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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

Hello.

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

 

 

Bug_in_integrate(floor_and_ceil).mw

Mariusz Iwaniuk

Hi,

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?

restart;
with(student);
assume(n::integer);
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"

code

Here is worksheet.mw

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
                                     0.

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!!

hi.i calculate this integral numerically but after use this result in my differential equation which is involve f3(x), i enconter with error

i found that this problem is due to answer of this integtal.may i use another way to calculte it?as shown power in result are very large!!!

``

restart; Digits := 15; g3 := theta^2*(theta-1)^2; beta := 100; chi := 5; kappa := 5; a := 0

0

(1)

with(Student[Calculus1]); -1; a31 := evalf(ApproximateInt(-beta^2*g3/((1-g3*f3(x))*ln(2*kappa*(1-g3*f3(x)))^2), theta = a .. 1, method = simpson)); -1; a32 := evalf(ApproximateInt(-chi*g3/(1-g3*f3(x))^4, theta = a .. 1, method = simpson))

-1.33333333333333*(-0.191597582462488e152*f3(x)-0.863011826221205e146*f3(x)^9-0.706378636532818e148*f3(x)^7+0.846944805095211e147*f3(x)^8+0.130151241709538e151*f3(x)^4-0.283379399043279e150*f3(x)^5+0.493554199542613e149*f3(x)^6+0.117104095781373e152*f3(x)^2-0.459581479474515e151*f3(x)^3+0.151119505345015e152+0.755561753647024e145*f3(x)^10+0.175123383079432e97*f3(x)^36-0.244416009548583e100*f3(x)^35+0.163803382818292e103*f3(x)^34-0.704973843312161e105*f3(x)^33+0.219681545650952e108*f3(x)^32-0.529211072065593e110*f3(x)^31+0.102639576866076e113*f3(x)^30-0.164658103849427e115*f3(x)^29+0.222629381398692e117*f3(x)^28-0.257124576224432e119*f3(x)^27+0.256176429040311e121*f3(x)^26-0.221804182616290e123*f3(x)^25+0.167834727658308e125*f3(x)^24-0.111474607475097e127*f3(x)^23+0.652137994527267e128*f3(x)^22-0.336921770775584e130*f3(x)^21+0.154034904897255e132*f3(x)^20-0.624063631993155e133*f3(x)^19+0.224251041670249e135*f3(x)^18-0.714943093769717e136*f3(x)^17+0.202175772822691e138*f3(x)^16-0.506683948932158e139*f3(x)^15+0.112368506172683e141*f3(x)^14-0.220033360070981e142*f3(x)^13+0.379271084508473e143*f3(x)^12-0.573179869691401e144*f3(x)^11)/((-160000.+9801.*f3(x))^4*(-16.+f3(x))^4*(-256.+9.*f3(x))^4*(-10000.+441.*f3(x))^4*(-160000.+8281.*f3(x))^4*(-625.+36.*f3(x))^4*(-160000.+361.*f3(x))^4*(-10000.+81.*f3(x))^4*(-160000.+2601.*f3(x))^4*(-625.+16.*f3(x))^4)

(2)

``

 

Download integral.mwintegral.mw

thanks..

>assume(x>0, n>0):int(x^n*ln(x)^n,x);

I_Mariusz

I mean

J := int((x^2+2*x+1+(3*x+1)*sqrt(x+ln(x)))/(x*sqrt(x+ln(x))*(x+sqrt(x+ln(x)))), x);

Of course, with Maple.

I want to plot a function given different value combinations of parameters. I used the following code and it doesn't work. Could anybody please help?

The attached worksheet shows a small selection of new and improved results in integration for Maple 2016. Note that integration is a vast topic, so there will always be more improvements that can be made, but be sure that we are working on them.

Maple2016_Integration.mw

A selection of new and improved integration results for Maple 2016

New answers in Maple 2016

 

 

Indefinite integrals:

 

int(sqrt(1+sqrt(z-1)), z);

(4/5)*(1+(z-1)^(1/2))^(5/2)-(4/3)*(1+(z-1)^(1/2))^(3/2)

(1.1)

int(arctan((-1+sec(x))^(1/2))*sin(x), x);

-arctan((-(1/sec(x)-1)*sec(x))^(1/2))/sec(x)+(1/2)*(-1+sec(x))^(1/2)/sec(x)+(1/2)*arctan((-1+sec(x))^(1/2))

(1.2)

int(((1+exp(I*x))^2+(1+exp(-I*x))^2)/(1-2*c*cos(x)+c^2), x);

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

(1.3)

int(x^4/arccos(x)^(3/2),x);

(1/4)*(-x^2+1)^(1/2)/arccos(x)^(1/2)-(1/4)*2^(1/2)*Pi^(1/2)*FresnelC(2^(1/2)*arccos(x)^(1/2)/Pi^(1/2))+(3/8)*sin(3*arccos(x))/arccos(x)^(1/2)-(3/8)*2^(1/2)*Pi^(1/2)*3^(1/2)*FresnelC(2^(1/2)*3^(1/2)*arccos(x)^(1/2)/Pi^(1/2))+(1/8)*sin(5*arccos(x))/arccos(x)^(1/2)-(1/8)*2^(1/2)*Pi^(1/2)*5^(1/2)*FresnelC(2^(1/2)*5^(1/2)*arccos(x)^(1/2)/Pi^(1/2))

(1.4)

 

Definite integrals:

int(arcsin(sin(z)), z=0..1);

1/2

(1.5)

int(sqrt(1 - sqrt(1+z)), z=0..1);

((4/5)*I)*(2^(1/2)-1)^(3/2)*2^(1/2)+((8/15)*I)*(2^(1/2)-1)^(3/2)

(1.6)

int(z/(exp(2*z)+4*exp(z)+10),z = 0 .. infinity);

(1/20)*dilog((I*6^(1/2)-3)/(-2+I*6^(1/2)))-((1/60)*I)*6^(1/2)*dilog((I*6^(1/2)-3)/(-2+I*6^(1/2)))+(1/20)*dilog((I*6^(1/2)+3)/(2+I*6^(1/2)))+((1/60)*I)*6^(1/2)*dilog((I*6^(1/2)+3)/(2+I*6^(1/2)))+((1/120)*I)*6^(1/2)*ln(2+I*6^(1/2))^2-((1/120)*I)*6^(1/2)*ln(2-I*6^(1/2))^2+(1/40)*ln(2+I*6^(1/2))^2+(1/40)*ln(2-I*6^(1/2))^2+(1/60)*Pi^2

(1.7)

simplify(int(sinh(a*abs(x-y)), y=0..c, 'method'='FTOC'));

(1/2)*(piecewise(x < 0, 0, 0 <= x, 2*exp(-a*x))+piecewise(x < 0, 0, 0 <= x, -4)+2*piecewise(c <= x, -cosh(a*(-x+c))/a, x < c, (cosh(a*(-x+c))-2)/a)*a-exp(-a*x)+piecewise(x < 0, 0, 0 <= x, 2*exp(a*x))+4-exp(a*x))/a

(1.8)

int(ln(x+y)/(x^2+y), [x=0..infinity, y=0..infinity]);

infinity

(1.9)


Definite integrals with assumptions on the parameters:

int(x^(-ln(x)),x=0..b) assuming b > 0;

(1/2)*erf(ln(b)-1/2)*Pi^(1/2)*exp(1/4)+(1/2)*Pi^(1/2)*exp(1/4)

(1.10)

int(exp(-z)*exp(-I*n*z)*cos(n*z),z = -infinity .. infinity) assuming n::integer;

undefined

(1.11)


Integral of symbolic integer powers of sin(x) or cos(x):

int(sin(x)^n,x) assuming n::integer;

` piecewise`(0 < n, -(Sum((Product(1+1/(n-2*j), j = 1 .. i))*sin(x)^(n-2*i-1), i = 0 .. ceil((1/2)*n)-1))*cos(x)/n+(Product(1-1/(n-2*j), j = 0 .. ceil((1/2)*n)-1))*x, n < 0, (Sum((Product(1-1/(n+2*j+1), j = 0 .. i))*sin(x)^(n+2*i+1), i = 0 .. -ceil((1/2)*n)-1))*cos(x)/n+(Product(1+1/(n+2*j-1), j = 1 .. -ceil((1/2)*n)))*ln(csc(x)-cot(x)), x)

(1.12)

int(cos(x)^n,x) assuming n::negint;

-(Sum((Product(1-1/(n+2*j+1), j = 0 .. i))*cos(x)^(n+2*i+1), i = 0 .. -ceil((1/2)*n)-1))*sin(x)/n+(Product(1+1/(n+2*j-1), j = 1 .. -ceil((1/2)*n)))*ln(sec(x)+tan(x))

(1.13)

int(cos(x)^n,x) assuming n::posint;

(Sum((Product(1+1/(n-2*j), j = 1 .. i))*cos(x)^(n-2*i-1), i = 0 .. ceil((1/2)*n)-1))*sin(x)/n+(Product(1-1/(n-2*j), j = 0 .. ceil((1/2)*n)-1))*x

(1.14)

Improved answers in Maple 2016

 

int(sqrt(1+sqrt(x)), x);

(4/5)*(1+x^(1/2))^(5/2)-(4/3)*(1+x^(1/2))^(3/2)

(2.1)

int(sqrt(1+sqrt(1+z)), z= 0..1);

-(8/15)*2^(1/2)-(8/15)*(1+2^(1/2))^(3/2)+(4/5)*(1+2^(1/2))^(3/2)*2^(1/2)

(2.2)

int(signum(z^k)*exp(-z^2), z=-infinity..infinity) assuming k::real;

(1/2)*(-1)^k*Pi^(1/2)+(1/2)*Pi^(1/2)

(2.3)

int(2*abs(sin(x*p)*sin(x)), x = 0 .. Pi) assuming p> 1;

-2*(sin(Pi*p)*signum(sin(Pi*p))*cos(Pi/p)-p*sin(Pi/p)*cos(Pi*(floor(p)+1)/p)+sin(Pi*(floor(p)+1)/p)*cos(Pi/p)*p-sin(Pi*p)*signum(sin(Pi*p))-sin(Pi*(floor(p)+1)/p)*p+sin(Pi/p)*p)/((cos(Pi/p)-1)*(p^2-1))

(2.4)

int(1/(x^4-x+1), x = 0 .. infinity);

-(sum(ln(-_R)/(4*_R^3-1), _R = RootOf(_Z^4-_Z+1)))

(2.5)


In Maple 2016, this multiple integral is computed over 3 times faster than it was in Maple 2015.

int(exp(abs(x1-x2))*exp(abs(x1-x3))*exp(abs(x3-x4))*exp(abs(x4-x2)), [x1=0..R, x2=0..R, x3=0..R, x4=0..R], AllSolutions) assuming R>0;

(1/8)*exp(4*R)-29/8+(7/2)*exp(2*R)-5*R*exp(2*R)+2*exp(2*R)*R^2-(5/2)*R

(2.6)

Austin Roche
Mathematical Software, Maplesoft

hi.i am a problem with calculate numeric integral.

please help me

thanks

Float(undefined).mw

I want to make sense of the expression

Int(t^2/ln(t)*exp(-t), t=0..infinity);

The denominator vanishes at t=1.  The singularity at t=1 is not integrable.  I want to see whether the integral is defined in the sense of Cauchy principal value.  Thus, I let

K := Int(t^2/ln(t)*exp(-t), t=0..1-a) + Int(t^2/ln(t)*exp(-t), t=1+a..infinity);

and wish to see whether the following limit exists:

limit(K, a=0, right);

Maple cannot evaluate this.  Nor can I.  Alternatively, we may try:

series(K, a=0);

or

series(K, a=0) assuming a>0, a<1;

In both cases Maple says that it is unable to compute the series.

So my question is: Does the Cauchy principal value exist, and can Maple help one to determine that?

 

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