convert(X4, rational);
simplify(%): expand(%): 
X4:=simplify(%, size);

This shows, that s1, s2 can not be taken arbitrary small because of log

plot3d( X4, s1= - 200 .. -1, s2= -500 .. -1, axes=boxed);

gives an impression and looking at (0,0) lets one guess a singularity:

limit(X4, s1=0); expand(%);
limit(%, s2=0, left);
                               infinity

Moreover the plot suggests a minimum for s1=0 and s2 in -100 ... -10:

eval(X4, s1=0);
diff(%, s2):
fsolve(%=0, s2= -100 .. -10);

                           -42.903624799046

You also may look up Wikipedia:

en.wikipedia.org/ ("Solution in terms of Chebyshev radicals" towards the end)
and en.wikipedia.org/wiki/Chebyshev_cube_root
 

Let be z = 1. Then you have 6 values to satisfy 1= (z^6)^(1/6), the 6 unit roots.

If you want to ignore subtilities:

(z^6)^(1/6);
simplify(%, symbolic);

                                  z

First: please do not post pictures, post code which can be copied.

arcsin(z): 
diff(%, z): 
convert(%, FormalPowerSeries,z=0):
op(1,%):
a:=unapply(%, k,z);
                                         (-k)  (2 k)
                                 (2 k)! 4     z
                  a := (k, z) -> -------------------
                                            2
                                        (k!)

Then you go:

Int(a(k,z), z=0 ..z): 
value(%) assuming (k::nonnegint):
Sum(%,k=0 .. infinity):
value(%):
simplify(%);

                              arcsin(z)

Besides what Joe Riel told you:

Int(exp(-a^2 + x),a) = exp(x) * Int(exp(-a^2),a)

Which says: you are fitting the exponential function * constant, constant =  Int(exp(-a^2),a = 0 .. 5)

use exp(1) if you mean that

Let S be the list of solutions for the equation.

  restart;
  assume(0 < a); # getassumptions(a);
  
  x^4-a^(1/1)*x-1;
  solve(%, x);
  S:=[allvalues(%)]: #nops(%);

Following Robert take the real part of the 2 conjugated complex solutions
and solve for the parameter a:

  mu:='Re(S[3])'; #``=%; #(S[3]+S[4])/2;
  'solve(A=mu,a)';
  %;

There are 2 of them, we take the positive one and from that the parameter
can be rediscovered:

  # a=4*(1+4*A^4)^(1/2)*A; isolate(%, A); # wrong sign
  a=4*(1+4*A^4)^(1/2)*A; 
  solA:=[solve(%, A)]:

  map(Im, solA);

This shows, that the first two of the 6 solutions have imaginary part = 0.
They have opposite signs, so take the positive one, which is the desired one:

  [0=Im(solA[1]), 0<=solA[1]]; 
  map(is, %);

                             [true, true]
 
One can see, that this is an unique assignment:

  0 <= 'diff(4*(1+4*A^4)^(1/2)*A, A)'; is(%) assuming 0 <= A;

                                 true

Thus change is given by

  A := solA[1];

Now start over by replacing the original a by our new parameter A:

  restart;
  a:=4*(1+4*A^4)^(1/2)*A;

  x^4-a*x-1;
  solve(%, x);
  [allvalues(%)]: 
  T:=expand(%);

Now this looks quite simple, up to signs the 4 terms are like the first one

  T[1];
                            2           4 1/2 1/2
                     A + (-A  + (1 + 4 A )   )

It is easy to see, that this is a positive Real if A is real.

And it solves the equation

  'map( 'x -> x^4-a*x-1', T)';
  expand(%); 

                             [0, 0, 0, 0]

By the first step one easily can express A in terms of the original parameter a,
it is

A = 1/12*(6*(108*a^2+12*(768+81*a^4)^(1/2))^(1/3)-
  288/(108*a^2+12*(768+81*a^4)^(1/2))^(1/3))^(1/2)

Fine, than one can show (S = list of the roots for x^4 - a*x -1) 

  'S[1]+S[2] = -Re(S[3])*2';
  is(%);
                                 true

Writing mu:='Re(S[3])' and S[1] = -mu + nu (defining nu as the difference)
we get S[2] = -mu - nu and Maple even determines the positive root:

  '0 < -mu+nu';
  is(%);
                                 true

(where I do not post my ugly hacks for all that - especially finding it -
and nu certainly even can be simplified in terms if mu).


As an idea/question: 

If the 4th root is the conjugate, then all is determined the by first 3,
(having that fact in mind). 

But the first 3 define a circle and some Moebius will map them onto the
unit circle and that can be done to give +-1 and I as the image points c. 

Thus I ask, whether it will map the 4th to -I (being afraid finding the
transform may be complicated). 

Or simply: can it give xi^4 = -1, xi = MoebiusTransform(x) ?

I used 

  convert(%, FormalPowerSeries,a); 
  subs(a=a^(1/4), %):
  combine(%):
  value(%):
  eval(%,a=a^4):
  h:=combine(%);

to get your result, but just with negative sign and plotting shows that
as being negative for a in [0 ... infinity[ and plotting again it lives
in [-1 ... 0[ (which is the 'x', for which it stands in the equation).

Graphical evidence is given by

  x^4-a*x-1;
  subs(x=h, %): combine(%);
  plot(%, a = -2 .. 2, numpoints=10, style=point);

(so also for negative a one gets a solution, it tends to -infinity).

Now I take the 'algebraic' solutions

  x^4-a^(1/1)*x-1;
  solve(%, x);
  S:=[allvalues(%)]:

Then I compare by

  map( 't -> h - t', S):
  map( 't -> taylor(t, a=0, 3)', %);

                2      4      4
  [ - 2 + 1/16 a  + O(a ), O(a ),

                                            2      4
         - 1 - I + 1/2 a + (1/32 - 1/32 I) a  + O(a ),

                                            2      4
         - 1 + I + 1/2 a + (1/32 + 1/32 I) a  + O(a )]

In my case it seems, that h = S[2] and the S[1] should by real,
if h is real. S[1] resp S[2] gives the positive resp negative
real root (where the numeration may depend on the 'state' of
the Maple session).

Have not tried Sturm chains or similar, I use it to seldom

Writing x^4-a*x-1 with 0 < a and using Robert Israel's way the solution h
(in terms of 3F2) seems to give always a negative x (have no proof, just
'evidence') living in the open range -1 ... 0.

But that equation is -1 in x=0 and +infinity for x=+-infinity, so your way
(what ever it is) should work, if you know what the negative x has to be.

Otherwise said: you can devide out a factor (x-x1)*(x-x2), where both the
x_i are real of opposite sign.

If you set up your equations, then do not use square brackets, if you mean the usual grouping done in Math.

This will give it a totally different meaning.

PS: for all that the 'Classical' interface is more safe, as you are forced to type in everything completely

May be I am oldish in contrast to Alejandro, but I hate to work with rtables (though they are efficient):

  Matrix(3,shape=identity);
  convert(%, listlist); # for 'export' or copy or whatever

                  [[1, 0, 0], [0, 1, 0], [0, 0, 1]]

  convert(%, Matrix); # gives me back the Matrix (but with other id)

if you multiply your Matrix by a column vector from the right and that vector consist of 1's then you get what you are looking for

Sorry for my my cryptic statement.

'evalc(argument(exp(y*I)))': '%'=%;

                    evalc(argument(exp(y I))) = y

Thus it would take any (real) value.

Bbut argument is at most Pi in magnitude and real, 
so evalc of it is the same and bounded by Pi.

  restart;interface(version);
  tstData:=[x=0, y=2*Pi];
  z:=exp(x+y*I);

  L:=argument(z);

    Classic Worksheet Interface, Maple 12.02, Windows, Dec 10 2008 Build ID 377066


  evalc(argument(z)); # evalc assumes real variables, see the help

                                  y

For the test data z equals 1, so the argument is 0. But using evalc
even produces values beyond Pi.


  argument(z) assuming (x::real, y::real);

                        arctan(sin(y), cos(y))

Which is correct.

Note that one could have set x=0 from beginning, since argument does
only care for the angle, not the magnitude and abs(z) = abs(exp(y*I))
for real y. It gives the same troubles.