Try to get used to write exp(z), not e^z (or use exp(1)^z if you really need it)

May be you have an error in coding?

If I take Denom(s, 2*M0*m, g, m) from your poleR for your data

'Denom(s, 2*M0*m, g, m)';
#subs(M0=M0, g=g, m=mK, %); indets(%, symbol);
subs(M0=M0, g=.935, m=mK, %); 
H:=unapply(%, M0, s);
indets(%%, symbol);

then you want to solve for s, if a specific M0 = 0.93 is given.

'H(0.93,s)';
plot(%, s=0.9 .. 1);

indicates that there is no real solution for your expected s ~ 0.95.

'H(0.93,x+y*I)';
Re(%):
plot3d((%), x=0.9 .. 0.98, y=-0.1 .. 0.1, axes=boxed);

indicates that there is no complex solution.

0.1*num/den; subs(x[3]=1,x[4]=2,%); simplify(%);
                                                            
                          0.0999999999999999

Note that using floating point numbers you may not get exact
results, they are approximations.

But you may use 1/10 instead of 0.1

Edited. Or using other wordings: for numerical calculations
the usual mathematical rules are only valid up to an error
(for example the 'route' of calculation or input matters).
And that happens for any (computer) system

Is this similar to your older question at http://www.mapleprimes.com/questions/201767-Fsolve-Cannot-Solve-This-Equations-And but now with log(-exp(...)) ?

I am not sure whether 3 1/2 is a good idea in Maple (but I always do use multiplication sign).

However the following gives something close

3 + ``(1/2); 
                              3 +  (1/2)

expand(%);

                                 7/2

There are various ways. The inner integral is Int(f1, y=0 .. h(x)).
Compute an anti-derivative and name ist V. To apply the fundamental
theorem V has to be continuous along the curve h(x). Substitude y =
ymax and plot real and imaginary part

  V;
  eval(%, y=ymax);
  plot([Re(%),Im(%)], x=0 .. Pi, color=[red,blue]);

You 'see' one will split in x=Pi/2 (and moreover the imaginary part
will drop out). Finally I get 3.13316121793696 (15 decimals)

PS: What is the result given by Matlab?

evalf(Int(u(t)^2, t = 0..1)); # upper case

    .448698034001023

Note that your 1st integral needs 0 < k (take t=0, k= -1, lambda= -1,
then it is not finite).

Changing by k*s = x, abbreviating k=lambda/a one finally has the task

task:=Int(exp(-exp(-x)+(a-1)*x),x = -infinity .. t); # a new t

Changing by x = -ln(-ln(y)) this is

task1 := Int((-ln(y))^(-a),y = 0 .. exp(-exp(-t)))

The exp term is between 0 and 1 (plot it, for example), so y is between
0 and 1. Hence z = ln(y) is between -oo and 0 (smells like a pdf, no?)

We have (-z)^(-a) = z^(-a)*(-1)^(-a)*exp(2*I*Pi*a) and for z=ln(y):

task1 = faktor * task2, task2 := Int(ln(y)^(-a),y = 0 .. exp(-exp(-t))),
faktor := (-1)^(-a)*exp(2*I*Pi*a)

In that form one can cheat asking Mathematica (or perhaps Gradsteyn?),
Anti := GAMMA(-a+1,-ln(y))*ln(y)^(-a)/((-ln(y))^(-a)) is an antiderivative

eval(Anti, y= exp(-exp(-t))) - MultiSeries:-limit(Anti, y=0, right);
my:=simplify(%) assuming t::real;

is an intermediate solution (no check for continuity) and then

faktor*my; 
simplify(%) assuming a::real;
evalc(%); 
My:=simplify(%);

                     My := GAMMA(-a + 1, exp(-t))
 
is a solution, I think.

Check 1: in integers

task = My;
'[seq(%, a= - 3 .. 3)]'; value(%): expand(%): simplify(%); map(is, %);

              [true, true, true, true, true, true, true]

 
Check 2: in ugly floats

Digits:=15;
task = My;
subs(t=1, %);
'[seq(%, a= - 3 .. 3, 1 + evalf(1/Pi))]'; evalf(%); #map(is, %);

  [5.99658218952224 = 5.99658218952226,

        1.50284712032528 = 1.50284712032527,

        0.737383845311871 = 0.737383845311871,

        0.750709217778417 = 0.750709217778418,

        1.29447312393282 = 1.29447312393282]


So 'task' is an integral for the incomplete Gamma function and knowing
that there should be a more direct way.

From that your initial 1st task will follow. The 2nd is just a product, no?

PS: Where does your series of questions come from?

The inner integral is of the form Int(f(x), x=0.. phi(y)) and using
IntegrationTools[Change](%, x=z*phi(y), z) this becomes 

   phi(y)*Int(f(z*phi(y)),z = 0 .. 1)


Now combine with the original task one can write a double integral
Int(K, [z=0..1, y=0..Pi], epsilon=1e-8) and evalf gives a good result:

      3.13316050306181 + .647491e-17*I

I did it a bit differently:

Maple finds an antiderivative w.r.t. z. Call it val, "starring" at it
result this seems to be continous in for those y, so I get a formal
solution for the inner integral Int(K, z=0 .. 1).

val:=int(K, z);
S:=eval(%,z=1) - eval(%,z=0)

Now Int(Re(S), y=0..Pi, method = _d01ajc); evalf(%);

      3.13316050516358

= -.666758443186807e-2 + (11/2*Ei(1,11/2)-exp(-11/2)+1)*Pi

For Im(S) like the above it is almost vanishing.

Plotting K's imaginary part a point-symmetry is seen in y=Pi/2, and
plotting left side + right side gives me a zero plot (ok, no proof),
so I would only consider the real part.

I used Digits:=15

The most easy way ist Compiler:-Compile, it comes with Maple.

Else you may search this site for "DLL", this should bring up most of stuff (may be it does not dicuss issues of 32 vs 64 bit DLL). And of course you can can look into the examples coming with Maple (+ coding manual), subdirectory "samples" of your installation.

I would suggest to use Compiler:-Compile - except you have _good_ reason to take another route.

I think it is roughly  exp( 0.3 E16331116 ), could that be?

I made an error, sorry. I get Carl's result as .4666594270e-3: converting to rationals the inner integral can be done anlytically (assuming 1 < eta) and then the overall integral can be computed with relativ error of 1E-10 quickly using method = _d01ajc

It "reduces" to 0 =  -2^(-1/4+1/4*c^2+c)+2^(-1/4-c+1/4*c^2)-c, c=cos(2*x)
by using 'combine'

Maple can not show "only for c=0", neither can (online-) MMA, where I mean
c (or x) to be complex.

From that it follows x = 1/4 *(2 *Pi *n+Pi), using Maple, as MMA asserts.

cos(x)^(n-1)*sin(x)+n*(Int(cos(x)^(n-2), x))-(Int(cos(x)^(n-2), x)); 
combine(%, Int); 
simplify(%, size);

           /
          |        (n - 2)                    (n - 1)
          |  cos(x)        (n - 1) dx + cos(x)        sin(x)
          |
         /

Your recursion formula is at Wikipedia as well (with ugly proof), http://en.wikipedia.org/wiki/Integration_by_reduction_formulae#Examples

Using Maple you can "verfiy" it:

F(n) = cos(x)^(n-1)*sin(x)/n +(n-1)/n*F(n-2);
eval(%, F= 'n -> Int(cos(t)^n, t=0 .. x)' ); # substitute the integral
'diff(%, x)';
simplify(%); # so up to constants it is ok 
 
eval(%%%, x=0);

Now one can use 'rsolve' to solve the recursion, giving a noisy result

Have not tried cos(x)^n = Sum in lower terms seriously (that sum is what is given by the hypergeometric 2F1 and the correcting factors)

CAmax:=(-k2+(k2*k1)^(1/2))*CA0/(k1-k2);
CAmaxgiven:= CA0*(k2/k1)^(1/2)/(1+(k2/k1)^(1/2));

CAmax - CAmaxgiven:
simplify(%) assuming 0<k1,0<k2;

                                  0

Edited: or even more simple:

CAmax - CAmaxgiven; 
subs(CA0=1, %); subs(k1=1,%);  # justify that !
normal(%);

                                  0