No, you start with : \begin{align}\mathsf E(Y) &=\mathsf E(\mathsf E(Y\mid X)) \\ &=\int_\Bbb R \mathsf E(Y\mid X=x) ~f_{X}(x)~\mathsf d x\\ & = \int_0^\infty \frac 1{\surd x}\lambda e^{-\lambda x}~\mathsf d x\end{align} ...

Wolfram interprets the input in a different way, not what you intended. Below the series expansion, there is a "(Laurent series)", which means Wolfram outputs the Laurent expansion of the function ...

Fix an enumeration of S as S=\{s_n:n\in\mathbb{N}\}. Let A_m=\bigcup_{n\geq m}s_n and A=\bigcap_n A_n. Note that \#(x)=\infty iff x\in A. I will show there are ways in which A is ...

I would say the relationship is weak at best. There are two easy ways to append infinite values to the set of reals numbers: affinely , where we append \{+\infty , -\infty\}; or projectively , ...

In the Lebesgue theory, one generally deals with equivalence classes of functions, rather than with individual functions; maybe a better way of saying this is that when one appears to be referring to ...

First of all, differential entropy almost can't be the right definition, and in my opinion it's mostly a historical accident. There are a number of reasons, but the clearest one is that it is ...