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Motivation & Question

So I can theoretically build a "computer" to calculate the exact anti-derivative of a particular function.

Using classical calculations and the Robin boundary condition I show that one calculate the anti-derivative of a function within time $2X$ (I can compute an integral below)

$$\frac{2 \alpha}{\beta} e^{ \frac{\alpha}{\beta}X} \int_0^{X} e^{- \frac{\alpha}{\beta}\tau} \tilde f_0(\tau) d \tau$$

where $\tilde f_0$ is an arbitrary function whose integral above converges.

Of course this assumes I measurements infinitely accurate. But I was curious if my measurement has accuracy up to $z$ decimal places could it out do a computer simulation trying to obtain the same result up to $z$ decimal places? ($\alpha$ and $\beta$ are arbitrary constants)

What is the quickest algorithm to numerically integrate the above function? For completeness $\Phi$ is a function which obeys the massless Klien Gordon equation in 2 dimensions

$$ (-\partial_t^2 +\partial_x^2)\Phi(x,t)=0\tag{1}$$

The solution to the Mass-less Klein-Gordon Equation is $$\Phi(x^+,x^-) =\frac{1}{2}\Big(\tilde f_0(-\sqrt 2 x^-) + \tilde f_0(\sqrt 2 x^+) + \int_{-\sqrt 2 x^-}^{\sqrt 2 x^+} f_1(z) dz \Big) \tag{2}$$

and

$$x^\pm = \frac{1}{\sqrt 2} (t \pm x) \tag{3}$$

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  • $\begingroup$ You have $\overline f_0,f_0,\Phi$ appearing at different places and seemingly unrelated. This is quite confusing. $\endgroup$ Commented Jun 1, 2022 at 13:20
  • $\begingroup$ @YvesDaoust $\Phi$ is defined by $(1)$ and the relation between $\Phi$ and $f_0$ is given by $(2)$ via equation $(3)$. Also I (previously) accidently used the notation $\tilde f_0 = f_0$ (now edited). Hopefully this works? $\endgroup$ Commented Jun 1, 2022 at 13:28
  • $\begingroup$ Anything about the asymptotic behavior of $f_0$ ? (There is also this mysterious $f_1$...) $\endgroup$ Commented Jun 1, 2022 at 13:30
  • $\begingroup$ @YvesDaoust Nope. Here's where $f_1$ is coming from if this helps: en.wikipedia.org/wiki/D%27Alembert%27s_formula $\endgroup$ Commented Jun 1, 2022 at 13:33
  • $\begingroup$ No better answer than Gauss-Legendre integration, then. $\endgroup$ Commented Jun 1, 2022 at 13:47

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