I have ran into the following integral equation as part of my research.

For $\xi = (\alpha\theta)^{1/\alpha}$ and for all $\theta>0$.

I have the following equality $$\int\limits_0^\infty g(\kappa) \exp{\left[-\frac{\left(\xi + \kappa \right)^2}{2\alpha\theta}\right]} d\kappa % = \exp{\left[-\frac{\left(\xi - x_0\right)^2}{2\alpha\theta}\right]}%. $$ This condition can also be presented as follows $$ \int\limits_0^\infty g(\kappa)\exp{\left(-\kappa^2/(2\alpha \theta) -\xi\kappa/(\alpha\theta)\right)} d\kappa = \exp{\left(-\frac{x_0^2 - 2\xi x_0}{2\alpha\theta}\right)}$$ I assume $\kappa$ to depend on $\theta$, Is there a way to find $g(\kappa)$ here ? Can it be interpreted as some form of convolution ? Thank you.

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    $\begingroup$ I don't think so. You can take $g(k)$ to be basically any function then compute the integral and rescale $g$ such that the result matches. For example the (constant) function $$g(k) = \frac{2 e^{-\frac{\zeta -x_0}{2 \alpha \theta }}}{\sqrt{2\pi \alpha \theta } \cdot \text{erfc}\left(\frac{\zeta }{\sqrt{2\alpha \theta }}\right)}$$ works. If you restrict $g$ to be, say, a smooth function that only depend on $k$ and $x_0$ then it might be possible to find a unique solution (but I don't know). $\endgroup$
    – Winther
    Commented Jun 1, 2016 at 1:01
  • $\begingroup$ @Winther, thank you, i will think about that. $\endgroup$ Commented Jun 1, 2016 at 1:12
  • $\begingroup$ @Winther, I wanted to get your opinion about something, assuming g was a function of $\kappa$ and $x_0$. Can en.wikipedia.org/wiki/… be applied to find $g$. $\endgroup$ Commented Jun 1, 2016 at 3:33


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