I think that formula is only valid, if the denominator has no multiple roots
(so it is a 'generic' answer):

It is summing over all roots alpha of the denominator for the summands as

  numer(f)/diff(denom(f),x):
  subs(x=alpha, %)* 1/(x-alpha); # = constant * 1/(x-alpha)
  int(%, x);                     # = constant * log

The constant (w.r.t. variable x) however is singular, if alpha is multiple.

Following the recipe at http://eqworld.ipmnet.ru/en/solutions/lpde/lpdetoc1.htm
(download the pdf there) I get the following solution (0 < t)

U := 1/2/(t*Pi)^(1/2) * 
  Int(arctan(xi)*exp(-1/4*(x-xi)^2/t),xi = -infinity .. infinity) +1-exp(-t)

-diff(U,t)+diff(U,x,x)+exp(-t); combine(%): simplify(%);

  0


For the ic I only have numerical evidence (using small values of t and
test values for x as well), no direct proof.


However have no idea for the limit.


Edited: I am not sure, whether interchanging limit and integral is valid

  U;
  IntegrationTools[Change](%, (x-xi)/2/sqrt(t) = u,u)  assuming 0 <t;
  eval(%, arctan = 'u -> -Pi/2'); # limit x = infinity
  value(%);

    1/2*Pi+1-exp(-t)

Did it in Maple 12.

If not:

You may search the forum, if I remember correctly there was some script
or sheet (by Joe Riel?) to repair sheets.

PS: you may wish to consider to use 'clasiscal sheets', they are more easy
to handle for beginners (but would not handle Maplets)

uss_if_2_M12.mw

e*N[a]* xd *(1-x/ xd )/epsilon[s] will work for integrating w.r.t. x

e*N[a]*x[d]*(1-x/x[d])/epsilon[s] does not work

In case you want to see indexed parameters you change the integration
variable (in this case I use xi, but you can take another one, just not x):

  e*N[a]*x[d]*(1-xi/x[d])/epsilon[s];
  int(%, xi);

                                 /           2 \
                                 |         xi  |
                     e N[a] x[d] |xi - 1/2 ----|
                                 \         x[d]/
                     ---------------------------
                             epsilon[s]

They use separate plots and combine them, like

p1:=plot(sin(x)   +  0, color=black):
p2:=plot(sin(x^2) +  5, color=blue):
p3:=plot(cos(x)   + 10, color=red):

plots[display]({p1,p2,p3});

I get the error "Error, missing operator or `;`". And can not read off what you want.

Suggestion: throw ir away and start from scratch.

And whenever you want (nested) IFs then first write (and use manual idents!)

if ... then
  abc
elif ... then
  abc
else
  abc
end if;

Which means: type in the 'body' first, then fill up and use new lines and identation.
And end your statements by some ';'

You can do by writing the following in you ini file

  option1:= 1.0; # use some ':' instead of ';' to avoid displaying it

But better avoid it: after some weeks or month you will hate it.


If really needed I do only define pathes, something like

  myLib:="D:\\_Work\\Maple_Work\\z_Packages\\myLib\\":   # in the ini file

and put all my my stuff in that directory as *.mpl files (=text files with code)

Then for example I use (like an 'include' in a C program)

  read cat(myLib, "theFileWithCodeForVeryTask.mpl");

Or just append 'libname' with that directory.

Not very smart - but I am aware of what I do. And only have to care for that
file, and do not have to care for the ini file. 


Within the ini file I only have stuff which I suppose is needed often and
will never change (well ... almost). Other and very experienced users try
to avoid any 'hidden' use through ini files. 

Do not use the same name for integration range and variable (what should it be?).
That is what the error message says.

Int(1/((1-a*sin(phi)^2)^(1/2)),phi = 0 .. Phi), a = (1-1/(kappa^2)) ~ 0.99.
Maple wants to know that there is no singularity and something about periods.

As long as you have 0 <= Phi < Pi/2 you can use

  Int(1/((1-a*sin(phi)^2)^(1/2)),phi = 0 .. Phi);
  IntegrationTools[Change](%, phi=arcsin(t),t); # there is an inverse
  value(%);
  eval(%, a = (1-1/(kappa^2)));

    EllipticF(sin(Phi),(1-1/(kappa^2))^(1/2));

However it is always worth to crosscheck results by numerical integration.

It is a bit disapointing that the originator fires a question and does
not feel any need to specify the question due to the discussion.

Anyway, here is an example, where minimize works and the variant
Optimization:-Minimize fails (with low degrees - strange).

m:= x^4*y^2 + x^2*y^4 - 3*x^2*y^2 +1; # Motzkin

PS: was looking for E. Artin, Uber die Zerlegung definiter Funktionen
in Quadrate, Abh. Math. Sem. Univ. Hamburg, 5 (1927), but only can
find Springer's access for a distgusting price of 35 € for 16 pages,
off copyright they are, bastards ( = Hilbert's 17th problem, rational)

You solve for the variables {a0, a1, a2, c, k} and that
depends on {a, b, mu, lambda} as parameters, no?

sys:={eq1,eq2,eq3,eq4,eq5};

indets(sys, symbol);
% minus {a0, a1, a2, c, k};

                             {a, b, mu, lambda}

So your (polynomial) system has 4 parameters, on which the
solution would depend.

Ok, that is too brute (so I would reply that having such a high degree it is
likely to have numerical troubles evaluating).

Let us assume, the polynomial has integer or rational coefficients and that it
is not too extreme in degrees (avoiding to much rounding errors or such ... or
just to simplify the task or to avoid to set Digits=very high).

Using parts of the idea one can try to use a random list for x an y (or loop
around that, starting in x=0=y).

Then one can use the command 'sturm' (or something similar) to look for the
existence of real roots.

I think, that it is very unlikely, that root = minimum (i.e. just touching the
axis) and in case that (just from sign changes) one root exists one can look
closer after finding it (say by fsolve with maxsolutions=1).

For the above example

  subs(x=0,y=0, p): 
  sturm(%,x,-infinity,+infinity);

    4

subs(x=0,y=0, p): fsolve(%,z, maxsols=1 );

                             -1.605634375

convert(%, rational);
subs(x=0,y=0, z=%, p):

                                -39497
                                ------
                                24599


If no roots are indicated then shoot some new test values.


Likewise one can test on arbitrary triples directly.

Not sure, what is more reasonable (having that degree of 100 in each variable)

But that is no proof it is non-negative, it is a kind of test that may fail 
(i.e. can only have a chance to dis-proof it).

In all variables the polynomial function must be of even degree (else it would reach +inf and -inf).

Then you may try to find the minima and check the function values.

@Markiyan Hirnyk 

We had sun, rain and hail this day - no spring, just a typical day in April :-(

Find the sheet appended, perhaps one of those local minima which are mentioned
by Sergey (errors are larger, yes, ~ 10^(-2) or about ~ 0.5 %).

Time for a patriotic (wheat) beer and some non-patriotic spaghetti + wine later.

MP_deep_fit.zip

M1:=Matrix([[1,0],[0,1]]):
M2:=Matrix([[1,0],[0,1]]):

LinearAlgebra:-Equal(M1,M2);

                                 true