrob]

thank so so much for your effort .. thats what i want

[Aalaa Abu alrob]

if i change the interval of interval.. for example from -1 to 1 do you believe that i can get answer!!

[John D'Errico]

CAN you do it then? There are still issues you need to consider.

pretty(vpa(g,5))
             34                                        34
exp(2.6418 10   log(sqrt(x + 2.0) + 1.6923) - 1.3209 10   sqrt(x (x + 2.0)))

We can rewrite this as

g(x) = exp(1.3209e34*H(x))

where H(x) is:

H = @(x) 2*log(sqrt(x+2.0)+1.6923) - sqrt(x.*(x+2.0));
ezplot(H,[-1,1])

Now, you need to recognize that nothing was plotted for negative x. Why not? Because the result is complex for negative x.

H(-.1)
ans =
       2.2438 -    0.43589i

Just as bad is the fact that this function is of the general form exp(1.3209e34*H(x)), where H(x) is on the general order of 1.

Do you have any clue how large exp(1.3209e34) is?

By way of comparison, the mass of the sun, measured in grams, is a smaller number.

So, can you compute that integral? Sigh. Does it mean anything?

[Aalaa Abu alrob]

that is a wonderful doctor .. thank so much

what about the integral of g(x) from -1 to 1 ,, how i can do it

can you give me more explanation about it ,,

Answer 2

[Torsten]

Google "Gaussian Integration & matlab"