So assume $Y$ and $X$ are exponentially distributed with parameters $y_1$, and $x_1$ respecitively. assume c is a constant.
I am having huge trouble to understand the integration of the following expression.
$P(Y<c/u(X))$
$=\int_{t}^{\infty}f_X(x)\int_{0}^{c/u(x)}f_Y(y)dydx +\int_{0}^{t}f_X(x)\int_{c/u(x)}^{\infty}f_Y(y)dydx$
where t is the cross-point that $u(x)$ change sign
here $c/u(x)$ is given by the plot below, t is the point crossing the zero:
Confusion:
I don't understand the second integration of the second term "$\int_{c/u(x)}^{\infty}f_Y(y)dydx$", this isn't right because Y an X only defined for y>0, and x>0. So it's the first quadrant in this plot.
Actually, one is not interested in $P(Y<c/u(X))$ but in $P(Yu(X)\lt c)$, and this is $$P(Yu(X)\lt c)=\int_{t}^{\infty}f_X(x)F_Y(c/u(x))dx +\int_{0}^{t}f_X(x)dx.$$ Thus, the factor $$ \int_{c/u(x)}^{\infty}f_Y(y)dy $$ in the second part of the RHS is not useful. Note that, if $x\lt t$, $u(x)\lt0$ hence $c/u(x)\lt0$ and, since $f_Y(y)=0$ for every $y\lt0$, $$ \int_{c/u(x)}^{\infty}f_Y(y)dy=\int_0^{\infty}f_Y(y)dy=1. $$ And it happens that this factor is rightfully replaced by $1$ in the second line of (75) in the paper you linked to.
Note finally that $$P(Y<c/u(X))=P(Yu(X)\lt c,X\gt t), $$ while $$P(Yu(X)\lt c)=P(Yu(X)\lt c,X\gt t)+P(X\lt t),$$ hence replacing the latter by the former was misleading.