Feynman parametrization for $\frac{1}{\prod_{i=1}^N A_i^{\lambda_i}}$

Question

I want to derive the general formula for Feynamn trick in the usual one(the integral region is $[0,1]$) and the alternative one(Integral region is $[0,\infty]$).

Starting from Wiki, for the usual one, one finds

\begin{align} \frac{1}{AB} = \int_0^1 \frac{du}{(uA+(1-u)B)^2} = \int_0^1 dx dy \delta(x+y-1) \frac{1}{(xA+yB)^2} \end{align} Alternative version can be obtained by $x= \frac{y}{1-y}$, $dx = \frac{dy}{(1-y^2)}$, so \begin{align} \frac{1}{AB} = \int_0^1 \frac{dx}{(xA+(1-x)B)^2} = \int_0^1 \frac{dx}{(1-x)^2} \frac{1}{(\frac{x}{1-x} A +B)^{2}} = \int_0^{\infty} \frac{dy}{(yA+B)^2} \end{align} For the general exponent, I can do and obtain \begin{align} \frac{1}{A^{\lambda_1} B^{\lambda_2}} = \frac{\Gamma(\lambda_1+\lambda_2)}{\Gamma(\lambda_1) \Gamma(\lambda_2)} \int_0^{\infty} dx \phantom{1} \frac{x^{\lambda_2 -1}}{(A+xB)^{\lambda_1 + \lambda_2}} \end{align}

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How about the general $n$ products? For the usual one, I know the following formula \begin{align} &\frac{1}{D_1 D_2} = \int_0^1 dx \frac{1}{(xD_1+ (1-x) D_2)^2} \\ &\frac{1}{D_1^{k_1} \cdots D_n^{k_n}} = \frac{\Gamma(\sum_i k_i)}{\prod_i \Gamma(k_i)} \int_0^1 dx_1 \cdots \int_0^1 dx_n \frac{\delta(\sum_i x_i -1) \prod_{i} x_i^{k_i-1}}{(\sum_i x_i D_i)^{\sum_i k_i}} \end{align} How one can obtain the alternative version, i.e., integration with $\int_0^{\infty}$ form?

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The followings are my trials

\begin{align} \frac{1}{A^{k_1} B^{k_2} C^{k_3}} &= \int_0^1 \int_0^1 \int_0^1 \frac{\delta(x_1+x_2+x_3-1) dx_1 dx_2 dx_3 x_1^{k_1 -1} x_2^{k_2-1} x_3^{k_3-1}}{(x_1 A + x_2 B + x_3 C)^{k_1+k_2+k_3}} \\ &= \int_0^1 \int_0^1 \int_0^1 \frac{\delta(x_1+x_2+x_3-1) dx_1 dx_2 dx_3 x_1^{k_1 -1} x_2^{k_2-1} x_3^{k_3-1}}{(1-x_1-x_2-x_3)^{k_1+k_2+k_3} (\frac{x_1}{1-x_1-x_2-x_3} A + \frac{x_2}{1-x_1-x_2-x_3} B + \frac{x_3}{1-x_1-x_2-x_3} C)^{k_1+k_2+k_3}} \end{align}

Answer 1

I found answer for this

\begin{align} \frac{1}{\prod_{i=1}^N A_i^{\lambda_i}} = \frac{\Gamma(\sum_{i=1}^N \lambda_i)}{\prod_{i=1}^N \Gamma(\lambda_i)} \int_0^{\infty} \left[ \prod_{i=1}^N dx_i x_i^{\lambda_i -1} \right] \frac{\delta\left(1 - \sum_{i=1}^N x_i \right)}{\left[ \sum_{i=1}^N x_i A_i \right]^{\sum_{i=1}^N \lambda_i}} \end{align}