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I've got the following integral to resolve:

$$\int_{-\infty}^\pi x\cdot \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx$$

If it were from $-\infty$ to $\infty$, is a first moment of gaussian distribution and i know how to solve. But with this domain of integration I have some doubts. Thank you in advance.

callculus42
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Diego
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1 Answers1

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With $z:=\frac{x-\mu}{\sigma}$ we can rewrite your integral as $\int_{-\infty}^a(\mu+\sigma z)\Phi^\prime(z) dz$ with $\Phi$ the $N(0,\,1)$ CDF and $a:=\frac{\pi-\mu}{\sigma}$. We can express this as $$\mu\Phi(a)+\sigma\int_{-\infty}^az\Phi^\prime(z)dz=\mu\Phi(a)-\frac{\sigma}{\sqrt{2\pi}}\exp -\frac{a^2}{2},$$where we have used $z\Phi^\prime=-\Phi^{\prime\prime}$.

J.G.
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  • J.G, are u sure i cannot simplify term near sigma? If i change domain of integration from a to infinite, i will have same result right? – Diego Apr 10 '19 at 18:24
  • @Diego If you integrate over $[a,,\infty)$ instead, you'll get $\mu\Phi(-a)+\frac{\sigma}{\sqrt{2\pi}}\exp-\frac{a^2}{2}$ (because the integral on $\Bbb R$ is $\mu$). (My comment concerns integration in $z$-space, not $x$-space.) – J.G. Apr 10 '19 at 18:25
  • J.G, thanks this will semplify all integral to zero, this was only half (first moment of skew normal distribution) – Diego Apr 10 '19 at 18:30