@janhardo 

It's a standard iterated integral. Use:

int(1/(1-x), [x=0..y, y=0..z]) assuming -1<z,z<1;

[strangely, abs(z)<1 does not work now].
You can also compute int(int(1/(1-x),x),x) + c1*x+c2; and adjust the constants such that f(0)=f'(0)=0.
 

@janhardo 

@janhardo 

It is not "included"; it can be "deduced from".

Actually, the standard metod is the following:

You have the series S:=Sum((-1)^(n+1)/(n*(n+1)),n = 1 .. infinity):

Now you associate to it the power series

f := x -> Sum(x^(n+1)/(n*(n+1)),n = 1 .. infinity):

So, S = f(-1).

Now observe that f''(x) = Sum(x^(n-1), n = 1 .. infinity):

But this is a geometrical series, so f''(x) = 1/(1-x).

It remains to integrate twice (and obtain f in terms of ln) and then make x=-1.

@janhardo 

This makes sense if you are doing the computations by hand (but then you don't need Maple), of if the Maple version is not able to compute the series directly (which is not your case -- Maple 2020).

BTW, the series can be computed by hand easier starting from sum((-1)^(k-1)/k, k=1..infinity) = ln(2)
or from limit(sum(1/k, k=1..n) - ln(n), n=infinity) = gamma

@Carl Love 

I don't agree:

1. It should be obvious (Leibniz)
2. Why? F could be anything. The constant does not matter.
3. Why? The hint was to detect S (somehow) in the series of F.

@radaar  It is not complicated to find the saddle points (numerically of course) for f:
1. Compute stationary points i.e. solve the system grad(f) = 0. I'd recommend DirectSearch which works well (download it from the Application Center).
2. Compute the Hessian at these points and select those which are saddle points.

@DanishMapleFan 

For this linear equation isolve gives a parametric solution. Use:

restart;
iS:=isolve({169 = 10*x + 3*y});
S:=solve({%[],x>=1,y>=1})[];
seq( iS, _Z1 = ceil(lhs(S[-2])) .. floor(rhs(S[-1])) );

@Sam2231 
A short procedure for this.

PSubs:=proc(eq::`=`, e)
local f;
uses Physics;
f:=proc(e)
  local i,j;
  for i to nops(e) do for j from i to nops(e) do
    if is(`*`(op(i..j,e)) = lhs(eq)) then return `*`(op(1..i-1,e), rhs(eq), op(j+1..-1,e)) fi
  od od;
  e
end:
subsindets(e, `*`, f)
end:

with(Physics):
Setup(op = {A, B, S, omega, v, x, x_, `&omega;_`}):
ex:=(((Iota[c][i]*Dagger(diff(x[i](t), t))) . (Dagger(A[i]))) . (A[i])) . (diff(x[i](t), t))/L[i](t)^2;
PSubs( Dagger(A[i]) . (A[i]) = -A[i]^2, ex );

@acer 

Yes, I see. On the other hand, using seq(a, ...) when a is local does not make much sense, a having only 1 level evaluation.

@Carl Love 

This seems to be equivalent to seq (simple form only).

SEQ:=proc(e::uneval, eq::uneval)
   local j, jold:=eval(lhs(eq)), S:=NULL;
   assign(lhs(eq),j);
   for j from lhs(rhs(eq)) to rhs(rhs(eq)) do S:=S,eval(e) od;
   assign(lhs(eq),jold);
   S;
end:

a:=i^2: i:=10:  
seq(a, i=1..3); i;
                            1, 4, 9
                               10
SEQ(a, i=1..3); i;
                            1, 4, 9
                               10

@Axel Vogt 

It seems that the following well known fact applies: Statisticians can prove almost anything.

@Carl Love 

Your heuristic test looks OK to me (but I am not a statistician :-)).

@Carl Love 

I think you used a heuristic approach. In the cited paper the computed probabilities are Prob(there are j runs of length i), but only for i=1,2,3. For i>3 (the most interesting for our case)  the probabilities are considered too hard to compute.

(Actually in the paper are constructed 3 tests, for each i=1,2,3).

I don't know what P(run length = k)  means; in general a string has several runs. [BTW, the expected number of runs is (n+1)/2].

@Carl Love 

I see that you have not used the exact probabilities for runs of lengths 1,2,... (from the cited paper) but the approximations seem to be good enough for a simple statistical test.

The OP's expression is in the title (!)