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I feel so stupid for asking this question, but I want to understand how this integral:
\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]}
can be transformed into this:
\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)}
The change of variables are:
\alpha =xy
\beta = x(1-y)
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:
d\alpha =xdy+ydx
and
d\beta = (1-y)dx -xdy
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?
\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]}
can be transformed into this:
\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)}
The change of variables are:
\alpha =xy
\beta = x(1-y)
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:
d\alpha =xdy+ydx
and
d\beta = (1-y)dx -xdy
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?
