@ecterrab 

If I do:

Fundiff(g_[~mu,~nu](X),g_[alpha,beta](Y))

then I get:

Dirac([x1-y1, x2-y2, x3-y3, x4-y4])*g_[`~alpha`, `~mu`]*g_[`~beta`, `~nu`]

which is wrong. I should get an extra minus sign. But I might be mistaken. This remains even after setting: usephysicsevaluator=false.

Therefore I guess that even explicitly writing down Ricci scalar in terms of g_mu nu it would give a wrong result.

By the way, physics package looks promising, but if it is not possible even to evaluate the Einstein equation from the Einstein-Hilbert action, including sqrt{-g}, then I find it still should get exciting new features [fundiff applied to Ricci, Riemann, Ricci-scalar, sqrt{-g}, etc.]

But the first steps are interesting at least in flat space.

@Alejandro Jakubi 

Thanks a lot first of all. Everything worked as you said. Just great.

Question: what if my integrand is

Q := int(a(t)*f(x)^2, x = -infinity .. infinity)

If I set

f := proc (x) options operator, arrow; Int(fk(k1)*exp(I*k1*x), k1 = -infinity .. infinity) end proc

Then I get

and the function GetIntegrand seems not be working any more.

Thanks in advance.

@Alejandro Jakubi 

Thanks a lot first of all. Everything worked as you said. Just great.

Question: what if my integrand is

Q := int(a(t)*f(x)^2, x = -infinity .. infinity)

If I set

f := proc (x) options operator, arrow; Int(fk(k1)*exp(I*k1*x), k1 = -infinity .. infinity) end proc

Then I get

and the function GetIntegrand seems not be working any more.

Thanks in advance.

@Alejandro Jakubi 

Thanks a lot. I will try it. That looks like it may work.

Thanks.

@Alejandro Jakubi 

Thanks a lot. I will try it. That looks like it may work.

Thanks.

@Alejandro Jakubi 

@Alejandro Jakubi 

Let us suppose we have

Q1 := f(x)*g(x)

then define

f := proc (x) options operator, arrow; int(f1(k1)*exp(I*k1*x), k1 = -infinity .. infinity) end proc

g := proc (x) options operator, arrow; int(g1(k2)*exp(I*k2*x), k2 = -infinity .. infinity) end proc

If I ask to integrate Q1 in x from -infinity to infinity, although possible (the integral in x can be performed), Maple does not integrate. Now strictly speaking I cannot take the limit for k1=-k2 as they are part of integrals...

So the limit solution should not work. Sigh. :( Maybe there is a way to do so with the inttrans package. Maybe.

The applyrule could work if I would be able to change int(Q1,x=-infintiy .. infinity) into

int( int( f1(k1)*g1(k2) * int(exp(I*(k1+k2)*x),x=-infinity..infinity)))

but I have no idea how to change the order of integration, as originally I have

int( int(f1(k1)*exp(I*k1*x),k1=-infinity..infinity) * (int(g1(k2)*exp(I*k2*x),k2=-infintiy..infinity), x=-infinity..infinity),

and collect the term in the exponential.

Let us suppose we have

Q1 := f(x)*g(x)

then define

f := proc (x) options operator, arrow; int(f1(k1)*exp(I*k1*x), k1 = -infinity .. infinity) end proc

g := proc (x) options operator, arrow; int(g1(k2)*exp(I*k2*x), k2 = -infinity .. infinity) end proc

If I ask to integrate Q1 in x from -infinity to infinity, although possible (the integral in x can be performed), Maple does not integrate. Now strictly speaking I cannot take the limit for k1=-k2 as they are part of integrals...

So the limit solution should not work. Sigh. :( Maybe there is a way to do so with the inttrans package. Maybe.

The applyrule could work if I would be able to change int(Q1,x=-infintiy .. infinity) into

int( int( f1(k1)*g1(k2) * int(exp(I*(k1+k2)*x),x=-infinity..infinity)))

but I have no idea how to change the order of integration, as originally I have

int( int(f1(k1)*exp(I*k1*x),k1=-infinity..infinity) * (int(g1(k2)*exp(I*k2*x),k2=-infintiy..infinity), x=-infinity..infinity),

and collect the term in the exponential.

Thanks a lot!

Using the limit sounds great to me.

Thanks a lot!

Using the limit sounds great to me.

Yes, but the point is that for example, many times, especially in Quantum Mechanics, when we use Fourier space we need that integral to be recognized as a dirac delta. I was wondering if Maple has a way to interpret that undefined integral (which, by the way, is Maple's definition for the function Dirac) as Dirac. I guess the answer is no.

Honestly, it is a pity. I guess i have to do a substitution... something lke subs...

:(

Yes, but the point is that for example, many times, especially in Quantum Mechanics, when we use Fourier space we need that integral to be recognized as a dirac delta. I was wondering if Maple has a way to interpret that undefined integral (which, by the way, is Maple's definition for the function Dirac) as Dirac. I guess the answer is no.

Honestly, it is a pity. I guess i have to do a substitution... something lke subs...

:(