Items tagged with integration

Hey guyz, I am in trouble with calculation attached integral. it is a simple function but a bit long. I can't solve it with maple, Do U have any idea?

 

 

intg.mw

 

 

Hi I try this integral:

m2 := int(exp(-(1/2)*z^2)*((exp(B*J*sqrt(q)*z))^2-1)/(sqrt(Pi)*sqrt(2)*((exp(B*J*sqrt(q)*z))^2+1)), z = -infinity .. infinity)

But not resolve.

How can i do?

Regards.

Dear all,

I would like to evaluate a double integral numerically. The integrand is a complicated function of the variables beta and s, with complex values. The computation lasts for decades without obtaining a result.

I was wondering whether there exists subroutines / methods / tricks that could be helpful to accelerate the integration process. I have attached a Maple script of the double integral of interest. Rough precision would be fine (4 or 5 digits).

Any help would be highly appreciated.

Thanks

Federiko

Question.mw

i could have sworn that when itegrating a gaussian maple will write it in terms of the erf functions... but i end up with:

gg:=A * exp( - ( (t - t0) / (tau) )^2 );
val1:=int(gg, t=-x0..x1) assuming t0::real, tau::real, x0<x1, t0>x0, t0<x1, x0::real, x1::real;  #or with no assumptions

 

the results is just gg unchanged... Doing:

convert(val1, erf)

does not help. I can set t0 (or transform it away), and it works, but I was hoping maple would not require this. 

Any thoughts how to help maple with this?

Mathematiaca can read my mind without issues:

 

Good evening sir.

 

I request your valuable support with regard to the above cited query.

 

 

With thanks & regards.

 

Mr.M.Anand

Associate Professor in Mathematics

Hi,
   I just finished a math quiz. I needed to find the length of the curve and area of the function, r=3*cos(theta)-2*sin(theta) bounded between 0<=Pi<=2*Pi.
   On the quiz I used Area=int(1/2*(r^2)) dtheta. For the length of the curve I used L=int(sqrt(r^2 + r'(theta)) dtheta.

How do you plug this into Maple and get an answer?
I came up with 20.42.... sq units for the area and 22.65.... for the length of the curve.

Thank you,
Jay.

I have 

dZ(x)=−xdlog(z(x))

where d is the exterior derivative. I would like to recover the function Z(x) by integrating both sides of the equation. How would I compute this in Maple?

int(a(t)*b(t)+2*(diff(a(t), t))*(diff(b(t), t)), a(t));
Error, (in int) integration range or variable must be specified in the second argument, got a(t)
 

do not understand this error message,

how to integrate it?

 Dear All ! 

I really need to solve this problem as soon as possible, As you know the downside equation is not exact, but I can not find its integration factor, blease help me !

                                                                 ∫{ ( ωx + σy ) d x + (ωy −σx)dy}=0

 Regards ,

 

 

so I'm trying this:

restart;

sigma := 0.143e-18;

l_0 := 1.87;

l0 := 1.87;

roll := rand(0 .. 25.0);

f_gauss := proc (x) options operator, arrow; exp(-(1/2)*x^2/`&sigma;_x`^2)/sqrt(2*Pi*`&sigma;_x`^2) end proc;

f_norm := proc (dx) options operator, arrow; int(f_gauss(x), x = -(1/2)*dx .. (1/2)*dx) end proc;

sol_gauss := proc (mix) options operator, arrow; evalf(eval(-ln((int(f_gauss(x)*exp(-2*sigma*N2O*sqrt((1/4)*l_0^2-x^2)), x = -(1/2)*dx .. (1/2)*dx))/f_norm(dx))/(sigma*N2O), [N2O = 0.25e20*mix/100])) end proc;

for ii to 10 do

a := roll();

eval(sol_gauss(a), [dx = l_0, `&sigma;_x` = l0])

end do

After several attempts on this question,

Int(x*sqrt(2*x^4+3),x) with substitution u = sqrt(2)*x^2,

I don't seem to find the solution. Can you guys help me?

So I have an integral that computes perfectly in wolfram alpha but not in maple...

I will post it here

int(1/((4.532055545*10^9/f^4.14-2.311250000*10^5/f^2+(111*(1-0.2163331531e-4*f^2+2.340001656*10^(-10)*f^4))/(1+0.1081665766e-4*f^2)))*(6*10^(-21)*abs(1/f^(4/3)))^2, [f = 50 .. 1500])

the answer should be 3.05364*10^-46

If you try that exact line of code in maple, it will not compute (is stuck on evaluating)


Best Regards to all,
Zeus

Dear all,

I would like to compute numerically using Maple the following improper integral

``

Integrand := (1/4)*(((((6*I)*beta-3-6*C+(6*I)*C*beta)*s^4+((24*I)*C*beta-24*C-12)*s^2+(24*I)*(1+C)*beta)*BesselK(0, s)+12*BesselK(1, s)*(C+1/2)*s^3)*BesselI(1, s)^3+6*BesselI(0, s)*(-(2*(I*beta*C*s^2+(2*I)*beta*C+(2*I)*beta+4*C+2))*s*BesselK(0, s)+((I*beta*C+I*beta-C-1/2)*s^4+((4*I)*C*beta+4*C+2)*s^2+(4*I)*(1+C)*beta)*BesselK(1, s))*BesselI(1, s)^2-(12*(-(1/2*((C+1/2)*s^2+I*beta*C+I*beta+8*C+4))*s*BesselK(0, s)+((I*beta*C+2*C+1)*s^2+(2*I)*(1+C)*beta)*BesselK(1, s)))*s*BesselI(0, s)^2*BesselI(1, s)+6*s^2*((-2*C-1)*s*BesselK(0, s)+BesselK(1, s)*((C+1/2)*s^2+I*(1+C)*beta))*BesselI(0, s)^3)/((BesselI(0, s)^2*s-BesselI(1, s)^2*s-2*BesselI(1, s)*BesselI(0, s))^2*(C+1/2)*s*Pi):

``


However, Maple does seem to give a result for this integral. I have tried to compute from e.g. 0.001 as an approximation but it turns out that the integrand diverges as s goes to zero. I have also tried some options such as method = _d01amc but I get Error, (in evalf/int) powering may produce overflow.

 

I would appreciate it if someone here could provide with some help with regards to the computation of such improper integrals. Thank you.

 

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