Question

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 !

Answer 1

hard to tell, its not parsing

Answer 2

\[\int_{R^2} e^{-(x+y)^2}~e^{-(x-y)^2}dxdy=\frac{\pi}{2}\]
you either gad an extra } in it or something

Answer 3

do we know the little domain region of dx dy?

Answer 4

The integral is over \[{\mathbb R}^2\]

Answer 5

hmmm, so kinda similar to a normal curve being over the whole of R ....

Answer 6

show us your calculus ....

Answer 7

It is so long, I used Substitution for multiple variables
\[(u,v)=(x+y,x-y)\]

Answer 8

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.

Answer 9

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:
\[\frac{1}{2} \int\limits_{\mathbb{R}}e^{-u^2}du \int\limits_{\mathbb{R}}e^{-v^2}dv=\frac{1}{2} \left( \sqrt{\pi} \right)^2=\frac{\pi}{2} \text{qed}\]

Answer 10

typo: \(e^{-(u^2+v^2)}\) instead of \(e^{-(u^2+y^2)}\).

Answer 11

Its a 5-6 liner depending on the jacobian.

Answer 12

another inaccuracy, I called \(I=\pi\) the gaussian integral. actually this \(I\) is the square of it. as @malevolence19 shows.

Answer 13

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:
\[\sqrt{\frac{\pi}{2}} \sqrt{\frac{\pi}{2}}=\frac{\pi}{2} \text{ qed}\]

Answer 14

Thanks for all of you !