Consider two continuous random variables $X,Y$ and scalars $b>a,$ and $t_1$, $t_2$. I want to compute expected value of $X,$ that is $E[X]$.
Let a realization of $X$ be $x$ and $Y$ be $y$. I know that $X=x t_1$ if $a+x<y<b+x$, $X=x t_2$ if $b+x<y$ and $X=0,$ otherwise. Hence, we can write \begin{equation} E[X]=E[X|a+X<Y<b+X] P(a+X<Y<b+X) + E[X|b+X<Y] P(b+X<Y). \end{equation}
Now, my question is how can we express these conditional expectations with integrals. That is, can we say \begin{equation} E[X|a+X<Y<b+X] P(a+X<Y<b+X)= \int_{x=-\infty}^{\infty}ta_1 \left[ \int_ {y=a+x} ^{b+x}f_{Y|X}(y|x)dy \right]f_X(x)dx? \end{equation} If this is true, then, \begin{equation} E[X|a+X<Y<b+X] = \frac{ \int_{x=-\infty}^{\infty}ta_1 \left[ \int_ {y=a+x} ^{b+x}f_{Y|X}(y|x)dy \right]f_X(x)dx}{\int_{x=-\infty}^{\infty} \left[ \int_ {y=a+x} ^{b+x}f_{Y|X}(y|x)dy \right]f_X(x)dx}? \end{equation}
This second expression looks very weird to me that makes me think that there is a problem with the first one as well.
No. When $X$ realises a value of $x$ then $X=x$. That's what it means.
However, perhaps you mean to find the expectation of a function $g(X,Y)$, where $g(x,y)=xt_1$ when ... and so on.
Then indeed:
$$\mathsf E(g(X,Y)\mid a+X<Y<b+X) = \dfrac{\int\limits_{-\infty}^{\infty} x t_1 f_X(x)\left(\int\limits_{a+x}^{b+x} f_{Y\mid X}(y\mid x)\operatorname d y\right)\operatorname d x}{\int\limits_{-\infty}^{\infty} f_X(x) \left(\int\limits_{a+x}^{b+x} f_{Y\mid X}(y\mid x)\operatorname d y\right)\operatorname d x}$$