# Tag Info

## Hot answers tagged integral-transforms

19

You can use a trick to prevent Mathematica from taking your expression apart: LaplaceTransform[Abs[1 - Cos[t]]/t, t, s] (* 1/2 Log[1 + 1/s^2] *)

10

This is not an answer, but a hint how to possibly proceed. It is a pity to pass on some nice Wolfram Language abilities. Let me know if I should remove it. Idea is: I think YOU should be the first to derive this integral - if it is possible. Consider only rational numbers. If this integral behaves smoothly - which it obviously does: ...

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 + ...

5

Here is a series-expansion way of doing your integral. Expand the integrand as Sum[(-1)^n Sin[θ/2]^(n β), {n, 0, ∞}] (* 1/(1+Sin[θ/2]^β) *) Integrate each term in the sum using Integrate[Sin[θ/2]^(n β), {θ, 0, π}] (* ConditionalExpression[(Sqrt[π] Gamma[1/2 (1 + n β)])/Gamma[1 + (n β)/2], Re[n β] > -1] *) Fold it all together to obtain ...

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

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 = ...

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 ...

3

This is a way which will give you a the final solution in terms of $\beta$. The idea is generate a Table of solution for different $\beta$ value and then find a fitting function. I choose here $-10<\beta<10$. data = Table[{\[Beta],NIntegrate[1/(Sin[\[Theta]/2]^\[Beta] + 1), {\[Theta],0,\[Pi]}]}, {\[Beta], -10, 10, .5}]; ListPlot[data] sol1 = ...

3

I'm not a mathematician so this may be "smoke and mirrors." You are looking for the inverse Laplace transform of g[s_] = (1 - Exp[-Sqrt[1 + s] x])/(1 + s); Looking at the simpler problem InverseLaplaceTransform[g[s - 1], s, t] // Simplify[#, x > 0] & 1 - Erfc[x/(2*Sqrt[t])] With x > 0, let f[t_] = E^-t (1 - Erfc[x/(2*Sqrt[t])]); g[s] == ...

3

This works in V9.0.1: Assuming[0 < a < b, Integrate[ArcCos[x/Sqrt[(a + b) x - a b]], {x, a, b}, GenerateConditions -> False]] // Timing (* {3.835651, ((a - b)^2 π)/(4 (a + b))} *)

3

There are several issues here with the way you are proceeding: FIRST, if $X$ is a random variable with pdf $f(x)$, then the moment generating function (mgf) is defined as: $$E[e^{t X}] = \int_{-\infty }^{\infty } e^{t x} f(x) \, dx$$ where the expectation is carried out over the full domain of support, which for a Normal distributed random variable ... is ...

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

Another quick way is to use the following trig identity In[13]:= TrigFactor[(1-Cos[2t])] Out[13]= 2 Sin[t]^2 If t==2*p, then we have: LaplaceTransform[2 Sin[p]^2/p, p, s] (* 1/2 (-2 Log[s]+Log[-2 I+s]+Log[2 I+s]) *)

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

\$Version "9.0 for Mac OS X x86 (64-bit) (January 24, 2013)" ft[w_] = FourierTransform[Sinc[t], t, w] (1/2)Sqrt[Pi/2](Sign[1 - w] + Sign[1 + w]) Plot[ft[w], {w, -2, 2}, PlotStyle -> Directive[Red, Thick]] InverseFourierTransform[ft[w], w, t] == Sinc[t] // FullSimplify True

1

From definition so f[z] is evaluated first, if known, and then the ztransform is computed. So what you get back is the value of of the function, not the function itself. In this case, Mathematica simplified Floor[x] to x, which I think due to the discrete assumption in the definition. I tested this in Maple, and Maple did the same assumption. f = ...

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 *)

1

Try : Integrate[BesselJ[1, t]/t t BesselJ[0, x t], {t, 0, Infinity}, Assumptions -> { x > 0}]

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