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 Hello every one,how do i integrate from this expression?



ode := (1+B*T(x))*(diff(T(x), x, x))-M^2*T(x)^(n+1)+B*(diff(T(x), x))^2;

(1+B*T(x))*(diff(diff(T(x), x), x))-M^2*T(x)^(n+1)+B*(diff(T(x), x))^2








i have two problem in maple file, that is attached..

one of them is RootOf...note that i suppose that [varepsilon := -2.3650203724313] for i can going on following calculation

and second is  Float(undefined) in calculation integral

please help me



Hello Everyone


I have an expression which I would like to integrate from x=0 to x=L. The expression is 



Here, beta, m, alpha are constant. However I want the result in terms of these quantities.


I will be grateful if you could help mw in this regard.

Thanks a lot.



I want to write the functional Z of J Z = exp(Int(Int(J(x)*Delta(x-y)*J(y), x), y))with Delta(x) = Int(I*exp(-I*k*x)*(1/(k^2-m^2)), k) in terms of the fourier transform of J: J(x) = Int(J(p)*exp(-I*p*x), p).

Actually I'm in Minkowski space and all the integrals should be over 4 dimensions, x,y,k,p should all be four-vectors, but I wanted to keep things short. (The only way I have found to express a 4D integral is using Physics-Intc with the singleparameters of the four vector. Is there a more convenient way to get d^4x?) But still in 1D I cannot solve it.

To find the solution, an exponential of only one integral, is actually pretty easy, since there are integrals over e. g. exp(-I*x*(p-k)) deliver a delta distribution, but I cannot reproduce this in Maple since he doesn't perform the integral over x.

I have found that I can/have to use the command inttrans-fourier to gain the delta distribution, but when I try to use it for the problem mentioned above I run into all kinds of problems. Not to mention that I cannot manage to perform a fourier transformation in 4D.

Does anybody know how to solve this problem? Thanks!

While performing integration of some expessions I bumped into a strange problem. My expession consists of quite a lot of terms, but here I present the susbset of only 2. If I integrate it as a whole maple does not want to solve it, and leaves it as there was no closed expression to this integral. But if I split this sum and integrate the parts, it all works fine. What is happening here? What do I miss? I used simplify and allvalues but didn't change a thing...

Is there a way to split my terms into list, integrate one by one and they recreate the solution by summing the parts? Its a bit of a workaround, but surely better that doing it manually (I have around 50 terms). I use Maple 2015



I am trying to use Maple to discretize a model by solving an integral analytically. In my textbook the problem is solved by using Maple and is according to them "These calculations are quite straightforward to compute with Maple".

In my textbook there is an example where the non-linear system with state vector x(t) = [x1(t), x2(t), v(t), h(t), w(t)]' is described as

dx/dt=[v*cos(h), v*sin(h), 0, w, 0]'=a(x(t)).

To discretize this model the sampling formula x(t+T) = x(t) + int(a(x(tau)), tau=t..t+T) is used

Firstly, we can integrate the expressions row by row, the first row would then be

x1(t+T)=x1(t)+int(v(tau)*cos(h(tau)), tau=t..t+T).

The result when the sampling formula is applied to the first row in the example is


I have tried using int() and dsolve() but have only managed to get results that involve integrals. I would appreciate if someone more experienced could help me with this and maybe shed some light on if it actually is possible to do this?



Dear all


If its possible in  Maple to change the integral of the sum to  the sum of integrals when I calucle the integral of a function series


Thank you

Hi all

If I have an expression:

B := int(A(x), x = -infinity .. infinity)

How can I use a statement to extract A(x) when A(x) is quite complecated during calculation. thanks


int(floor(x^2), x = 0 .. 2) = 6 , but it should be 1.85374.




Mariusz Iwaniuk


For the paste several years, I have occasionally tested Maple on a simple integral,

simplify(int( exp(-I*k*x)/cosh(x), x=-infinity..infinity) assuming k,real);

The integral is returned unevaluated. There is no problems related to the complex exponential,

since also e.g.

simplify(int( cos(k*x)/cosh(x), x=0..infinity) assuming k,real);

and other variants are not evaluated.

Turning to the inbuilt fourier function (inttrans package) also does not work.

Evaluating laplace(1/cosh(x),x,s), and subsequently substituting s=I*k and s=-I*k and summing WORKS,

but gives (unnecessarily) an ugly result which it cannot simplify.

Finally, subtracting the (manually simplified) correct expression and simplifying gives 0 as

verification. I can not see any technical explanation for this bug.

Hello everyone,

I am trying to solve numerically int( f(t,z) , t=0..T ) = 0 , in z for a cumbersome f.

I tried z1=fsolve( int( f(t,z) , t=0..T ) = 0 , z). But then I tried int( f(t,z1) , t=0..T ) and the result is clearly not zero nor anything small.

It looks like Maple evaluates analytically the integral, and does it wrong (check this for more details) so fsolve uses the wrong equations.

Anyone knows how I can force Maple to evaluate numerically the integral at each step of the fsolve function?

Thank you!

hi,i am studying the maple most recent.But when calculating function integral,I ran into trouble.I hope to get your help.Here is the code I wrote, but it runs a very long time. How to effectively reduce the integration time?

Fourierf := proc (sigma, a, b, N) local A, A0, B, T, S, Ff; T := b-a; A0 := int(sigma, t = a .. b); A := int(sigma*sin(n*Pi*t/T), t = a .. b); B := int(sigma*cos(n*Pi*t/T), t = a .. b); S := sum(A*sin(n*Pi*t/T)+B*cos(n*Pi*t/T), n = 1 .. N)+(1/2)*A0; Ff := unapply(S, t) end proc;

f := proc (t) options operator, arrow; piecewise(t < .13*2.6 and 0 <= t, 100*t/(.13*2.6), .13*2.6 <= t and t < 2.6, 100, 2.6 <= t and t < 2.6*1.1, 0) end proc;

sigma := f(t);
a := 0;
b := 1.1*2.6;
s1 := unapply((Fourierf(sigma, a, b, 500))(t)/uw0, t);

s2 := unapply((Fourierf(sigma, a, b, 500))(t)/ua0, t);
A1 := (2*n+1)^2*Pi^2*(C3+1+sqrt(4*C1*C2*C3+C3^2-2*C3+1))/(8*C1*C2-8);
A2 := (2*n+1)^2*Pi^2*(C3+1-sqrt(4*C1*C2*C3+C3^2-2*C3+1))/(8*C1*C2-8);
g := -C2*Cww*(diff(s1(x), `$`(x, 2)))+Caa*(diff(s2(x), `$`(x, 2))+(n+1/2)^2*Pi^2*(diff(s2(x), x)));
f1 := -(1/2)*(n+1/2)^2*Pi^2*sqrt(4*C1*C2*C3+C3^2-2*C3+1)+C2*Cww*((D@@1)(s1))(0)-Caa*((D@@1)(s2))(0)+(n+1/2)^2*Pi^2*(C2-(1/2)*C3+1/2);

CN := ((2*(int(exp(-A1*x)*g, x = 0 .. t)-f1))*exp(A1*t)-(2*(int(exp(-A2*x)*g, x = 0 .. t)-f1))*exp(A2*t))/((n+1/2)^3*Pi^3*sqrt(4*C1*C2*C3+C3^2-2*C3+1));
ua := sum(CN*sin((n+1/2)*Pi*z), n = 0 .. 100);



I want to solve one equation with one variable and the variable is also in definite integral delimiter. When trying fsolve, I get the error:

"Error, (in fsolve) Can't handle expressions with typed procedures"


Here is

How can I obtain solution with method other from simple manual testing Te values?

Hello everyone

I have a problem with the code below in maple:

> restart;
> f := signum(a*cos(t+b)+sin(t+c));
> res := int(f, t = 0 .. 2*Pi);
                       2 Pi signum(a cos(b) + sin(c))
> a := 0.01; b := 0; c := 0;
> res; int(f, t = 0 .. 2*Pi);
                                    2 Pi

As you can see, results are not consistent with one another...

If I set a := 0.01; b := 1.35; c := 4.4; It is even worse, the computation of int(f, t = 0 .. 2*Pi) lasts more than 10 minutes (when I interrupt computation)

I tested it in Maple 12 and Maple 2015 and I get exactly the same results. I also replaced signum with csgn and I get the same problem.

I really need your help!!

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