Symmetry of an Integral: Is it Zero?

[LagrangeEuler]

If function is $f(-x,-y)=f(x,y)$, is then

$\int^{a}_{-a}\int^{a}_{-a}f(x,y)dxdy=0$?
Thanks for answer.

 

[CompuChip]

No, consider for example
$$\int_{-a}^a \int_{-a}^a x^2 y^2 \, dx dy$$

 

[LagrangeEuler]

Could you tell me some explanation why this is valid only for one integral?

 

[jgens]

Could you tell me some explanation why this is valid only for one integral?

The result does not hold for 1-dimensional integrals either. If f(x) = x², then f(x) = f(-x) and integrating over any interval of the form [-a,a] (where a ≠ 0) gives you a non-zero number.

If you have a function that satisfies f(-x,y) = -f(x,y) then the integral over [-a,a] × [-a,a] should be zero. So you should probably look for a condition like this.

 

[LagrangeEuler]

If function is $f(-x,-y)=f(x,y)$, is then

$\int^{a}_{-a}\int^{a}_{-a}f(x,y)dxdy=0$?
Thanks for answer.

Sorry I thought about $f(-x,-y)=-f(x,y)$

Is it always valid?
Also is it valid in case $f(-x,-y)=f(x,y)$ that
$\int^{a}_{-a}\int^{a}_{-a}f(x,y)dxdy=2\int^{a}_{0}f(x,y)dxdy$

 

[lurflurf]

For double integrals there are four cases
f(x,y)
f(-x,y)
f(x,-y)
f(-x,-y)

f(-x,-y)=-f(x,y)
implies that
f(x,-y)=-f(x,-y)
thus
the integral would be zero

f(-x,-y)=f(x,y)
implies
f(-x,y)=f(x,-y)
$\int^{a}_{-a}\int^{a}_{-a}f(x,y)dxdy=2\int^{a}_{-a}\int^{a}_{0}f(x,y)dxdy$

 

[LagrangeEuler]

Tnx.