# Distributing integral into product of integrals

I would like to know if $$\int_{x_1} \left[ \left[ \int_{x_2} g_1(x_1,x_2) dx_2 \right] \cdot \left[ \int_{x_3} g_2(x_1,x_3)dx_3 \right] \right] dx_1 \stackrel{?}{=} \left[ \int_{x_1} \int_{x_2} g_1(x_1,x_2) dx_2 dx_1 \right] \cdot \left[ \int_{x_1} \int_{x_3} g_2(x_1,x_3) dx_3 dx_1 \right]$$ I don't need a proof, but maybe a pointer to what rule of integration would show this to be true (or not true). I have tried to expand the left hand side (LHS) of the equation above as follows $$\text{LHS} = \int_{x_1} \left[ \int_{x_3} \int_{x_2} g_1(x_1,x_2) g_2(x_1,x_3) dx_2 dx_3 \right] dx_1$$ Simplifying further, we get $$\text{LHS} = \int_{x_1} \int_{x_3} \int_{x_2} g_1(x_1,x_2) g_2(x_1,x_3) dx_2 dx_3 dx_1$$

Again, in summary, are there properties of integrals, or is there a proof which shows that the LHS equals the right hand side (RHS)? The RHS is repeated below for clarity.

$$\text{RHS} = \left[ \int_{x_1} \int_{x_2} g_1(x_1,x_2) dx_2 dx_1 \right] \cdot \left[ \int_{x_1} \int_{x_3} g_2(x_1,x_3) dx_3 dx_1 \right]$$

It is safe to assume that the functions $g_1(x_1,x_2)$ and $g_2(x_1,x_3)$ are integratable over the regions specified.

Thanks!

• Essentially, your first line reads $\int_{x_1} G_2(x_1) G_3(x_1) \, \mathrm{d}x_1$. Now use partial integration. – Ritz Feb 18 '15 at 14:51
• Your comment made me realize I specified the question incorrectly. I've respecified the question properly. The $g$ functions are not necessarily the same, so they should be indexed. Thanks for the hint on partial integration, I need to understand whether that still applies in light of me correcting the question. – Kiran K. Feb 18 '15 at 14:58
• If you had finite integration intervals and the identity as integrands, it would turn into something like $\Delta x_1 \Delta x_2 \Delta x_3 = (\Delta x_1 \Delta x_2)(\Delta x_1 \Delta x_3) = (\Delta x_1)^2 \Delta x_2 \Delta x_3$ so I think your equation does not hold. Alteratively I could argue with units which would not match ($L^3$ vs. $L^4$). – mvw Feb 18 '15 at 15:07
• @mvw Good point, I didn't think about that. I suppose then that if the LHS was $\int_{x_1} \int_{x_1} \int_{x_2} \int_{x_3} g_1(x_1,x_2) g_2(x_1,x_3) dx_2 dx_3 dx_1 dx_1$ then the RHS would equal the LHS? I guess that doesn't prove it works for all $g_1$ and $g_2$, just when $g_1$ and $g_2$ is identity, but its a start. – Kiran K. Feb 18 '15 at 15:20
• @mvw What do you mean by the units not matching? Referencing $L^3$ vs $L^4$? I'm not tracking there. – Kiran K. Feb 18 '15 at 15:31