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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] *)


17

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 ...


17

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: ...


10

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 ...


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 ...


7

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, ...


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 & ...


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 ...


5

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 ...


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

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) ...


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

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))} *)


3

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, ...


2

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


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 + ...


2

It seems to be a branch cut issue. It depends when the substitution is done, that causes the integrator to go one way vs. the other. This below shows difference ClearAll[x, a]; sol = Integrate[1/Sqrt[-1 + a^2*Sech[x]^2], x] The integrand at x=0 is always zero, so we can ignore this, The issue is with the upper limit low = Limit[sol, x -> 0] (*0*) ...


2

I would use a slightly different derivation but the concept is basically the same. Here $I_q:=4\pi \int_0^{\infty}P(r)\frac{\sin qr}{qr}dr$. Hence $qI_q=4\pi \int_0^{\infty}P(r)\frac{\sin qr}{r}dr$. The derivative of $qI_q$ with respect to $q$ is \begin{equation} \frac{d}{dq}(qI_q)=4\pi\int_{0}^{\infty}P(r)\cos(qr)\,dr. \end{equation} This relation can be ...



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