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I've got the following double integral over a region A:

e^(1/x*y)/(y^2)*(x+1)^2 where A={(x,y):1/2<=x*y<=2,1<=x<=3}

to evaluate this:

I've tried :

int6:=int(int(e^(1/x*y)/((y^2)*(x+1)^2),x=1..3),y=(1/2)..(2/3));

since the largest lower bound and smallest upper bound for y based on 1/2<=xy<=2 are 1/2 and 2/3 respectively.

This statement however, only evaluates the inner integral; is my approach correct?

I want to solve a Laplace integral transform step-by-step its as follows:

 

f(t):=e^at

IntTutor(Int(e^-st f(t),t=0..infinity)

 

but it says that the expression can't be integrated, and secondly it does not work with 'infinity'. How I can solve this for step-by-step solution?

Thanxs

 

I had to use integration by parts to get the answer I need, but I do not think it is really needed for this. compare the output from:

r:=int(tan(x)^(n-2)*sec(x)^2,x);

(which is pages of very complicated output), with this:

r:=Int(tan(x)^(n-2)*sec(x)^2,x):
simplify(value(IntegrationTools:-Parts(r,tan(x))));

Does one really have to tell Maple to do integration by parts for this? I would have expected int() to do it as is.

Please compare to Mathematica:

Maple 18.02

added

The reason I think this should have been done by int() directly is that it is direct  application of the power rule of integration. I can't write Latex here, so I wrote it on my computer, and I copy the screen shot to explain for those interested:

Maple is so good in differential equations, but does not seem to be as good as it should when it comes to integration.

I solve numerically very simple boundary value problem for the following ordinary diff. equation:

-1.2*y''(t)+0.8*y(t)=2,

y(0)=1, y(1)=0.

So := dsolve([-1.2*diff(y(x), x$2)+0.8*y(x) = 2, y(0)=1, y(1)=0], y(x), 'numeric', 'output' = listprocedure);

Solution looks as needed

u := unapply(rhs(So[2])(x), x): plot(u(t), t=0 .. 1);
pic

and can be numerically integrated in usual way:

evalf(int(u(t), t = 0..1));

0.6041717543123311

But integral of u^2(t)  (evalf(int(u(t)^2, t = 0..1))) returns:

pic1

How to avoid this issue?

In Maple help on int, it says

"If Maple cannot find a closed form expression for the integral, the function call itself is returned"

But then, what is the correct way to check for this in the code? How do I know that the result returned is the call I made? (it would have been easier if these functions throw an error, or set some status code that one can check for success or failure).

Hello everybody,

I am working on a somewhat complicated analytical problem, which includes the evaluation of the following (finite and singularity free) double integral (the stated numerical parameters may vary):

Int(exp(18.1818*(Int((0.579160e-1*sqrt(x)*Ei(1., 0.500000e-4*x)+(1072.23*(.999950-1.*exp(-0.500000e-4*x)))/sqrt(x))/sqrt(x), x = 1. .. eta))-9.10000*eta)/eta, eta = 1. .. 100.)

There are no unevaluated variables inside (all the e's are real exp()s). However, when putting it through evalf(), Maple returns it (after calculating for some time) unevaluated. I have no idea why, as a simple box sum should succeed.

Any input would be appreciated. Best regards,

  K. Reuther

 

P.S.: Raw expression for further deliberations

Int(exp(18.1818*(Int((0.579160e-1*sqrt(x)*Ei(1., 0.500000e-4*x)+(1072.23*(.999950-1.*exp(-0.500000e-4*x)))/sqrt(x))/sqrt(x), x = 1. .. eta))-9.10000*eta)/eta, eta = 1. .. 100.)

Dear all;

I need your help to display the value of phi[jj] in the following loop. 

phi[1]:=(x,y)->exp(x-y);

for jj from 1 by 1 to 2 do
phi[jj+1]:=(x,y)->int(phi[1](x,s)*phi[jj](s,y),s=y..x);
end do;

 

but the index jj in the integral does not change. I have any idea what is the problem. 

Thank you very much for your help. 

 

 

Why this simplifes:

z1:=n*(Int(cos(x)^(n-2), x))-(Int(cos(x)^(n-2), x));

simplify(z1);

But when adding an extra term to z1, it no longer simplfies the above any more:

z2 := cos(x)^(n-1)*sin(x)+n*(Int(cos(x)^(n-2), x))-(Int(cos(x)^(n-2), x));

simplify(z2);

You can see the second term, which is z1, was not simplfied any more.

Why? And how would one go about simplifying z2 such that the second term gets simplfies as with z1, but while using z2 expression. It seems simplify stopped at first term and did not look ahead any more?

Maple 18.02, windows.

I am trying to find the root of an equation that involves a procedure and a definite integral (solved numerically). Of course, I don't need the root to be found symbolically, but numerically would be fine. The problem is, I keep getting the error

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

whenever I try to solve it. Anyone have any ideas? My worksheet is here: Table-1-duplication-mapleprimes.mw 

 

Excuse me,

 

  I have the following input

***

fff:=x1^k;
int(fff,x1=0..x2);

***

I would like to obtain

1/(k+1)* (x2^(k+1)), k<>-1

 

but maple gives me just the same integral. Is there any way to get the expression?

 

Thank you!

 

 

 

hi everyone , i need your point of view in my question,any help would be appreciated in advance .

we have a discrete function named g(t) and a continous function f(t) in in convolution integral just like this :
int(f(t-x)*g(x),x=0..t) ; 
we have just g(x) in some special points int the interval (0..t) , thus i need to convert this integral to a series.
how should i do this ? can anyone help or any idea ? i need at first a mathematical solution or idea about how to do this and then, how to do this in a software ?
tnx again.

I'd like to pay attention to the recent article "The Misfortunes of a Trio of Mathematicians Using Computer Algebra Systems. Can We Trust in Them?"

In particular, the authors consider the integral

int(abs(exp(2*Pi*Ix)+exp(2*Pi*I*y)),[x=0..1,y=0..1]),

stating "Both Mathematica and Maple return zero as the answer to this calculation. Yet this cannot be correct, because the integrand is clearly positive and nonzero in the indicated region". Unfortunately, they give only the Mathematica command to this end.

Of course, the integral under consideration is complicated so the the simple-minded trials

int(evalc(abs(exp((2*Pi*I)*x)+exp((2*Pi*I)*y))), [x = 0 .. 1, y = 0 .. 1]);

and

VectorCalculus:-int(evalc(abs(exp((2*Pi*I)*x)+exp((2*Pi*I)*y))), [x,y]=Rectangle( 0 .. 1, 0 .. 1));

fail. However,this can be found with Maple (I think with Mathematica too.) in such a way.

 

A := evalc(abs(exp((2*Pi*I)*x)+exp((2*Pi*I)*y)))

((cos(2*Pi*x)+cos(2*Pi*y))^2+(sin(2*Pi*x)+sin(2*Pi*y))^2)^(1/2)

(1)

NULL

B := simplify(A, trig)

(2*cos(2*Pi*x)*cos(2*Pi*y)+2+2*sin(2*Pi*x)*sin(2*Pi*y))^(1/2)

(2)

op(B)[1]

2*cos(2*Pi*x)*cos(2*Pi*y)+2+2*sin(2*Pi*x)*sin(2*Pi*y)

(3)

combine(op(B)[1], x)

2*cos(2*Pi*x-2*Pi*y)+2

(4)

C := eval(B, op(B)[1] = combine(op(B)[1], x))

(2*cos(2*Pi*x-2*Pi*y)+2)^(1/2)

(5)

int(C, [x = 0 .. 1, y = 0 .. 1])

4/Pi

(6)

``

 

Download int.mw

 

 

Hello,

I wanted to get the following function integrated with Maple 17:

I used the following command to get a numerical solution for my choise of limits:

Unfortunately, I just get the Integral itself back. Also, taking pi/2 as the upper limit for x does improve anything. I also tried the AllSolutions option but it does not help as well as first trying to integrate only over y and leaving x as it is in order to get at least one integration done.

I hope someone can help. I do not necessarily need to solve it for myself, so if someone gets the result and posts it here, I'm okay with that. Anyway, being able to calculate it myself would be even better.

1) Consider the followin Integral:

int(cosh(a*x)*cos(b*x),x)

How can I tell maple to give the results in terms of hyperbolic and harmonic funtions?
The maple returns the answer in term of exponentials with imaginary powers.
Also, I could not use the "simplify" command to turn the answer returned by maple into an answer in terms of what I want! Please see the code below.


2) Is there a way that I can update the maple integration database manaully?

 

Example.mw

Below is the function that I have.

 

f := (t-1)^(1/3)

p:=2;

b[n] := 2/p*(Int(f*sin(2*Pi*n*t/p), t = 0 .. p))

 

I also included a picture below to show what it is doing. Some help would be greatly appreciated. All I need to know is why maple doesn't want to evaluate bn?

 

Maple Code

 

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