Why should that worry you?  The answer is clearly 0, because the integrand is an odd function of z and the interval for z is symmetric around 0.

More specifically, the problem seems to be the interaction between alias and the parsing of a Vector or Matrix entered using the palette.  Here is a simpler example.  This is a Maple 12 worksheet: Maple 11 will complain, but will open it OK.

View 4541_alias.mw on MapleNet or Download 4541_alias.mw
View file details

I strongly agree with the advice to use 1-D (and not just for novices: it's almost always preferable, as far as I'm concerned, if you want to get something done rather than just make something that looks nice).

More specifically, the problem seems to be the interaction between alias and the parsing of a Vector or Matrix entered using the palette.  Here is a simpler example.  This is a Maple 12 worksheet: Maple 11 will complain, but will open it OK.

View 4541_alias.mw on MapleNet or Download 4541_alias.mw
View file details

I strongly agree with the advice to use 1-D (and not just for novices: it's almost always preferable, as far as I'm concerned, if you want to get something done rather than just make something that looks nice).

For a physicist, the existence theorems don't have any sense - the differential equations that they have, come from real life examples, and there is absolutely no doubt that the solutions exist.

For the most part, this is indeed true.  On the other hand, there's fluid mechanics, where the Euler and Navier-Stokes equations in three dimensions have no good existence theorems, and there is real doubt that the solutions exist.  See www.claymath.org/millennium/Navier-Stokes_Equations/navierstokes.pdf

Well, isn't that enlightening...

I think German must be one of the harder languages for automatic translators to work with.  I tried a random mathematical quotation from math.furman.edu/cgi-bin/randquote.pl:

In an examination those who do not wish to know ask questions of those who cannot tell.

Google translated this into German as

In einer Untersuchung, die nicht wissen wollen, stellen Sie Fragen, von denen nicht sagen kann.

and then back into English as

In a study not want to know, ask questions, which can not say.

For comparison, when translated into French and then back to English, it's almost right:

In an examination those who do not want to know ask questions to those who can not say.

When translated into Chinese (simplified) and back to English, it's

In the examination of those who do not want to ask those who know can not tell.

When translated into Korean and then back to English:

Those who do not wish to question the test does not tell you to know.

When translated into Hindi and then back to English:

An examination of them do not want to know what questions to ask can not tell who those people are.

Assign a value to A.

> A := 2;

> de:= (D@@2)(v)(t) + D(v)(t) + v(t) = sin(A*t);

> ics:= v(0) = 0, D(v)(0) = 0;

> with(DEtools): DEplot(de, v(t), t = 0 .. 10, [[ics]], linecolour=blue);

Assign a value to A.

> A := 2;

> de:= (D@@2)(v)(t) + D(v)(t) + v(t) = sin(A*t);

> ics:= v(0) = 0, D(v)(0) = 0;

> with(DEtools): DEplot(de, v(t), t = 0 .. 10, [[ics]], linecolour=blue);

Note that Maple is case-sensitive: if it's v in the equation, it has to be v, not V, in the initial conditions.

> dsolve({(D@@2)(v)(t)+D(v)(t)+v(t)=0, v(0)=a, D(v)(0)=0});

v(t) = 1/3*a*3^(1/2)*exp(-1/2*t)*sin(1/2*3^(1/2)*t)+a*exp(-1/2*t)*cos(1/2*3^(1/2)*t)

Note that Maple is case-sensitive: if it's v in the equation, it has to be v, not V, in the initial conditions.

> dsolve({(D@@2)(v)(t)+D(v)(t)+v(t)=0, v(0)=a, D(v)(0)=0});

v(t) = 1/3*a*3^(1/2)*exp(-1/2*t)*sin(1/2*3^(1/2)*t)+a*exp(-1/2*t)*cos(1/2*3^(1/2)*t)

The Euler-Maclaurin formula gives the asymptotic series of

sum(1/n^p, n=x..infinity) as

x^(1-p)/(p-1) + Sum((-1)^k*bernoulli(k)*binomial(p+k-2,k)/(p-1)*x^(1-p-k), k=1..infinity)

The Euler-Maclaurin formula gives the asymptotic series of

sum(1/n^p, n=x..infinity) as

x^(1-p)/(p-1) + Sum((-1)^k*bernoulli(k)*binomial(p+k-2,k)/(p-1)*x^(1-p-k), k=1..infinity)

As I said:

x = k1*z for 0 <= z <= L1,
k1*L1 + k2*(z-L1) for L1 < z <= L1+L2

k = k1 for 0 < z <= L1, i.e. 0 <= x <= k1*L1

k = k2 for L1 < z <= L1+L2, i.e. k1*L1 <  x <= k1*L1+k2*L2

Perfectly straightforward, no recursive nightmare.

As I said:

x = k1*z for 0 <= z <= L1,
k1*L1 + k2*(z-L1) for L1 < z <= L1+L2

k = k1 for 0 < z <= L1, i.e. 0 <= x <= k1*L1

k = k2 for L1 < z <= L1+L2, i.e. k1*L1 <  x <= k1*L1+k2*L2

Perfectly straightforward, no recursive nightmare.

It looks interesting.  But how does Cadabra handle the ambiguities of TeX notation?  E.g. is a(b+c) the function a applied to b+c or the variable a multiplied by b+c?

You need exp(t), not exp^(t) or exp(^t) or e^t.