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 May18 comment Computing real integrals using the Residue Theorem where singularities are on the real line You have $\int_0^\inf \frac{\sin(x)/x}{x^2+a^2}$. The function $\sin(x)/x$ does not have a pole at $x=0$ (strictly speaking it's undefined at $x=0$ but it has a holomorphic extension that includes the origin). So there are only two poles, $ia$ and $-ia$, both of which are not on the real line. Sep11 comment Under what conditions is expectation value distributive? Thanks to everyone who tried to answer my question. I have now found a partial answer to my question. Unfortunately there is no guarantee that given marginal or conditional distributions for $X,Y$ and $X,Z$, there exists a joint distribution for $X,Y,Z$. This is why the distributive law does not always hold. The paper dealing with conditional distributions can be seen here. However, if anyone can lucidate the matter with a simple condition, the question is still open. I have also editted the question to make it clearer what I want. Sep10 comment Under what conditions is expectation value distributive? Thanks for your answer Michael. Here is what I don't understand. Suppose $X,Y$ follow some measure $dp(x,y)$ (where $x,y$ are particular values of $X,Y$) and $X,Z$ follow $dq(x,y)$, where you can assume that $p$ and $q$ are densities. Now of course $\int dy p(x,y) = \int dz q(x,z)$. It is not obvious to me that $E(XY) + E(XZ) = int dx dy p(x,y) x y + \int dx dz q(x,z) x z = \int dx dy dz x(y+z) r(x,y,z) = E\bigl(X(Y+Z)\bigr)$. Does such extension of $p$ and $q$ always exist? Sep10 comment Under what conditions is expectation value distributive? Thanks mjqxxxx, editted. I think it can't always be true because the violation of this property is needed to arrive at the violation of Bell inequalities in quantum mechanics.