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So the equation that I was trying to solve was:

$$I=\int\limits_{a}^\infty \frac{\exp(x)}{2\pi\lambda}\int\limits_0^\infty \frac{1}{\tau^2}\exp\left(-\frac{(x-\mu)^2}{2\tau^2}-\frac{(\log(\sigma/\tau))^2}{2\lambda^2}\right)d\tau dx$$.

Where $a\in\mathbb{R}$, $\mu\in\mathbb{R}$, $\lambda\in\mathbb{R}^+$ and $\sigma\in\mathbb{R}^+$.

Now for the entire integration-range the integrand is positive so if I change the order of integration, then I should get the same solution for this one. If I do that this gives:

$$I=\frac{1}{2\pi\lambda}\int\limits_{0}^\infty \frac{\exp(-\frac{(\log(\sigma/\tau))^2}{2\lambda^2})}{\tau^2}\int\limits_a^\infty \exp\left(-\frac{(x-\mu)^2}{2\tau^2}+x\right)dx d\tau$$.

The integral over x is doable and gives: $$\int\limits_a^\infty \exp\left(-\frac{(x-\mu)^2}{2\tau^2}+x\right)dx=\exp\left(\mu+\frac{\tau^2}{2}\right)\sqrt{\frac{\pi}{2}}\tau\left(1+\text{erf}\left(\frac{-a+\mu+\tau^2}{\sqrt{2}\tau}\right)\right).$$

Where erf($x$) denotes the error function. Now with the exponential going as $\exp(\tau^2)$ the integral over $\tau$ will diverge! (I also checked the limit $\tau\rightarrow\infty$ and this yielded that the integrand indeed went to $\infty$.

This seems strange to mee since I've done numerical integration of the distribution:

$$P(x) = \frac{1}{2\pi\lambda}\int\limits_0^\infty \frac{1}{\tau^2}\exp\left(-\frac{(x-\mu)^2}{2\tau^2}-\frac{(\log(\sigma/\tau))^2}{2\lambda^2}\right)d\tau$$, which is a mixture variance model, where the normal distribution is broadend by a lognormal distribution of the standarddeviation. And this one seemed to converge. I've also been able to calculate the integral over x for the case where $\lambda \rightarrow 0$ and this even gave me some analytical results! So the integral over $\tau$ should definetely be finite (for as far as I know) and the integration over x for $\exp(x)$ shouldn't blow up the intire thing because the $\exp(-x^2)$ should normally compensate for that. Does anyone know what I might be doing wrong ?

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