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Libniz

  • 2 years ago

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

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  1. Libniz
    • 2 years ago
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    @myininaya ,@amistre64

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

  3. Libniz
    • 2 years ago
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    show the work; I am not interested in solution

  4. Libniz
    • 2 years ago
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    @zarkon

  5. Libniz
    • 2 years ago
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    @Jemurray3

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

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

  8. Jemurray3
    • 2 years ago
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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).

  9. Libniz
    • 2 years ago
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    thanks, man

  10. Jemurray3
    • 2 years ago
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    Sure

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