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I'm trying to compute the following integral: $$ \lambda m(x) =\frac{1}{2\varepsilon} \int_{\max(\alpha x - \varepsilon, -1)} ^{\min(\alpha x + \varepsilon , 1)}m(y) \mathrm{d} y, $$ with $\alpha$ and $\varepsilon$ positive constants.

Assuming $\int_{-1}^{1} m(x) \mathrm{d} x = 1$ I should be able to solve for $\lambda$, which is the quantity I'm actually interested in.

So far I've only been able to solve for the trivial case where $\varepsilon > 1 + \alpha$, for which I obtain $\lambda = 1/\varepsilon$. Also, for $\alpha > 1$ and $\varepsilon < \alpha - 1$ I can get $\lambda = 1/\alpha$, while for $\alpha \leq 1$ and $\varepsilon < 1 - \alpha$, I have that $\lambda = 1$.

All other cases:

  1. $\alpha > 1$, $\varepsilon \in (\alpha - 1, \alpha + 1)$

  2. $\alpha < 1$, $\varepsilon \in ( 1 - \alpha, \alpha + 1)$

I can't find a solution to.

Do you know if there a "relatively easy" way to compute such integrals or a reference that could point me in the right direction? Thanks!

BB3C
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  • $\lambda$ can be interpreted as an eigenvalue of a certain integral operator. Have you explored this way ? – Jean Marie Oct 23 '23 at 14:10
  • The origin of the problem is indeed an eigenfunctional equation. In this setting (cases 1,2) it ensures that $\lambda \in (0,1)$ but I can't get anything else from it. – BB3C Oct 23 '23 at 14:30
  • I think it would be good that you give more context : what has brought you to be interested by this expression ? A personal intuition : isn't it amenable to a box-spline issue ? – Jean Marie Oct 23 '23 at 17:44
  • Can you assume $m$ an even function ? – Jean Marie Oct 23 '23 at 21:14
  • Yes, $m$ is an even function, I should have mentioned that. The question arises from computing quasi-stationary distributions in the theory of conditioned random dynamics. I'll look carefully at the box-spline problem, thanks! – BB3C Oct 24 '23 at 09:42

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