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Let $f \colon \mathbb{R} \to \mathbb{R}$ be an exponential function defined by $f(x) = \exp (x)$.

Let $\pi$ be a normal distribution.

How to show that the integration $\int f(x) \pi( \mathrm{d}x)$ of unbounded real-valued function $f$ under Gaussian distribution is always finite?

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I'll show you how to do it in the case of the standard normal distribution ${\rm N}(0,1)$, leaving the general case to you.

The integral we wish to evaluate is $$ \int_{x=-\infty}^{x=\infty}dx \ e^{x}. \tfrac 1 {\sqrt{2\pi}} e^{-\frac 1 2 x^2}.$$ The first term is $f(x)$, and the second term is the probability density function for ${\rm N}(0,1)$.

We can combine the two exponentials as $e^{-\frac 1 2 x^2 + x}$, and then complete the square, i.e. write $- \frac 1 2 x^2 + x = -\frac 1 2 (x - 1)^2 + \frac 1 2$. Now, if we substitute $u = x-1$, the integral becomes $$ \int_{u=-\infty}^{u=\infty}du \ \tfrac 1 {\sqrt{2\pi}} e^{-\frac 1 2 u^2 + \frac 1 2} = e^{\frac 1 2} \int_{u=-\infty}^{u=\infty} du \ \tfrac 1 {\sqrt{2\pi}} e^{-\frac 1 2 u^2 } = e^{\frac 1 2}. $$

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