not sure whether I understand you correctly, but it is defined in the interval ]-1,+1[, 
it is just complex

1)  You could use int(pf,x); evalc(%); convert(%,piecewise,x);  % assuming (-1<x,x<1);and see that 
in 0 it is -1/4*I*Pi. The reason is that indefinite integration is unique up to a constant 
(which may be complex, since Maple does not restrict to the Reals).

2) You may therefore try int(pf,x=0..t) assuming (-1<t,t<1), since your sum is 0 in 0.

in that case try to concatenate "file" + convert(i,string) + ".txt" to get a string as name

Hm ... what is the reason you used the first version for changing variables?

Seems that Inta is a Laplace transform and essentially this is to find
laplace(t*y(t), t, s) for y(t) = exp(1/(50*t^(9/20)-10*t^(69/100)+t))^61
for Inta = -8570/97*(s-100/857)*(1/(s^2)-J)*s with J = the Laplace, where
I use 'rationalize', substituted Pa = 1000/97/s-8570/97 (and expanded to
evaluate)

i would prefer combine(%,exp) ... no idea, why M does not use it

beside what Georgios Kokovidis said (there are better methods for solving a numerical system than inverting)

i would check the data input, either in some coordinates MatrixA[1,2] or for all by using convert(MatrixA, listlist) to see whether all
are floating point numbers, sounds as if  you may have some non-numerical values (it should only need milli seconds) ...

as a remark only: i would avoid blanks in Excel sheet names (even if it works) and may be you should say which Excel version
(not sure whether it runs with the latest shits [sorry, actually i like Excel])

John means that const + linear term is automatically simplified here (through converting to a polynomial I guess)

Type in 2 + (x-5)/4 and you should get 3/4+1/4*x ...

A brute work around (+ showing the effect):

taylor( f(x),x=5,4); eval(%, O=0); # difficult to see: set the letter 0 to the number zero  
convert(%,polynom);
                                  2                3            4
    2 + 1/4 (x - 5) - 1/64 (x - 5)  + 1/512 (x - 5)  + O((x - 5) )


                                         2                3
           2 + 1/4 (x - 5) - 1/64 (x - 5)  + 1/512 (x - 5)


                                      2          3
                               (x - 5)    (x - 5)
                   3/4 + x/4 - -------- + --------
                                  64        512

i second that, it would be helpful to re-install that (even if older user data can not be recovered) ... was just using to 'flag' this (as non-English i would read it as 'mark it'), but it would mean something different :-)

when i worked with non-technical users i always tried to fix requirements in tables, not through
lists of logical statements: for the programmers it is more easy to implement and for the users
it is easier to check for completeness or consistence (but i am not any more in that business ...)

the basic questions are 'what do you actually want' and 'why', in words, one page at _most_ ...

anthony: how about?

i always get dizzied by that ... what may be the reason to distinguish between type and property (here) ?

You may use the following in Maple:

# differentiation w.r.t. to an integral bound

  int(f(s)*(x-s),s=0..x):
  'diff(%,x)': '%'=%;


                  /   x                \      x
                  |  /                 |     /
               d  | |                  |    |
               -- | |   f(s) (x - s) ds| =  |   f(s) ds
               dx | |                  |    |
                  |/                   |   /
                  \  0                 /     0

# now use integration having an anti-derivative

  D(F)(x): int(%,x=0..t) = F(t) - F(0);


                       t
                      /
                     |
                     |   D(F)(x) dx = F(t) - F(0)
                     |
                    /
                      0

# but you have an anti-derivate: this is your integral at
# the rhs of your equation, substitute it, it has F(0)=0

  eval(%, F = 'xi -> int(f(s)*(xi-s),s=0..xi)');


                t    x                 t
               /    /                 /
              |    |                 |
              |    |   f(s) ds dx =  |   f(s) (t - s) ds
              |    |                 |
             /    /                 /
               0    0                 0

Just use the Reasonable Interface ... ähh ... I mean the Classical Interface ... and type it in

L:=[alpha = 2.494476546, beta = 1.555603312, gamma = .4046124214, delta = 1.272183039];
K := (1-alpha*r*cos(t)+beta*r^2)/(1+gamma*r^2-delta*r*cos(t))/t^2; # his new integrand

Making the bounds rational (still working with that ugly 10 Digits ...) now use

 Int(K,t = 610605461/15882 .. 3875151341/100000, method=_d01akc);
 eval(%,L);
 plot(%,r=5*1e-4 .. 5*1e-3, numpoints=400);

to guess from the graphics (and its symmetry) that the minimum is the middle of
0.003 and 0.0018

The different result may be caused by

'eval(K, t = t + 2*Pi)/K': '%'=%;

                     K|                    2
                      |t = t + 2 Pi       t
                     -------------- = -----------
                           K                    2
                                      (t + 2 Pi)

Now start over, slicing (and using better exactness) should fill some gap in the thesis.