I do not have MMA and do not want to type its result.

But why MMA stops below x = 6? The result given by Maple in terms of an integral seems to correct (just apply 'laplace' again with 0 < x to verify it).

And that integral gives a real value for any positive x.

I would write

U=-(s^2/(s^2+1)+1/(s^2+1)-1)*exp(sqrt(s)*x/sqrt(phi))/(s^2+1)+exp(-sqrt(s)*x/sqrt(phi))/(s^2+1);
eval(%, phi=r^2); simplify(%) assuming 0<r;

     U = exp(-s^(1/2)*x/r)/(s^2+1)

Now one sees that this depends on a single parameter x (by combining r into it, most simple by setting r=1), for which on needs some range. Essentially it is to change from exp(1) to exp(1)^x and I doubt that x can arbitrary.

Having 0<x (as Preben did) is quite natural and gives something which can be delivered numerically (just change upper bound to 1 and plot to see)

an euphemism indeed :-)

It means Software Change Request ... in SW industry that term is used to indicate that the SW works as designed/agreed and the supplier expects to be payed for changing it ...

One can observe that this converges to exp(1) and in double precision that
limit is approached for n = 17. So best is to compute the values into an
array and call it (for that note that Excel has option base 0 as default
and also note that Excel cuts down to 15 decimals, you want 18 ...) and
take that as exp(1.0) beyond.

  Sum(1/(i!), i=0..n)/exp(1); 
  s:=value(%);
  [seq( s, n=0 .. 17)];
  evalf[18](%);

What is the reason for your series of questions towards that (may be you already said it and I missed that)?

Every CAS uses coded rules, tables and pattern abilities. 
So what. I just had a gut feeling 'fourier or residues?'. 
The point is: a special case. But acer did the very work. 

Here you go without inttrans, finally giving the same, using M 18:

Int(-1/5*sin(w)^3/w^(3/5)*exp(-w*t),w = 0 .. infinity); 
value(%) assuming 0<t;
limit(%, t=0); simplify(%);
evalf(%);


                          1/2     3 Pi    (3/5)
                      Pi 2    cos(----) (3      - 9)
                                   10
                 1/30 ------------------------------
                                          1/2 1/2
                         GAMMA(3/5) (5 + 5   )


                          -0.153562126953524


BTW: do you have a reference for the task?

Int( D(f)(x), x=a..b, continuous); value(%);

Note that Maple does not assume/conclude that f is continuous by writing down a differentiation. That is by design.

As an example try signum(x) between -3 and +3.

@frajola 

The attached file should answer your questions

MP_Meijer_Integral_question.mws

MP_Meijer_Integral_question.pdf

f:=z -> (310*(z+0.5)^0.2)+70; 
L:= [-0.1, 0.0, 0.2];

map(f, L);

         [328.091494294581, 339.870674621798, 358.656473679400]

f~(L);

         [328.091494294581, 339.870674621798, 358.656473679400]

Nice kidding. Switching to n much higher than 100 will certainly break in the original task. The point is: apply log, check situation and solve. Or in educated terms: preprocessing a problem for a solver.

For such a task involving exponentials I always try log first (yes, one may 
need to care for complex log(negative)), but here it seems simple to me:

x*ln(100) - 100*ln(x);
[Re(%), Im(%)];
plot(%, x=-50.. 150, color=[red,blue]);
#fsolve(%%[1]);

I think it is too complicated (and my impression is, you try some brute force)

You can simplify a bit, multiply by r^4 to get (formally) rid of denominators, it
will 'look' like a polynomial in r.

I have not looked into the method, however the folloing example shows some
basic approach:

tmp:=(64*a^2+a)*r^(1*a+1)+4*r^(2*c+1)*c;
#tmp:=(64*a^2+a)*r^(1*a+1)+4*r^(-(2*c+1))*c;
s:=[solve( identity(%, r))];

    [                 /    -3      -3 \    /    -1       \ ]
    [{a = 0, c = 0}, { a = --, c = --- }, { a = --, c = 0 }]
    [                 \    64      128/    \    64       / ]

What does that mean? For that look at the 2 summands of tmp

[op(tmp)];

             [/    2    \  (1 + a)     (2 c + 1)  ]
             [\64 a  + a/ r       , 4 r          c]

So the solution s[1] and s[3] are just found by setting the 'coefficients'
to zero. While s[2] is such that the exponentents become the same - thus
they can be factored out - and the remaining coefficients sum up to zero.


Now for your task there are a lot of conditions for that route. And it is
not so clear what is needed to satisfy them. It is neither clear that there
is any solution (depending on the parameters), nor is it clear that there
are unique, isolated tuples - or complete varieties.

Note that all the possibilities for the exponents have to be checked.

You can get an impression using 'asympt' in r for the expression, the
error message is what you want to understand for that.

I think the result for f is correct, it is sqrt(3):

Lame sketch (I am rusty): write F(n)=sqrt(n^2+2*n/3)*Pi for the input of tan.

Then monotonical without jump in tan, large values (alternating series in oo or 'is')

F(x) = (x*Pi+Pi/3) + epsilon(x);
tan(rhs(%));
expand(%);
simplify(%) assuming x::integer;
eval(%, epsilon=0);


                          1/2
                         3    + tan(epsilon(x))
                      - ------------------------
                         1/2
                        3    tan(epsilon(x)) - 1


                                  1/2
                                 3

I accept the second example, h = function is a product in the variable as upper bound = Prod( 1.. infinity)
PS: is this a classical identity?

And I can imagine that similar problems may occure for series or integrals

For the first one, f = tan, I am not sure.

Writing the sequence "as sum" and taking the numerical value of the series it turns out to be sqrt(3) as well. I think the 'lame way' = "large n, so n^2 dominates, so ~ tan(sqrt(n^2)*Pi) = tan(n*Pi) = 0" fails, because tan is not continous thus does not commute with limits.

I think it is - abs(sin(t)), t Real (just plot)

And for n=1 it gives + (cos(t)-1)^(1/2)*(cos(t)+1)^(1/2)