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Given the weight function

$w: (-\infty, \infty) \rightarrow \Bbb R$,

$w(x) = e^{-|x|}$,

I need to show that

$\int_{-\infty}^{\infty} w(x) dx = 2$.

Unfortunately, I still have a lack of understanding about what the weight function actually is. I believe that this should work by Gauss quadrature somehow, but I don't know how to apply it.

Any kind of help is appreciated.

Julian
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  • By symmetry : $$ \int_{-\infty}^\infty e^{-|x|} dx = 2 \int_0^\infty e^{-x} dx = 2\cdot 1 $$ – Zubzub Jun 19 '17 at 09:02
  • Sometimes it is good not to overthink things. Clearly $\int_{-\infty}^{\infty} e^{-|x|} dx = 2 \int_{0}^{\infty} e^{-x} dx$... – Martigan Jun 19 '17 at 09:03
  • Would this be the same for something like $\int_{-\infty}^{\infty} x^3 w(x) dx = 0$? These excercises are part of an exam in Numerical Analysis, so I doubt that the goal is to solve them by calculating the integral directly. – Julian Jun 19 '17 at 09:07
  • Why not solve exactly the integrals that can easily be solved exactly, even in numerical analysis? Numerical analysis is for those integrals that cannot be solved exactly (or takes hours of work, or a once-per-year stroke of genius insight). – Arthur Jun 19 '17 at 09:09
  • But how would I solve this in a "numerical" kind of sense? – Julian Jun 19 '17 at 09:15
  • I doubt that I have to solve something like \int_{-\infty}^{\infty} x^3 e^{-|x|} dx directly. This would be way too complicated for an exam in Numerical Analysis since one has to use several integration techniques. – Julian Jun 19 '17 at 09:19
  • @Julian, notice that $x^3\exp(-|x|)$ is odd function, hence $\int_{-a}^{a}x^3\exp(-|x|)\text{d}x=0$ – Galc127 Jun 19 '17 at 09:27

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