# An integral formula for the multiplication of inverses.

I've tried to find a similar formula of following.

I've heard about this formula \begin{align} \prod_{i=1}^n \dfrac{1}{a_i} = \Gamma(n)\left[ \prod_{i=1}^n \int_0^1 \mathrm{d }x_i\right]\dfrac{\delta(1-\sum_{i=1}^n x_i)}{(\sum_{i=1}^n a_i x_i)^n} \end{align} in a class. ($$\delta(x)$$ is a Dirac-delta function. I use this for convenience.)

I checked that it's okay when $$n=2$$.

I tried to prove this but I've noticed when $$n=3$$, it holds that \begin{align} \int_0^1 \int_0^1 \mathrm{d}x_1 \mathrm{d}x_2 \dfrac{1}{(m_1 x_1 + m_2 x_2 + m_3)^3} &= \int_0^1 \int_{m_1 x_1 + m_3}^{m_1 x_1 + m_2 + m_3} \dfrac{\mathrm{d}x_1 \mathrm{d} u}{m_2 u^3}\\ &= -\dfrac{1}{2m_2}\int_{0}^1 \dfrac{\mathrm{d}x_1}{(m_1 x_1 + m_2 + m_3)^2} - \dfrac{\mathrm{d}x_1}{(m_1 x_1 + m_3)^2} \\ &=\dfrac{1}{2}\dfrac{m_1 + m_2 + 2 m_3}{m_3 (m_1 + m_3)(m_2 + m_3)(m_1 + m_2 + m_3)}\\ \end{align} Obviously, it doesn't coincide with what I want to show.(When substitute $$m_1$$ to $$a-c$$, $$m_2$$ to $$b-c$$ and $$m_3$$ to $$c$$.)

Therefore, I think the formula suggested in the class is wrong. However, it seems to be used in several calculation and so, I think there is a similar formula rather than this.

Could someone tell me what it might be?

• Could you edit in your $n=3$ calculation?
– J.G.
Nov 21, 2019 at 13:53
• @J.G. When you substitue $m_i$ to what I suggested, then you can get the above formula. Nov 21, 2019 at 13:56
• Could you prove it?
– J.G.
Nov 21, 2019 at 13:58
• @J.G. I think I add some conditions for that formula. Anyway, I'll add the proof. Nov 21, 2019 at 14:00
• @J.G. Your feedback is right. I fix my formulas Nov 21, 2019 at 14:04

The formula you cite is correct; in fact, it can be generalized. Your error is assuming $$x_2$$ integrates from $$0$$ to $$1$$; it should be $$0$$ to $$1-x_1$$. So the correct result for$$a:=a_1,\,b:=a_2,\,c:=a_3,\,m_1:=a-c,\,m_2:=b-c,\,m_3:=c$$is$$\int_0^1dx_1\int_{m_1x_1+m_3}^{m_1x_1+m_2(1-x_1)+m_3}\frac{du}{m_2u^3}\\=\frac{1}{2m_2}\int_0^1dx_1\left(\frac{1}{(m_1x_1+m_3)^2}-\frac{1}{((m_1-m_2)x_1+m_2+m_3)^2}\right)\\=\frac{1}{2m_2}\left(\frac{1}{m_3(m_1+m_3)}-\frac{1}{(m_1+m_3)(m_2+m_3)}\right)\\=\frac{1}{2(m_1+m_3)m_3(m_2+m_3)}=\frac{1}{2acb}.$$