## 1491 Reputation

8 years, 260 days

## Syntax mistakes....

Error messages says: "system must be entered as a set/list of expressions/equations"

sys := ((D@@2)(theta1))(t) = t, ((D@@2)(theta2))(t) = t;#this is a list of equations.

# I assume a system of equations, because you did not give any.

sol1 := dsolve({sys, theta1(0) = (1/2)*Pi, theta2(0) = (1/4)*Pi, (D(theta1))(0) = 0, (D(theta2))(0) = 0}, {theta1(t), theta2(t)}, type = numeric, output = listprocedure);

sol1(0);

## Workaround....

Workaround is  changing  integration method to _d01ajc that gives 7 correct digits.

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## Workaround....

y := func^2; eval(diff(y, func), func = diff(tau(t), t));

#2*(diff(tau(t), t))

## Syntax....

Check yours syntax,because is wrong!

Probably you want as a example:

int(x*y, [x = 4 .. 16, y = 0 .. 2]);

#240

## Not exist....

Laplace transform of yours examples not exist. This is only my opinion.

Not sure what is expected from that input.

A correct syntax is:

```L1 := inttrans:-laplace(psi1(t)*(diff(z1(t), t)), t, s);
L2 := inttrans:-laplace((diff(psi1(t), t))^2, t, s);

#Return unevaluated```

## Simple mistake.....

You made a  simple mistake in the equation (18).You  wrote xi(1),but  it should be: 1/xi(1).

See attached file:

feijisuan.mw

## ....

You forgot to deliver a inital conditions to DEplot function.

with(DEtools):

DE3 := {diff(y(x), x) = y(x)-z(x), diff(z(x), x) = z(x)-2*y(x)};

DEplot(DE3, [y(x), z(x)], x = 0 .. 3, y = 0 .. 2, z = -4 .. 4, [[y(0) = 1, z(0) = 1]], arrows = medium, numpoints = 200);

## ....

On Maple 2018.1  definition doublefactorial is the same as 2016 nothing has changed.

From Maple help: ?factorial;

"Note: In Maple, !! is used for repeated factorials and so it does not indicate the double factorial."

then use: doublefactorial(n);

Maybe you can use alias function:

alias(`bi-factorial` = doublefactorial);

`bi-factorial`(15);

#2027025

doublefactorial(15);

#2027025

## Works fine on Maple 2018.1...

What version of Maple do you use?.

If you are using an older version, the latest version of Maple int correctly solves.

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 Definite Integration:   Integrating expression on Z=0..infinity Definite Integration:   Using the integrators [distribution, piecewise, series, o, polynomial, ln, lookup, cook, ratpoly, elliptic, elliptictrig, meijergspecial, improper, asymptotic, ftoc, ftocms, meijerg, contour] IntegralTransform LookUp Integrator:   Integral might be a laplace transform with expr=32/(exp(1/5*Z)+3)*Z/(exp(3/10*Z)+3)^2 and s=-3/10. Definite Integration:   Integrating expression on T=0..infinity Definite Integration:   Using the integrators cook Cook LookUp Integrator:   but does not fit into the first class Cook LookUp Integrator:   returning answer from cook pattern 1a Definite Integration:   Returning integral unevaluated. Definite Integration:   Integrating expression on T=0..infinity Definite Integration:   Using the integrators cook Cook LookUp Integrator:   but does not fit into the first class Cook LookUp Integrator:   returning answer from cook pattern 1a Definite Integration:   Returning integral unevaluated. Definite Integration:   Integrating expression on T=0..infinity Definite Integration:   Using the integrators cook Definite Integration:   Returning integral unevaluated. IntegralTransform LookUp Integrator:   Integral transform returned unevaluated. LookUp Integrator:   unable to find the specified integral in the table Definite Integration:   Method ftoc succeeded. Definite Integration:   Finished sucessfully.
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Maybe you can try and see what hapens ?

```infolevel[IntegrationTools] := 3;
lambda[1] := 3/10; lambda[2] := 2*(1/10); alpha := 4:
int(2*alpha^2*Z*exp(lambda[1]*Z)/((exp(lambda[1]*Z)-1+alpha)^2*(exp(lambda[2]*Z)-1+alpha)), Z = 0 .. infinity, method = ftocms);```

## Another 2 ways....

Substitution x = t^2:

IntegrationTools:-Change(Int(1/(1+sqrt(x)), x = 0 .. 1), x = t^2);

value(%);

#2-2*ln(2)

Or using convert(expr,elementary):

Int(1/(1+sqrt(x)), x = 0 .. 1) = int(1/(1+sqrt(x)), x = 0 .. 1);

#Int(1/(1+sqrt(x)), x = 0 .. 1) = MeijerG([[0, 0, 1/2], []], [[1/2, 0], [-1]], 1)/Pi

Substitution: 1=x

convert(MeijerG([[0, 0, 1/2], []], [[1/2, 0], [-1]], x)/Pi, elementary);

#-ln(1-x)/x-2*arctanh(sqrt(x))/x+2/sqrt(x)

limit(%, x = 1);

#2-2*ln(2)

## Bug!...

It's a bug in int !

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## Only for: z>0,j>0,Re(z),Re(j)...

Only for: z>0,j>0,Re(z),Re(j)

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## Only workaround for interface & typesett...

```interface(typesetting = extended);
F := n-> ifactors(ithprime(n)-2)[2, 1, 1] ;
F(11);
print(F);

#                               29
#n -> ifactors(ithprime(n) - 2)[2, 1, 1]
```

On Maple 2018.1

## ....

I a little speed up computation.The plot is now gererated by Maple about 1 minute.

If N=0 then integral is probably singular ,I changed q = 0.1e-2 to q = 0.1e-1.

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## ....

A simple example than yours and  we convert to piecewise to understand more:

convert(signum(0, x), piecewise);

#piecewise(x < 0, -1, x = 0, 0, 0 < x, 1)

convert(signum(0, x,1), piecewise);

#piecewise(x < 0, -1, 0 <= x, 1)

Using derivatives:

diff(signum(0, x),x);

convert(%, piecewise);

#signum(1, x)

#piecewise(x = 0, undefined, 0)

diff(signum(0, x, 1),x);

convert(%, piecewise);

#signum(1, x,1)

#piecewise(x = 0, undefined, 0)

_Envsignum0 := 0; convert(signum(0, x), piecewise);

#piecewise(x < 0, -1, x = 0, 0, 0 < x, 1)

_Envsignum0 := 1; convert(signum(0, x), piecewise);

#piecewise(x < 0, -1, 0 <= x, 1)

_Envsignum0 := 2; convert(signum(0, x), piecewise);

#piecewise(x < 0, -1, x = 0, 2, 0 < x, 1)

_Envsignum0 := -2; convert(signum(0, x), piecewise);

#piecewise(x < 0, -1, x = 0, -2, 0 < x, 1)

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