In the first post, taking into account what you find in:

http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions

you are definitely left with this integral:

2*Pi*(int(sin(phi)*BesselI(0, sin(phi)), phi = 0 .. Pi))

wich Maple cannot do either.  I will post this in the beta test site.

 You are looking for this:

If I understand your question, you want to find (don't forget that
 you must use capitalize I for the imaginary part, not i)

> z = 625*exp(I*Pi*(1/2));

                                  z = 625 I
and you want to find
> z^(1/4);

                                    (1/4)
                                   z     
> r = simplify(625^(1/4));

                                    r = 5
now the angle will be
> theta := (1/2)*Pi;

                                    1   
                                    - Pi
                                    2   
> for k from 0 to 3 do z || k := 5*exp(I*(theta+2*k*Pi)*(1/4)); end do;

                                     /1     \
                                5 exp|- I Pi|
                                     \8     /
                                     /5     \
                                5 exp|- I Pi|
                                     \8     /
                                    /- 7     \
                               5 exp|--- I Pi|
                                    \  8     /
                                    / -3     \
                               5 exp|--- I Pi|
                                    \  8     /

 I thought I was helping myself by introducing "beta" when in fact, it hide that so simple expression of the exp( ).  Stupid beta :-(

Thanks Alec!

 if you give us a link since this is a web page in wich you must FIRST logon.  And since we don't know this information, we will never be able to go see your problem.

My guest to solve this problem if you want us to help you is to log to the web page corresponding at you problem and then do  File -> Save as.  Give it a name and upload this file in MaplePrimes using the green up arrow.  But check carefully.  Usually, when a file with the extension *.htm is save, there will be a sub-repertory called Image containing all the *.gif images of that web page.  So be sure to upload the wep page AND the repertory Image (with all the *.gif in it) that come with it.

I hope you will make sense of all this and succeed!

Thanks for analysing the région where the function exist.  But, I need a proof of the identity, not simple say it is true because whent Maple simplify, the two sides agree.  Le me put it in the situation.

My exploration is one of Ramanujan solution of:

arctan(2/n^2)=arctan(n+1)-arctan(n-1);
                       /2 \                                
                 arctan|--| = arctan(n + 1) - arctan(n - 1)
                       | 2|                                
                       \n /                                
S:=Sum(arctan(2/n^2),n=1..infinity);
                             infinity          
                              -----            
                               \               
                                )          /2 \
                               /     arctan|--|
                              -----        | 2|
                              n = 1        \n /
SN:=Sum(arctan(2/n^2),n=1..N)=Sum(arctan(n+1)-arctan(n-1),n=1..N);
            N                  N                                  
          -----              -----                                
           \                  \                                   
            )         /2 \     )                                  
           /    arctan|--| =  /    (arctan(n + 1) - arctan(n - 1))
          -----       | 2|   -----                                
          n = 1       \n /   n = 1                                
SN := arctan(N+1)+arctan(N)-arctan(1)-arctan(0);
                                                  1   
                      arctan(N + 1) + arctan(N) - - Pi
                                                  4   
S:=limit(SN,N=infinity)assuming N>1;
                                    3   
                                    - Pi
                                    4   

You can see the advantage of using this identity.  So what I need is to prove:

arctan((x-y)/(1+x*y))=arctan(x)-arctan(y);
                         / x - y \                        
                   arctan|-------| = arctan(x) - arctan(y)
                         \1 + x y/                        

 In graph theory,  I saw a large variety of Snark  wich are birdgeless cubic graph:

http://mathworld.wolfram.com/Snark.html
 

My question is does someone can tell me if, in french, you use "Graphe cubique sans ponts" or just "Snark."?

 Where are the terms "Gaussian elimination"?  The word "from " make it clearer, I agree.  But it seems to me that it is like asking to find "The lower triangular matrix from the PLU decomposition by Gaussian elimination"

Is it mathematically correct.?

if I write the following:

"Get  the lower triangular Matrix of the PLU decomposition using Gaussian elimination"

Do you agree?

 ...it would be this way:

> with(inttrans);

> alias(Y = laplace(y(t), t, s));

> f := 10*exp(-3*t);
> df := diff(f, t);

                             df := -30 exp(-3 t)
> DV := diff(y(t), t, t)+3*(diff(y(t), t))+2*y(t) = df;

                /  2      \                                        
                | d       |     / d      \                         
          DV := |---- y(t)| + 3 |--- y(t)| + 2 y(t) = -30 exp(-3 t)
                |   2     |     \ dt     /                         
                \ dt      /                                        
> ci := y(0) = 0, (D(y))(0) = 0;

                         ci := y(0) = 0, D(y)(0) = 0
> eval(laplace(DV, t, s), {ci});

            /                                                30  \
           =|+(+(*(^(s, 2), Y), *(*(s, Y), 3)), *(Y, 2)), - -----|
            \                                               s + 3/
> solve(%, Y);

                                      30         
                           - --------------------
                              3      2           
                             s  + 6 s  + 11 s + 6
> convert(%, parfrac);

                              15      15      30  
                           - ----- - ----- + -----
                             s + 1   s + 3   s + 2
> invlaplace(%, s, t);

                  -15 exp(-t) - 15 exp(-3 t) + 30 exp(-2 t)

 > eq1 := x^3+p*x = q;

                                     3          
                             eq1 := x  + p x = q
> eq2 := expand(subs(x = u+v, eq1));

                      3      2          2    3                
              eq2 := u  + 3 u  v + 3 u v  + v  + p u + p v = q
> eq3 := u^3+v^3+factor(3*u^2*v+3*u*v^2+p*u+p*v) = q;

                          3    3                          
                  eq3 := u  + v  + (u + v) (3 u v + p) = q
Since 
                                 u + v <> 0
 then

> eq4 := 3*u*v+p = 0;

                            eq4 := 3 u v + p = 0
> eq5 := u^3+v^3 = q;

                                     3    3    
                             eq5 := u  + v  = q
> sol := isolate(eq4, u);

                                            p 
                              sol := u = - ---
                                           3 v
> eq6 := expand(v^3*subs(sol, eq5));

                                  1   3    6    3  
                         eq6 := - -- p  + v  = v  q
                                  27               
> eq7 := lhs(eq6)-rhs(eq6);

                                  1   3    6    3  
                         eq7 := - -- p  + v  - v  q
                                  27               
> eq8 := eval(eq7, {v^3 = r, v^6 = r^2});

                                   1   3    2      
                          eq8 := - -- p  + r  - r q
                                   27              
> discrim(eq8, r);

                                 4   3    2
                                 -- p  + q 
                                 27        
For real solutions, we need that
> 4*p^3*(1/27)+q^2 > 0;

                                   4   3    2
                               0 < -- p  + q 
                                   27        
> sol := solve(eq8, r);

                                    (1/2)                          (1/2)
            1     1  /    2       3\       1     1  /    2       3\     
     sol := - q + -- \81 q  + 12 p /     , - q - -- \81 q  + 12 p /     
            2     18                       2     18                     
> v := sol[1]^(1/3);

                                                      (1/3)
                       /                        (1/2)\     
                       |1     1  /    2       3\     |     
                  v := |- q + -- \81 q  + 12 p /     |     
                       \2     18                     /     
> u := (q-v^3)^(1/3);

                                                      (1/3)
                       /                        (1/2)\     
                       |1     1  /    2       3\     |     
                  u := |- q - -- \81 q  + 12 p /     |     
                       \2     18                     /     
> x = u+v;

                                   (1/3)                                  (1/3)
    /                        (1/2)\        /                        (1/2)\     
    |1     1  /    2       3\     |        |1     1  /    2       3\     |     
x = |- q - -- \81 q  + 12 p /     |      + |- q + -- \81 q  + 12 p /     |     
    \2     18                     /        \2     18                     /     

Thank you all for your suggestions.

Hi Robert,

How do I tell Maple to give me only the real solution (integer or others)?  I remember vaguely that this is possible.  I have look in ?solve/details.

This problem force me to conclude that his formula is quite restricted as for the king of cubic you can solve because e1=2 et e2=11 are in fact of cubic form.  So it's a loop.  You have to understand that you can obtain the others two only if you find x=4 and then factorize.

Well, it was just a small look into the history of sqrt(-1)

 This is a known bug that was mentionned during beta testing.  It only happen is you are in document mode.  When this happen, just use a procedure like tatan2501 suggested.

> factorielle := proc (n) if n = 0 then 1 else factorielle(n-1)*n end if end proc;

> factorielle(30);

                      265252859812191058636308480000000

but add the option "remember" to have quicker calculation

> factorielle := proc (n) option remember; if n = 0 then 1 else factorielle(n-1)*n end if ;end proc;
> factorielle(30);

                      265252859812191058636308480000000
> factorielle(31);

                     8222838654177922817725562880000000

And my opinion is that in document mode (2-D Math) is easier to read and make a more beautiful look.   It's just to bad that the only way to paste the inputs and outputs in Maple primes is in 1-D Math.  But I don't want to start a tread, please!