# Tag Info

20

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] *)

19

You can realize a discrete Hilbert transform by convolving your discrete signal with a Hilbert kernel. The convolution is implemented with least effort in the frequency domain, where the spectrum of the Hilbert kernel is $$\sigma_H(\omega)=-i\cdot\mathrm{sgn}(\omega)$$ where $\omega$ is the angular frequency. Continuous case We define a function to perform ...

18

Here's a direct implementation of the formula $$\mathcal H(u)(t) = \frac1{\pi} -\hspace{-1.1em}\int_{-\infty}^\infty \frac{u(\tau)}{t-\tau}\, \mathrm d\tau$$ hilbertTransform[f_, u_, t_] := FullSimplify[Convolve[f, 1/u, u, t, PrincipalValue -> True]/π] Try it out: hilbertTransform[#, v, w] & /@ {Sin[v], Cos[v], 1/(1 + v^2), Sinc[v], ...

11

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: ...

11

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

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) ...

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 ...

9

InverseLaplaceTransform[(2 s^2 + s + 13)/((s - 1) ((s + 1)^2 + 4)), s, t] 2*E^t + (3/4)IE^((-1 - 2*I)t) (-1 + E^(4*I*t)) % // ComplexExpand // Simplify 2*E^t - (3*Cos[t]*Sin[t])/E^t % // TrigReduce ((1/2)*(4*E^(2*t) - 3*Sin[2*t]))/E^t In "one" step InverseLaplaceTransform[(2 s^2 + s + 13)/((s - 1) ((s + 1)^2 + 4)), s, t] // ...

8

For continuous signal, I think is easy, took this in a course. For discrete, hard for me, we did not study it at school (yet). But if the signal is continuous, this gives the Hilbert transform of the signal: f[t_] := Sin[t]; g = FourierTransform[f[t], t, omega]; InverseFourierTransform[I* Sign[omega]*g, omega, t] ==> -Cos[t] P.S. I just saw a ...

8

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, ...

7

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 ...

6

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 + ...

6

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 ...

6

f = a[x] + c1*b[x] + c2*c[x]; g = Expand[f*f]; (* we need to get the constants out of the integrals first*) h = Distribute@Integrate[g, {x, -Infinity, Infinity}] //. Integrate[q1___ r__ q2___, {v_, s__}] /; FreeQ[{r}, v] :> r Integrate[q1 q2, {v, s}]; s= Solve[And @@ Thread[D[h, #] & /@ {c1, c2} == 0], {c1, c2}] (* now we go to ...

6

Mathematica's FourierTransform can be a bit difficult at times, so I'll let someone else address whether your code can be tweaked to work, and instead show how to do it symbolically. Defining $$f(\mathbf{r})=\exp\left(-|W\mathbf{r}|\right)$$ where $$W=\left( \begin{array}{ccc} \sqrt{w_x} & 0 & 0 \\ 0 & \sqrt{w_y} & 0 \\ 0 & 0 & ... 5 There are several issues here with the way you are proceeding: FIRST, if X is a random variable with pdf f(x), then the moment generating function (mgf) is defined as:$$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 ... 5 Similar idea to belisarius, except in V10 we can inactivate Integrate to keep it from evaluating or even trying to evaluate: h = Inactive[Integrate][g, {x, -Infinity, Infinity}] It is not necessary in this example, as belisarius' answer shows, but one of its intended uses is to do algebra/calculus on integrals and derivatives. Inactive can be removed ... 5 Adding assumptions finds two of the cases. Assuming[{a, t, r, z} ∈ Reals && t > z, Integrate[ DiracDelta[ t - a - (r^2 + (z - a)^2)^(1/2)]/(4 Pi (r^2 + (z - a)^2)^(1/2)), {a, -Infinity, Infinity}] ] (* 1/(4 π t - 4 π z) *) Assuming[{a, t, r, z} ∈ Reals && t < z, Integrate[ DiracDelta[ t - a - (r^2 + (z - ... 4 Another quick way is to use the following trig identity In[13]:= TrigFactor[(1-Cos[2x])] Out[13]= 2 Sin[x]^2 If t==2*x, then we have the integrand becomes: LaplaceTransform[2 Sin[x]^2/x, x, 2*s] (* 1/2 Log[1+1/s^2] *) 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 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 = ... 4 To begin with, in the definition of fx include the parameter "a" as an argument and write x_ instead of x : fx[x_, a_] := Piecewise[{{(1/2) a*E^(-a*x), x >= 0}, {(1/2) a*E^(a*x), x < 0}}] If you now you tell Mathematica that "t" is real and that a>0 there is no problem integrating fx: Integrate[fx[x, a], {x, -Infinity, t}, Assumptions -> {a ... 4 It's much simpler than what you're describing, if I understand you correctly. Assuming the space index comes first, you can simply call Fourier /@ data, where data is your cube of data. For example: dat = RandomReal[{0, 1}, {3, 100, 200, 300}]; recipDat = Fourier /@ A; Otherwise, Transpose the indices so that the spatial index (ie, the one with length 3) ... 4 Update First I'd like to mention that after checking the definition of bilateral Laplace transform and Fourier transform carefully, I'm sure currently the formula for the relationship between them on the wikipedia page is wrong, the correct one should be:$$\mathcal{B}\{f(t)\}(s) = \sqrt{2\pi}\mathcal{F}\{f(t)\}(is)$$Then we can use the internal function ... 4 The functions Fourier, FourierTransform, InverseFourier, and InverseFourierTransform (possibly others that I am not aware of) all accept a parameter called FourierParameters to determine the flavor of transform performed. By default, the value of FourierParameters is {0, 1}, which gives the results you've observed. To run a 'traditional' Fourier transform, ... 3 This answer contains no Mathematica, only mathematics. Let P be extended to negative input via P(-x)=P(x), and let k_q(x) = \sin(qx)/qx = \operatorname{sinc}\bigl(\frac q\pi x\bigr). Both P and k_q are even functions. You have$$\begin{align} I_q &= 4\pi\int_0^\infty P(r)k_q(r)\,\mathrm dr \\ &= 2\pi\int_{-\infty}^\infty ...

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))} *)

2

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

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 + ...

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