highBandWidth
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 Jul2 awarded Curious Mar20 awarded Yearling Sep18 awarded Caucus Jul7 asked Assumptions with D Jul5 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. Jul5 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? Jul5 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}]? Jul4 awarded Supporter Jul4 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? Jul4 revised Can't compute definite integral added 5 characters in body Jul4 asked Can't compute definite integral Jun28 revised Normal[Series[ ]] does not give a normal expression added 342 characters in body Jun28 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. Jun28 asked Normal[Series[ ]] does not give a normal expression Jun28 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])]. Jun28 comment Inverse Laplace transform not obtained I agree, but is there an alternative to Laplace transforms for solving complicated equations? Jun27 asked Inverse Laplace transform not obtained Mar21 awarded Student Mar20 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 $a>l$. Mar20 awarded Editor