# Tag Info

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] *)

15

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 ...

15

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: ...

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

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 ...

7

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 ...

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 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 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 ... 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 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 ... 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 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 ... 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

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 + ...

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 $$\frac{d}{dq}(qI_q)=4\pi\int_{0}^{\infty}P(r)\cos(qr)\,dr.$$ This relation can be ...

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*) ...

1

Your syntax is wrong for Assuming Clear[fx]; fx[x_] := Piecewise[{ {(1/2) a*E^(-a*x), x >= 0}, {(1/2) a*E^(a*x), x < 0}}] Assuming[ {x \[Element] Reals && a \[Element] Reals && a > 0 && t \[Element] Reals}, Integrate[fx[x], {x, -Infinity, t}]] Piecewise[{{E^(a*t)/2, t <= 0}}, ((1/2)*(-1 + ...

1

First: There might be a typo in your expression as b2 and c2 do not appear. This is the solution for the expression like it is written in your question: When you have more information on parameters, you can pass it to the Integrate function in the Assumptions option: Integrate[..., Assumptions -> b1 > 0 && b2 > 0 && c1^2 >= 0 ...

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

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