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I'm trying to integrate $\int_{-\infty}^{\infty}\frac{1}{2\pi}e^{(-\frac{1}{2}(\frac{x^2}{4}+4y^2))} dy$ using the fact that the integral of any normal PDF is 1. But I'm having trouble completing the square for $(\frac{x^2}{4} + 4y^2)$. Can anyone help me please? Thanks!

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  • $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ – S. Kolassa - Reinstate Monica May 5 at 5:13
  • $\begingroup$ Hint: You don't need to complete a square. You only integrate over $y$, not $x$, so you treat $x$ as a constant. $\endgroup$ – S. Kolassa - Reinstate Monica May 5 at 5:13
  • $\begingroup$ The solution of this question is already posted, but the details of solving the integral are omitted. $\endgroup$ – qhy May 5 at 5:22
  • $\begingroup$ I think in the course we always use the completing square trick, and we can use that trick exactly because x is a constant. Otherwise integrating the pdf with respect to y does not yield 1. $\endgroup$ – qhy May 5 at 5:24
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    $\begingroup$ This is just a normal pdf multiplied by a constant. Pull out the parts that don't depend on $y$ and you should be able to solve it pretty quickly. $\endgroup$ – Reinstate Monica May 5 at 5:24
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You don't need to complete a square. You only integrate over $y$, not $x$, so you treat $x$ as a constant.

$$ \int_{-\infty}^{\infty}\frac{1}{2\pi}e^{(-\frac{1}{2}(\frac{x^2}{4}+4y^2))} dy = \frac{1}{2\pi}e^{-\frac{x^2}{8}}\int_{-\infty}^{\infty}e^{-2y^2} dy. $$

Multiply and divide by the same factor that turns the integral into an integral over the PDF of a normal distribution with mean $0$ and variance $\frac{1}{4}$, since this integral is one:

$$ = \frac{1}{2\pi}e^{-\frac{x^2}{8}}\sqrt{2\pi\times\frac{1}{4}}\underbrace{\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\times\frac{1}{4}}}e^{-\frac{y^2}{2\times\frac{1}{4}}} dy}_{=1} = \frac{1}{2\pi}e^{-\frac{x^2}{8}}\sqrt{2\pi\times\frac{1}{4}} = \frac{1}{2\sqrt{2\pi}}e^{-\frac{x^2}{8}}.$$

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