anonymous
  • anonymous
I have calculated this integral and I found it like this \[\int_{ R\times R}e^{-(x+y)^2}{e^{-(x-y)^2}dxdy=\frac{\pi}{2}\] I want to be sure if my calculus is correct !
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
hard to tell, its not parsing
amistre64
  • amistre64
\[\int_{R^2} e^{-(x+y)^2}~e^{-(x-y)^2}dxdy=\frac{\pi}{2}\] you either gad an extra } in it or something
amistre64
  • amistre64
do we know the little domain region of dx dy?

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anonymous
  • anonymous
The integral is over \[{\mathbb R}^2\]
amistre64
  • amistre64
hmmm, so kinda similar to a normal curve being over the whole of R ....
amistre64
  • amistre64
show us your calculus ....
anonymous
  • anonymous
It is so long, I used Substitution for multiple variables \[(u,v)=(x+y,x-y)\]
reemii
  • reemii
Good. then you get \[ \iint_{\mathbb R^2} e^{-(u^2+y^2)} \frac12 \,dudv = \frac12 I \] and \(I\) is known to the "Gaussian integral": http://en.wikipedia.org/wiki/Gaussian_integral The value is \(I=\pi\). Therefore you are correct.
anonymous
  • anonymous
Its not too awful though, making that substitution you find the jacobian. From there you sub in and you get and integral over u and v. So you have: \[\frac{1}{2} \int\limits_{\mathbb{R}}e^{-u^2}du \int\limits_{\mathbb{R}}e^{-v^2}dv=\frac{1}{2} \left( \sqrt{\pi} \right)^2=\frac{\pi}{2} \text{qed}\]
reemii
  • reemii
typo: \(e^{-(u^2+v^2)}\) instead of \(e^{-(u^2+y^2)}\).
anonymous
  • anonymous
Its a 5-6 liner depending on the jacobian.
reemii
  • reemii
another inaccuracy, I called \(I=\pi\) the gaussian integral. actually this \(I\) is the square of it. as @malevolence19 shows.
anonymous
  • anonymous
Even alternative path: \[\iint_{\mathbb{R} \times \mathbb{R}} e^{-x^2-y^2-2xy}e^{-x^2-y^2+2xy}dxdy=\int\limits_{\mathbb{R}^2}e^{-2x^2} e^{-2y^2} dxdy\] From there use: \[\int\limits_{\mathbb{R}} e^{- \sigma x^2}dx=\sqrt{\frac{\pi}{\sigma}} \text{ with } \sigma > 0\] Applying this on both integrals we get: \[\sqrt{\frac{\pi}{2}} \sqrt{\frac{\pi}{2}}=\frac{\pi}{2} \text{ qed}\]
anonymous
  • anonymous
Thanks for all of you !

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