Symmetry in Integrals: Peskin's Equation 6.43 & 6.44

[physichu]

In peskin p. 192, they says that the denominator (that is equation 6.43) is symmetric under x<--> y. Thay all so say that you can see it in equation 6.44.

But one of the terms in the denominetor is y*q which dose not have that symmetry!
Looking at (6.43) and removing the summetric parts leave me with
2yk⋅q +yq².
Whitch is not x<-->y symmetric.

 

[Einj]

Your denominator is what he calls D, which is completely symmetric under x <-> y. He obtains that expression from D starting from 6.43 and changing the integration variable.

 

[physichu]

I don't see how :(
We have:

$\int\limits_0^1 {dxdydz\delta \left( {x + y + z - 1} \right){{2{q^\mu }m\left( {z - 2} \right)\left( {x - y} \right)} \over {\left[ {{k^2} + 2k \cdot \left( {yq - zp} \right) + y{q^2} + z{p^2} - \left( {x + y} \right){m^2} + i\varepsilon } \right]}}}$.

Changing x<-->y gives:

$\int\limits_0^1 {dxdydz\delta \left( {x + y + z - 1} \right){{ - 2{q^\mu }m\left( {z - 2} \right)\left( {x - y} \right)} \over {\left[ {{k^2} + 2k \cdot \left( {\underline {xq} - zp} \right) + \underline {x{q^2}} + z{p^2} - \left( {x + y} \right){m^2} + i\varepsilon } \right]}}}$

It's tempting to say that the integration region is summetric under x<-->y, so that the "x" instead of a "y" doesn't metter, but i found thise exemple:

$\int\limits_0^1 {dxdy \cdot x = } \int\limits_0^1 {dxdy \cdot y = {1 \over 2}}$

and

$\int\limits_0^1 {\left( {x - y} \right)dxdy} = \int\limits_0^1 {\left( {y - x} \right)dxdy = 0}$

But

$\int\limits_0^1 {dxdy\left( {x - y} \right)x} = \int\limits_0^1 {dxdy\left( {y - x} \right)y} = {1 \over {12}}$

I expected it to be zero as a multiplication of symmetric and anti symmetric factors but it turned out to be summetric.
What do i miss?

 

[Einj]

You first have to make the change of variables explained after eq. 6.43. After that you'll be able to explicitely see the symmetry

 

[physichu]

O.K. got that :)
than'x a lot :):):)

 

[Einj]

You're welcome!