- #1

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## Main Question or Discussion Point

I feel so stupid for asking this question, but I want to understand how this integral:

[tex]\int^{\infty}_{0}d\alpha \int^{\infty}_{0} d\beta \frac{i}{[4 \pi i(\alpha + \beta)]^\frac{D}{2}} e^{[i \frac{\alpha\beta}{\alpha + \beta}p^2 - i(\alpha + \beta)m^2]} [/tex]

can be transformed into this:

[tex]\int^{\infty}_{0}dx \int^{1}_{0} dy \frac{1}{[4 \pi i x]^\frac{D}{2}} e^{[-i x (m^2 -y(1-y)p^2)} [/tex]

The change of variables are:

[tex]\alpha =xy[/tex]

[tex]\beta = x(1-y)[/tex]

My question is both these equations are in terms of two variables and I have never come across how to do this kind of integral. I am guessing that we just to substitute:

[tex]d\alpha =xdy+ydx[/tex]

and

[tex]d\beta = (1-y)dx -xdy[/tex]

into the original integral.

However, if I do this, I don't get the second expression... I get something quite messy.

Can anyone shed any light on this?

[tex]\int^{\infty}_{0}d\alpha \int^{\infty}_{0} d\beta \frac{i}{[4 \pi i(\alpha + \beta)]^\frac{D}{2}} e^{[i \frac{\alpha\beta}{\alpha + \beta}p^2 - i(\alpha + \beta)m^2]} [/tex]

can be transformed into this:

[tex]\int^{\infty}_{0}dx \int^{1}_{0} dy \frac{1}{[4 \pi i x]^\frac{D}{2}} e^{[-i x (m^2 -y(1-y)p^2)} [/tex]

The change of variables are:

[tex]\alpha =xy[/tex]

[tex]\beta = x(1-y)[/tex]

My question is both these equations are in terms of two variables and I have never come across how to do this kind of integral. I am guessing that we just to substitute:

[tex]d\alpha =xdy+ydx[/tex]

and

[tex]d\beta = (1-y)dx -xdy[/tex]

into the original integral.

However, if I do this, I don't get the second expression... I get something quite messy.

Can anyone shed any light on this?