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4

This is a start $Version "10.0 for Mac OS X x86 (64-bit) (December 4, 2014)" assume = Element[{x, y, z, x0, y0, z0}, Reals]; f[x_, y_, z_] = Simplify[ ({x, y, z} - {x0, y0, z0})/Norm[{x, y, z} - {x0, y0, z0}]^3, assume]; f1[x_, y_, z_] = f[x, y, z][[1]]; Simplify[ Integrate[f1[x, y, z], {x, -1/2, 1/2}, Assumptions -> assume], assume] ...


6

Method 1, using Laplace transform eq = y'[t] + Integrate[y[x], {x, 0, t}] == Exp[-t]; eq = LaplaceTransform[eq, t, s]; eq /. LaplaceTransform[y[t], t, s] -> U0 sol = Solve[%, U0] Simplify@InverseLaplaceTransform[U0 /. sol, s, t] % /. y[0] -> 0 Method 2, convert to second order ODE You can, but you are missing a second initial condition. This is ...


1

Integrate[1/(1 + (f^2/B^2)), {f, 0, Infinity}, GenerateConditions -> False] π/(2 Sqrt[1/B^2])


1

Just specify the Assumptions for the Integrate: Integrate[1/(1 + (f^2/B^2)), {f, 0, Infinity}, Assumptions -> B > 0] (* (B π)/2 *)


2

Integrate[f, x] integrates the function that is a constant (with value f), so the answer is f x. Then you are solving f x=x^4. The correct answer is, of course, f=x^3. It looks like what you are trying to do is to find the function f[x], which, when integrated, gives x^4. The way to solve this is to take derivatives of both sides of the equation. The left ...


6

Large t Approximation Because this is a neutral delay integral-differential equation, solving it may seem very difficult at first glance. However, the term 1/5^t makes the integral negligible for t greater than about 2. So, with the integral ignored for large t, the equation can be reduced to x[t] - x[t - 1]/(2 E) = c[2] t^2 + c[1] t + c[0] where the ...


2

The consistent behavior displayed by the OP's integral -- increase recursion, increase the magnitude of the integral -- normally is the result of a divergent integral. It's possible the OP seeks the principal value of the integral. There is one pole of order 3 in the interval of integration f = (5184 (-11 + k (98 + k (-16 + k (-40 + k (17 + 2 k (7 + ...


2

There are too many singularities in your integrand (according to Reduce there are infinity of them, but these are complex valued) and then there are 3 real valued poles {-1., -3.82843, 1.82843} (I did not check for zero/pole cancellations). So the only real pole in the range of the integration is 1.82843 or -1 + 2 Sqrt[2] integrand = 5184 ((-11 + k (98 + k ...



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