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I am working on fopurier analysis and below is what I have. The method I used worked fine when I had a piecewise continuous function, but now the bottom int function below is not calculating the integral.

restart;
with(plots);

f := 2*sin(Pi-Pi*exp(-t))

 

A[0] := (1/3)*(Int(f, t = 0 .. 3))

A[0] := (1/3)*(int(f, t = 0 .. 3))   I think this one is evaluating it correctly. The bold and italicized below is the answer from this calulation

A[0] := (2/3)*Si(Pi)-(2/3)*Si(exp(-3)*Pi)

 

a[n] := (2/3)*(Int(f*cos(2*Pi*n*t*(1/3)), t = 0 .. 3))  this prints the integral like I want it to.

a[n] := (2/3)*(int(f*cos(2*Pi*n*t*(1/3)), t = 0 .. 3))  this just reporints the integral, I want it to evaluate it. Am I doing something wrong? Even with the assume(n, integer); it still just prints the integral.

 

 

Is there any help that y'all can provide? My professor is not ver proficent in the Math softwares out there.

 

 While calculating an integral including complex numbers, I have encountered with the output "undefined if a+ib>0". What does this mean?A complex number bigger than zero???

Hi,

I read this thread, but it didnt really help.

test.mw (i noticed an error in the expression pj, now fixed!)

See this file above.

The function needs to be maximized is "loglik", with constraints,

f0>0, sigma>0 and mu is FREE. Only three variables.

 

Using some calculus and by changing variables, I was able to use Gauss–Hermite quadrature, and do some approximation, to get the values:

f0=105.9535

mu=-2.1587

sigma=3.5156

 

I am wondering if Maple can do this straight away. By that, I mean, without doing much further,  how could I get the values? (most efficiently and quickly)

 

Thanks,

 

casper

 

http://www.mapleprimes.com/questions/122813-Is-It-Possible-To-Optimize-This-Kind-Of-Integral

Hi all

I have a mathematical problem and I asked it in various sites but the answers till yet are not correct.

Assume that we have:

T[m]:=t->t^m:
b[n,m]:=unapply(piecewise(t>=(n-1)*tj/N and t<n*tj/N, T[m](N*t-(n-1)*tj), 0), t):

where n,N,tj are known constants. furthermore assume that we want to comute the following integral:

for following approximations:

I have written the following code but it seems to be incorrect:

V1:=Vector([seq(seq(b[n,m](t),m=0..1),n=1..3)]);
V:=evalf(V1.Transpose(V1));

the original program is :

taaylor.mws

I will be so grateful if any one can help me to solve it by maple

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

Maple evaluates

Int(ln(1+x)/(1+x^2),x=0..1);

as

1/2*I*dilog(1/2+1/2*I)-1/2*I*dilog(1/2-1/2*I)+1/4*Pi*ln(2)-Catalan

It would be nice if Maple could simplify this to 1/8*Pi*ln(2). If you evalf the integral and use "identify" then Maple does return 1/8*Pi*ln(2).

MMa directly returns 1/8*Pi*ln(2), which seems preferable.

can i compute this integral so that the answer does not have any complex part ? i just want the integral returns me real function or number .
 i also use evalf(int ... ) and evalf(Int ... ) and two different answers were given ! what is wrong ?
tnx for help in advance .

restart:

int(log(sec(x)+ tan(x)),x);

-I*ln(exp(I*x))*ln((-I*(exp(I*x))^2+I+2*exp(I*x))/((exp(I*x))^2+1))+(1/2)*Pi*ln(-I*(exp(I*x)-I))+I*dilog(-I*exp(I*x))+I*dilog(-I*(I+exp(I*x)))+I*ln(exp(I*x))*ln(-I*(I+exp(I*x)))

(1)

evalf(Int(log(sec(x)+ tan(x)),x=0..10));

3.493589070+16.61151993*I

(2)

evalf(int(log(sec(x)+ tan(x)),x=0..10));

7.035851984+23.35343547*I

(3)

 

 

Download integral.mw

This integral is defined as

Where

 

Now, How can I solve this integral?

tnx for your help.

 

 

hi,

i cant solve this 

int(1/1-exp(a*x)*erfc(a*x))

Can anyone help to compute the following integral in terms of the bessel functions in maple.

Help Please! :)
As it is seen in the picture, I can not integrate the power series. In contrast, the differentiation works!
what is wrong?

Hi,

I have a linear system to solve.

 

mm:=proc(a,x,h,i)
local A,Z1,Z2,Z,F,result;  # to declare the local variable
A:=array(1..2,1..2,[[1,1],[a,a+h]]);
Z1:=evalf(int(1/(abs(y-x)+.000000001),y=a..a+h));
Z2:=evalf(int(y/(abs(y-x)+.000000001),y=a..a+h));
Z:=array([Z1,Z2]);
F:=evalm(inverse(A)&*Z);
result:=F[i]
end:

My questions: 

1) My exact Z1 is Z1:=evalf(int(1/(abs(y-x)),y=a..a+h)); but I ask if can I put

Z1:=evalf(int(1/(abs(y-x)+.000000001),y=a..a+h));

the same for Z2.

2) Can I writte in a simple form the vector Z.  Because, later, il have a second system contains Z1,Z2, Z3, Z4,Z5.  The difference between Z1 and Z2 is the variable "y" added in the integral of Z2.

 

Many thinks.

 

Hi, My goal is to compute the coefficient beta_i, so i will solve a system and get the coefficient beta_i. But my code return an error. Any help please. Many thinks

coef_approx:=proc(a,N,i,d)
local Fredholm,eq2,eq3,Vct_basis,fct,sys,eq4,M,w,b,M1,V,Vect_beta,h,x,phi,Kernel,lambda;
# Fredholm Integral equation
Fredholm:=phi(x)=f(x)+lambda*int(Kernel(x,y)*phi(y),y=-a..a);
# stepsize
h:=a/N;
# First Approximation of integral
eq2:=int(Kernel(x,y)*phi(y),y=-a..a)=sum(int(Kernel(x,y)*phi(y),y=n*h..(n+d)*h),n=-N..N-d);
#Approximate the integral (Method used)
eq3:=phi->int(Kernel(x,y)*phi(y),y=n*h..(n+d)*h)=add(beta[i]*phi((n+i-1)*h),i=1..d+1):
eq4:=int(Kernel(x,y)*phi(y),y=n*h..(n+d)*h)=add(alpha[i](n,m)*phi((n+i-1)*h),i=1..d+1);
# Fct used to compute the coeffcient beta[i]
Vct_basis:=[seq(x^i,i=0..d+1)]:
fct:=[seq(unapply(Vct_basis[i],x),i=1..d+2)];
# system of equation must be solved
sys:=[seq(eq3(fct[i]),i=1..d+1)]:
x:='x';
x:=m*h:
w := [seq(beta[i],i=1..d+1)];
M,b := GenerateMatrix(sys,w);
M1:=-M: V:=-b:
Vect_beta:=(M1)^(-1).V:
return Vect_beta;
end proc;

Is it possible to numerically calculate  the integral

int((-12*y^2+1)*ln(abs(Zeta(x+I*y)))/(4*y^2+1)^3, [y = 0 .. infinity, x = 1/2 .. infinity])

in Maple?

The code

int((-12*y^2+1)*ln(abs(Zeta(x+I*y)))/(4*y^2+1)^3, [y = 0 .. infinity, x = 1/2 .. infinity],numeric,epsilon=0.1)

has been executed on my comp  without any output since this morning.

 

 

 

help evaluate the integral int(1/(-98*x+75*x*ln(1+2*x),x)

Hi all,

As we know that the indefinite integral in Maple is defined up to a piecewise constant.

For example,

 

Due to the indeterminate range of variable n, int returns the piecewise function.

But in the other similar case, we get a generic solution rather than a piecewise function,

 

Why not to returns the result like this,

 

Thanks for any help.

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