Thanks, that's it. I was doing the same with an "assume(a>0)" command before the limit command but it did not work, don't know why.

Thanks, that's it. I was doing the same with an "assume(a>0)" command before the limit command but it did not work, don't know why.

@Markiyan Hirnyk 

For instance the following sum: 

evalf( sum ( (int(r*R(n, r)*V(r)*u(r), r = 0 .. infinity))^2/(-1+1/n^2), n = 2 .. infinity));

where:

u(r) is the particular solution mentioned above: 

2*r*exp(-r)*ln(r)+2*r*exp(-r)*ln(2)+2*exp(-r)*r^2-2*r*exp(-r)

V(r)=1/r

R(n,r)=2*exp(-r/n)*LaguerreL(n-1, 1, 2*r/n)/(n^(5/2))

(hydrogen ground state wave functions)

@Markiyan Hirnyk 

For instance the following sum: 

evalf( sum ( (int(r*R(n, r)*V(r)*u(r), r = 0 .. infinity))^2/(-1+1/n^2), n = 2 .. infinity));

where:

u(r) is the particular solution mentioned above: 

2*r*exp(-r)*ln(r)+2*r*exp(-r)*ln(2)+2*exp(-r)*r^2-2*r*exp(-r)

V(r)=1/r

R(n,r)=2*exp(-r/n)*LaguerreL(n-1, 1, 2*r/n)/(n^(5/2))

(hydrogen ground state wave functions)

Thanks for commenting. Well, the problem is that you cannot really work with such a form, for instance calculating of infinite sums of integrals involving this and other functions did not work for me. With the solution obtained by Carl Love above, this is possible.

Thanks for commenting. Well, the problem is that you cannot really work with such a form, for instance calculating of infinite sums of integrals involving this and other functions did not work for me. With the solution obtained by Carl Love above, this is possible.

Thank you for your response. Sorry again for such an uclean formulation of my very first question on this forum. 

My orinal problem is:

-((D@@2)(y))(r)-2*y(r)/r+y(r) = (2*(2-1/r^2))*r*exp(-r)

(so 2 instead of 4/9 on the right side - but this of course does not change the nature of the problem)

I won't post again the original maple solution, the one from mathematica, however, should have been:

-E^r C[2] + E^-r r (-2 + 2 r + C[1] + 2 C[2] ExpIntegralEi[2 r] + 2 Log[r])

If I use the identity you mentioned, maple gives:

r exp(-r)*_C2+_C1*exp(r)-2*_C1*r*exp(-r)*Ei(2*r)+2*r*exp(-r)*ln(r)+2*r*exp(-r)*ln(2)+2*exp(-r)*r^2-2*r*exp(-r)

which differs from mathematica in the particular solution, there is on more term, namely:

2*r*exp(-r)*ln(2)

which is already covered in the term r exp(-r)*_C2 of the general solution.

What could be said to this?

Thank you for your response. Sorry again for such an uclean formulation of my very first question on this forum. 

My orinal problem is:

-((D@@2)(y))(r)-2*y(r)/r+y(r) = (2*(2-1/r^2))*r*exp(-r)

(so 2 instead of 4/9 on the right side - but this of course does not change the nature of the problem)

I won't post again the original maple solution, the one from mathematica, however, should have been:

-E^r C[2] + E^-r r (-2 + 2 r + C[1] + 2 C[2] ExpIntegralEi[2 r] + 2 Log[r])

If I use the identity you mentioned, maple gives:

r exp(-r)*_C2+_C1*exp(r)-2*_C1*r*exp(-r)*Ei(2*r)+2*r*exp(-r)*ln(r)+2*r*exp(-r)*ln(2)+2*exp(-r)*r^2-2*r*exp(-r)

which differs from mathematica in the particular solution, there is on more term, namely:

2*r*exp(-r)*ln(2)

which is already covered in the term r exp(-r)*_C2 of the general solution.

What could be said to this?

@Markiyan Hirnyk 

Thanks for notifying me and sorry for that. It is of course defined in my code. I forgot to copy here the definition of V(r) and u[0](r):

V := proc (r) options operator, arrow; 1/r^2 end proc;

u[0] := proc (r) options operator, arrow; 2 r exp(-r) end proc;

the result from Mathematica should be:

-E^r C[2] +  E^-r r (-2 + 2 r + C[1] + 2 C[2] ExpIntegralEi[2 r] + 2 Log[r])