<s>[INST] Issues with Feynman parameters As a sanity check, I have tried to evaluate a Feynman parameter integral, and have been unable to reproduce the textbook result. I wish to verify the identity
$$\frac{1}{ABC} = \int\limits_0^1\int\limits_0^1\int\limits_0^1dxdydz\frac{2\delta(x+y+z-1)}{[Ax + By + Cz]^3} ~\hat{=}~I.$$
We can use the delta function to do the integral over $dz$.
$$I = \int\limits_0^1\int\limits_0^1 dxdy \frac{2}{[C + (A-C)x + (B-C)y]^3}.$$
To do the integral over $y$ we introduce the substitution
$$w = C + (A-C)x + (B-C)y$$
This will give a factor of $(B-C)^{-1}$, and change the limits
\begin{align*}
I &= \int\limits_0^1 \frac{1}{B-C}dx \int\limits_{C + (A-C)x}^{B+(A-C)x}dw \frac{2}{w^3} \\
&= \int\limits_0^1 dx \frac{1}{B-C}\left[\frac{1}{[C+(A-C)x]^2} - \frac{1}{[B+(A-C)x]^2}\right]
\end{align*}
The first integral is the Feynman parameter identity for 2 terms in the denominator, which is much easier to verify and gives $1/AC$. The second can be solved by essentially the same substitution as above.
$$w = B + (A-C)x$$
which gives a factor of $(A-C)^{-1}$ in similar fashion to before.
\begin{align*}
I &= \frac{1}{(B-C)AC} - \frac{1}{(B-C)(A-C)}\int\limits_{B}^{A+B-C}\frac{dw}{w^2} \\
&= \frac{1}{(B-C)AC} - \frac{1}{(B-C)(A-C)}\left[\frac{1}{B} - \frac{1}{A+B-C}\right]\\
&= \frac{1}{B-C}\left[\frac{1}{AC} - \frac{1}{A-C}\frac{A-C}{B(A+B-C)}\right] \\
&= \frac{1}{B-C}\frac{B(A+B-C) - AC}{ABC(A+B-C)} \\
&= \frac{1}{ABC} + \frac{1}{AB(A+B-C)}
\end{align*}
I don't know where this extra term has come from, and I can't seem to figure out where I went wrong.
 [/INST] The problem is in the very first step sadly. When you resolve the $\delta$ function you are putting
$$
z = 1 - x - y\,.
$$
This will hold only when $x+y \leq 1$ because you are integrating only in the region $z \in [0,1]$. In other words, the zero of the $\delta$ function sometimes falls outside of the region of integration and thus gives no contribution. The correct way to resolve the $\delta$ function is
$$
I = \int_0^1 \mathrm{d}x \int_{0}^{1-x}\mathrm{d}y \frac{2}{[C + (A-C)x + (B-C)y]^3}\,.
$$
If you follow the same exact steps as you did in your post with the limits modified this way, you'll get the right answer.
</s>