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Libniz

\[\int_{-inf}^{inf} e^{-x^2}e^{-x y} dx\]

  • one year ago
  • one year ago

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  1. Libniz
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    @myininaya ,@amistre64

    • one year ago
  2. gezimbasha
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    I think this is the solution\[e^{\frac{y^2}{4}}\sqrt{\pi}\]

    • one year ago
  3. Libniz
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    show the work; I am not interested in solution

    • one year ago
  4. Libniz
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    @zarkon

    • one year ago
  5. Libniz
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    @Jemurray3

    • one year ago
  6. Jemurray3
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    \[ \large \int_{-\infty}^\infty e^{-x^2} e^{-xy} dx = \int e^{-(x^2+xy)}dx = e^{y^2/4}\int e^{-(x^2+xy+y^2/4)}dx\] \[ \large = e^{y^2/4}\int e^{-(x+y/2)^2}dx = \sqrt{\pi}\cdot e^{y^2 / 4}\]

    • one year ago
  7. Libniz
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    how did you get \[\int e^{-(x^2+xy)}dx = e^{y^2/4}\] ?

    • one year ago
  8. Jemurray3
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    I didn't. I multiplied the integrand by exp(-y^2/4) and to compensate multiplied the outside by exp(y^2/4).

    • one year ago
  9. Libniz
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    thanks, man

    • one year ago
  10. Jemurray3
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    Sure

    • one year ago
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