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1

See if this is getting close.. Kpart[A_?NumericQ][x_, y_] := (A/2)*(BesselJ[1, A*(x + y)] - BesselJ[1, A*Abs[x - y]] - I*StruveH[1, A*(x + y)] + I*StruveH[1, A*Abs[x - y]]) f2[A_?NumericQ, x_?NumericQ] := FredholmKind2[{a, b, lambda, Kpart[A], Gpart}, n, ...


4

There exist few typos in your kernel definition. This is how your kernel looks assuming a=2 (we denote it as A while defining the kernel as Kpart in the following). Please utilize the code from here to solve your problem. Below I changed the constants and functional arguments of FredholmKind2 to fit your particular problem. n = 20;(*number of ...


1

As I said in a comment I object to the downvaluing of my solution by somebody anonymus without giving the reasong for it. Now you can convince yourself here that my solution of the real problem is correct. And, consequently, that the second integral of "identy" (and sequel) is not. Here we go: I gather from your first lines that you wanted to calculate ...


0

I think what you mean in your second integral is the integral between y^2/2 and \[Infinity]: Integrate[(E^-t ArcCos[y/(Sqrt[2] Sqrt[t])])/\[Pi], {t, 1/2 y^2, \[Infinity]}] It is returned unevaluated by Mathematica: (* $\int_{\frac{y^2}{2}}^{\infty } \frac{e^{-t} \text{ArcCos}\left[\frac{y}{\sqrt{2} \sqrt{t}}\right]}{\pi } \, dt$*) Therefore, let's ...


3

Rahbar and Hashemizadeh's approach You could use the code provided by PlatoManiac in response to a similar question which implements the method from Rahbar and Hashemizadeh's paper A Computational Approach to the Fredholm Integral Equation of the Second Kind. NISolve.nb The Mathematica library includes code for Numerical Solution of One-Dimensional Linear ...


0

Sometimes it is good to find the general integration and then find the bounded one using substitution. int[y_] = Integrate[x/(x^2 + y^2)^(3/2), y]; int[a] - int[-a] (*(2 a)/(x Sqrt[a^2 + x^2])*)


3

You can throw your assumptions into Assuming as follows: Assuming[{y ∈ Reals, x > 0, a > 0}, Integrate[x/(x^2 + y^2)^(3/2), {y, -a, a}]]


2

The second case integrates analytically, so you can do a series expansion: series = Normal@Series[ Integrate[Exp[-alpha Sqrt[x]], {x, 0, y}], {alpha, Infinity, 3}] E^(-alpha Sqrt[y]) (-(2/alpha^2) + (2 E^(alpha Sqrt[y]))/alpha^2 - (2 Sqrt[y])/ alpha) Plot[{NIntegrate[Exp[-alpha Sqrt[x]], {x, 0, 1/2}] , series /. y -> 1/2 }, {alpha, 0 , 10}] ...



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