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visits member for 2 years, 7 months
seen Aug 25 at 21:42

Jul
2
awarded  Curious
Mar
20
awarded  Yearling
Sep
18
awarded  Caucus
Jul
7
asked Assumptions with D
Jul
5
comment Can't compute definite integral
@J.M., if you have a toy example where Mathematica does something non-optimal with the elliptic integrals, sharing it along with the better manual solution might shed light on the issue.
Jul
5
comment Can't compute definite integral
@J.M., perhaps that means Mathematica can improve it. Do they take suggestions like this? Does someone from Wolfram read this?
Jul
5
comment Can't compute definite integral
You have integrated along the circle boundary, rather than the circular disk. Perhaps we need a second integral Integrate[ . , {r,0,b}]?
Jul
4
awarded  Supporter
Jul
4
comment Can't compute definite integral
Thanks for the insight! Now if we go back to the problem however, the integrand is always finite, even if $r=x+b$ or $r=x-b$. So shouldn't the integral also be finite? I think the terms in red actually cancel with the other such terms in the limit. How should this be evaluated?
Jul
4
revised Can't compute definite integral
added 5 characters in body
Jul
4
asked Can't compute definite integral
Jun
28
revised Normal[Series[ ]] does not give a normal expression
added 342 characters in body
Jun
28
comment Normal[Series[ ]] does not give a normal expression
Plot[Evaluate@Normal[Series[Sin[x], {x, 0, 3}]], {x, -[Pi], [Pi]}] but if I try to do both as Plot[{Sin[x], Evaluate@Normal[Series[Sin[x], {x, 0, 3}]]}, {x, -[Pi], [Pi]}], it doesn't work while Plot[{Sin[x], x - x^3/6}, {x, -[Pi], [Pi]}] does.
Jun
28
asked Normal[Series[ ]] does not give a normal expression
Jun
28
comment Inverse Laplace transform not obtained
Well, I have an integral equation, specifically a Volterra equation like $y(x) = f(x) + \int_0^{x+a}y(t)K(t-x) dt$ where I have to solve for $y$. The answer needs InverseLaplace[Laplace[K]/(1-Laplace[K])].
Jun
28
comment Inverse Laplace transform not obtained
I agree, but is there an alternative to Laplace transforms for solving complicated equations?
Jun
27
asked Inverse Laplace transform not obtained
Mar
21
awarded  Student
Mar
20
comment Integral not simplifying
@partial81, In assumptions, we know that $a>l>0$, so the integral from $0$ to $a$ can be broken into $0$ to $l$ and $l$ to $a$. Then, $x<l$ for the first part and $x>l$ for the second part. The values you give don't satisfy $a>l$.
Mar
20
awarded  Editor