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1

The iteration can be performed as follows: A[t_] = {{t, 0}, {Cos[t], 1}}; B[t_] = {{Exp[t], 0}, {0, Exp[-t]}}; T[F_] := Expand[A[t] + Integrate[B[s].F /. t -> s, {s, 0, t}]]; iteration = NestList[T, {{t, 0}, {0, t}}, 7]; Column[Framed /@ MatrixForm /@ iteration[[1 ;; 3]]] Well, it's not immediately clear what the limits are, as it was over in this ...


0

Unless you have hidden functions and variables: n = 4; (*Put whatever term you want here*) x0 = 0; (*This is the point you want to expand around*) (*Define g[s] here*) g[t] = t + Integrate[f[s], {s, 0, t}] ; Series[g[t], {t, x0, n}]; Normal[%][[n]] (*This is the answer you want*)


3

I assume you wanted the Neumann series method. ClearAll[f, t, s]; f[0] = t; f[n_] := f[n] = Integrate[(f[n - 1] /. t -> s) s, {s, 0, t}]; data = Table[f[i], {i, 0, 10}] Total@data % /. t -> .5 (*0.543827*)


2

T[f_] := t + Integrate[f /. t -> s, {s, 0, t}]; NestList[T, t, 5] (* Out: {t, t + t^2/2, t + t^2/2 + t^3/6, t + t^2/2 + t^3/6 + t^4/24, t + t^2/2 + t^3/6 + t^4/24 + t^5/120, t + t^2/2 + t^3/6 + t^4/24 + t^5/120 + t^6/720} *) Looks like $f(t)=e^t - 1$ is a fixed point. T[E^t - 1] (* Out: -1 + E^t *)


2

Your integral is very unlikely to exist in terms of elementary functions. In particular, it involves terms of the form $$ \int\exp\left[-\frac12\sqrt{a\, \text{poly}(\xi)+b \xi^{0.998906}}\right]\text d\xi, $$ which is very unfriendly as regards symbolic integration. Note that in general symbolic integration is not possible; do you have some specific reason ...


0

You can also depict the expected value nicely as a function of Mu and Sigma using Plot3D. px = Simplify[PDF[LogNormalDistribution[Mu, Sigma], x], x > 0] E^(-((Mu - Log[x])^2/(2 Sigma^2)))/(Sqrt[2 \[Pi]] Sigma x) Plot3D[NIntegrate[Log2[1 + x] px, {x, 0, \[Infinity]}], {Mu, -5, 10}, {Sigma, .2, 2}, AxesLabel -> {Mu, Sigma, EY}] PS: Reminds me ...


3

If I understand your question correctly - as judged from the text typed - the answer should read (writing - 6x instaed of just - 6 as some others have done, put the function in brackets before multiplying with Boole, also use +- Infinity which is better than some arbitrary (?) finite value like +-2) Integrate[(y^2 - 2*x^2*y + 6*x^3 - 3*x*y + 2*y - 6*x)* ...


3

Amplifying on the answer by Chenmingi: Boole will automatically restrict the integral to the appropriate region so you can integrate from -Infinity to Infinity for each of the variables. int1 = Integrate[ (y^2 - 2 x^2 y + 6 x^3 - 3*x*y + 2 y - 6) * Boole[y >= 2*x^2 - 2 && y <= 3*x], {x, -Infinity, Infinity}, {y, -Infinity, Infinity}] ...


5

Clear["Global`*"] expr = y^2 - 2 x^2 y + 6 x^3 - 3*x*y + 2 y - 6; Integrate[ expr Boole[y >= 2*x^2 - 2 && y <= 3*x], {x, -Infinity, Infinity}, {y, -Infinity, Infinity}] -(36625/2688) Reduce[y >= 2*x^2 - 2 && y <= 3*x, y] (x == -(1/2) && y == -(3/2)) || (-(1/2) < x < 2 && -2 + 2 x^2 ...


1

I suspect there may not be a closed form expression (I have not looked at it hard enough). If the aim is to not analytical but numerical, I post the following for illustration (apologies if not the intent of question): f[x_, y_] := NIntegrate[ Log2[t + 1] Exp[-(x - Log[t])^2/(2 y^2)]/(Sqrt[2 Pi] t y), {t, 0, Infinity}] rv[a_, b_, n_] := ...



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