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 Dec 13 awarded Popular Question 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, $xl$ for the second part. The values you give don't satisfy $a>l$.