You must provide epsfunc for a concrete answer.

@student_md 

Actually omega should be >0, otherwise you do not have an inner product.

@Carl Love 

Should be Joachimsthal.

Edit. Actually, the OP proves partly another (easier) result due to Joachimsthal:

Let [a*cos(t[i]), b*sin(t[i])], i=1..4, be four distinct points on the ellipse x^2/a^2+y^2/b^2=1.
These points are concyclic iff t[1]+t[2]+t[3]+t[4] is a multiple of 2*Pi.

Anyway, the OP does not seem to be very interested in his post.

@acer 

A more unpleasant fact is that Maple is not able to compute e.g.

evalf[20](Int(LegendreP(-1/2 + y*I, 3)*exp(-y), y=0..infinity));

The integrand is "nice" but the user will have to chop the interval.

@MapleUser2017 

You still have typos: Mylib <> MyLib, Mylib <> MyMat. Why don't you use copy+paste?

Please open the attached file and execute it using the (!!!) icon.

Download libex.mw

@Rouben Rostamian  

And also:

with(MyMat);

@MapleUser2017 

I suppose that the OS is Windows.

Probably you forgot to "execute" the module definition. Press Enter in its execution group.

1. Do you have the directory C:/temp  ? If not, you must create it (or use an existing one).
2. Do you have MyLib.mla in this directory after executing Save?

You may execute

LibraryTools:-ShowContents(LibLocation);
to see the contents of the library

The builtin method seems to be marginally faster than a simple "square" recursive method.

restart;
pow:=proc(a,n)
     if n=0 then return 1 fi;
     if n::odd then a*thisproc(a,n-1)
     else thisproc(a,n/2)^2 fi
end:
a:=841117; n:=311012;
CodeTools:-Usage(pow(a,n), iterations=100):
CodeTools:-Usage(a^n, iterations=100):

@Carl Love 

- No, the integral is from 1 to -1. Our results are the same.

- The integral is independent of the path in {z: Im(z)<0}, the function being holomorphic here. The limit was needed just because of this.

@Christian Wolinski 

The problem is fprintf(output, "%f\t %Zf\t \n", x, b),  not evalf

@Carl Love 

The problem is that the assume facility is not reliable in nontrivial situations.

is(2^n+r, rational) assuming n::integer, r::rational;
                              true
is(2^n+r^2, rational) assuming n::integer, r::rational;
                              FAIL

@Christian Wolinski 

Yes, but this was the point: using the current implementation it's impossible to decide whether a real number is real!

@Christian Wolinski 

A clear and proper answer should be true.
FAIL means "I don't know" for an obvious question.

@Carl Love 

A nice find, vote up! When I saw the sequence I suspected that a closed formula should exist. But I am sure that a formal proof is not easy. You should try the Collatz conjecture too :-)

@Kitonum 

Then you probably have the 32 bit version.