# Tag Info

9

When Mathematica tries to pull the fraction apart, it gets $$\mathcal{L}_t\left[\frac{1-\cos(t)}{t}\right](s)=\mathcal{L}_t\left[\frac{1}{t}\right](s) - \mathcal{L}_t\left[\frac{\cos(t)}{t}\right](s)$$ While the cosine term has a Laplace-transform, $1/t$ doesn't have a transform. That might be the reason why Mathematica cannot solve it. The problem is, ...

9

Here is my attempt at an answer - I had to make up an example, and obviously much of what follows is dependent on details of this example. Edit However, what I believe this example shows quite clearly is that a finite set of tabulated data at discrete points does not suffice to guarantee a good inverse Laplace transform, because the analytic structure of ...

8

Finally I found the most promising algorithm proposed in this really good reference Manuel Guizar-Sicairos and Julio C. Gutiérrez-Vega, "Computation of quasi-discrete Hankel transforms of integer order for propagating optical wave fields," J. Opt. Soc. Am. A 21, 53-58 (2004). The authors call the algorithm pth-order quasi-discrete Hankel Transform (pQDHT) ...

6

Ok, I think I know why the integral worked, but not the Laplace transform. When using the integral, there is a pole at t=0 but this is a removable singularity. Series[1 - Cos[t], {t, 0, 6}] // Normal Now dividing by t (#/t) & /@ r So, the t in the denominator is gone. I do not know how Mathematica actually removed this pole at t=0 in the code, ...

5

just different forms: hilbertTransformV1[f_, u_, t_] :=Module[{fp = FourierParameters -> {1, -1}, x}, FullSimplify@InverseFourierTransform[-I (2 HeavisideTheta[x] -1) FourierTransform[f, u, x, fp], x, t, fp]]; hilbertTransformV2[f_, u_, t_] := FullSimplify[Convolve[f, 1/u, u, t, PrincipalValue -> True]/Pi] Looking at the 1/(1 + ...

4

If you don't want to resort to tricks, you can differentiate the transform over s first, which would bring -t downstairs and cancel 1/t. You can then take the transform for: LaplaceTransform[(1 - Cos[t]), t, s] (* 1/s - s/(1 + s^2) *) and then integrate this over s (with a negative sign, since differentiation produced an extra -1): -Integrate[1/s - s/(1 ...

4

If for pedagogical purposes you do not wish to use FourierSeries, here's one approach: ClearAll[x, y]; y[t_] = 110 Sin[120 Pi t] + 50 Cos[360 Pi t + Pi/3]; x[k_] := With[{T = 1/60}, Integrate[y[t]*Exp[-2 Pi I t k/T], {t, 0, T}] ] DiscretePlot[Abs[x[i]], {i, -10, 10}] You're not specifying T anywhere in your example. I don't actually know what you ...

4

I have two suggestions. You may: Establish a function that contains the integration $g(s)$. The bad news is that you may need a lot of time to evaluate the integration. In my opinion, in the inverse Laplace calculation, we don't have to give the analytical expression of $F(S)$. If I were you, I may write the function like: F[s_]:=(g = ...

3

Some integrals cannot be found in terms of a finite set of functions. I'm not an expert on what the limits of Mathematica are, but the following suggests to me that this integral is beyond them. The reason comes down, by a sequence of substitutions, to the fact that this integral returns unaltered: -Integrate[Erf[Sqrt[1 - w^2]], w] (* ...

2

I can get the integral by using the first quadrant (for positivity), placing an assumption on n, and recasting without Boole (I don't know why that was needed). in = 4 Integrate[(x^(2 n) + y^(2 n)), {x, 0, 1}, {y, 0, (1 - x^(2 n))^(1/(2*n))}, Assumptions -> n >= 1] (* Out[112]= (2^(2 - 1/n) Sqrt[\[Pi]] Gamma[1 + 1/(2 n)])/((1 + n) Gamma[(1 + ...

1

By manual inspection I got a closed formula for your integral: s[n_] := 2/(n + 1) Gamma[1/(2 n)] Gamma[(2 n + 1)/(2 n)]/Gamma[1/n] f[n_] := Integrate[(x^(2 n) + y^(2 n)) Boole[x^(2 n) + y^(2 n) < 1], {x, -1, 1}, {y, -1, 1}] For example s[16] == f[16] (* True *) Plot[s[x], {x, 1, 100}] So Limit[s[n], n -> Infinity] (* 0 *)

Only top voted, non community-wiki answers of a minimum length are eligible