If you use polar coordinates, you can smartly work outside the singularity.

We have x=R*cos(theta), y=R*sin(theta), and z=1/R^2.

plots[contourplot]([R*cos(theta),R*sin(theta),1/R^2],R=0.1..1,theta=0..2*Pi,labels=[x,y]);

The wikipedia page for mathtype indictates that latex can be entered directly into mathtype. So use Maple to generate latex code for your equations.

> latex(Int(x^x,x))

Of course Maple's latex command is in serious need of polish, but that's another story.

Any group G acts on any set X trivially. You map G to Aut(X) by sending each g in G to the identity automorphism. So you probably need to sharpen your question.

If G is a group of order 12 then the number of Sylow 3 subgroups is 1 or 4. If it is 4, then G acts transitively on this set of 4 subgroups, and it turns out G is isomorphic to A4. If there is one Sylow 3 subgroup, then it is normal so G acts on this subgroup by conjugation.  This gives a map from G to S3, which is a subgroup of S4, hence an action of G on a set of size 4--one of the four objects being acted on trivially.

It is straightforward to find a linear transformation (matrix) that takes the vectors A-D, B-D, C-D to the <1,0,0>, <0,1,0>, <0,0,1>. This transforms the tetrahedron to the one defined by x+y+z<1 with x, y, z positive.

Apply the transformation to the vector P-D to get vector P'. It is easy to check if P' is in the transformed tetrahedron.

Another approach is to convert the coefficients to rational numbers before integrating:

F:=convert(f(x),rational);

int(F,x=0..2); evalf(%);

0.1471400297 x 10^8

This is an old problem that does not get fixed

http://www.mapleprimes.com/posts/42882-CopyPaste-Problem--Mac-Integral-1003

Another possibility:

> Dirac(0):=a;

> h := sum(Dirac(i-1), i = 1 .. f);

> eval(subs(f=3,h));

Are u(0.25) and u(0.16) prescribed? If so, you could split this up into 3 separate boundary value problems: one on 0..0.16, another on 0.16..0.25, and another on 0.25..1 and then use dsolve(**, numeric). You should not expect the global solution to be differentiable at the two interior points.

If u(0.25) and u(0.16) are not prescribed, then your problem is not well-posed. Your ode is then u''+const+ uu'=0 and you need to prescribe the value of const before you can expect a unique solution. There is a simple solution when const=0.

You can solve your equation for Vb to get a function Vb(ph). You can then use the inverse function theorem to calculate diff(ph(Vb),Vb) since you can calculate diff(Vb(ph),ph).
 

Notice the second graph below reproduces the implicitplot of eq.

> F:=solve(eq,Vb);VB:=unapply(F,ph);
> VB(3.8):
> plot(VB(ph),ph=3.8..11.75);
> plot([VB(ph),ph,ph=3.8..11.75]);

It is not clear if you want n to be an integer. Assuming it is a continuous variable, you can go as follows:

f:=(t,n)->exp(-0.4*I*ln((2+t)/(2-t)))*exp(I*n*t)/sqrt(4-t^2);

data:=[seq([j*.1,evalf(1/Pi*Int(Re(f(t,j*.1)),t=-2..2))],j=-40..40)]:

plot(data);

You need

plot3d(phiX1[i](x1,x2),x1=x2..1-x2,x2=0.. 1);

not

plot(phiX1[i](x1,x2),x1=x2..1-x2,x2=0.. 1);

Now that I look back, I am pleased to read that this problem was reported to be "fixed" in 2007.

http://www.mapleprimes.com/posts/40437-Funny-Behavior-With-Int#comment74508

Note: Wolfram Alpha gets this definite integral correct.

You could get away with t_^3/t^2

"Assuming" never really seems to go very far.

For example:

> assume(y::prime);

> is(y,positive);

true

> is(sqrt(y)>0);

FAIL

> assume(a::positive);


> is(sqrt(a)>0);

true

I seem to recall that the documentation says to not load packages as you are doing 

with(Statistics):

Instead, you should use Statistics[RandomVariable](....)

You are doing this for "Fit" but not for "RandomVariable" and "Sample"