I am confused about the following integration prblem.
Suppose we define $$e(x)=\int_0^1 h\left(\frac{x}{y}\right)dy,$$ then I want to know whether the following equality holds
$$e(0)=\int_0^1 h(0)dy $$
Or under which condition it hods?
How about change $e(x)=\int_0^1 h\left(\frac{x}{y}\right)t(y)dy$, then can we claim that $e(0)=\int_0^1 h(0)t(y)dy $ ?
How about change $e(x)=\int_0^1 h\left(a+\frac{x}{y}\right)t(y)dy$, then can we claim that $e(0)=\int_0^1 h(a)t(y)dy $ ?
I'm treating your integral as an improper Riemann integral, because the integrand is undefined at $y=0$, and even more so when $x=y=0$. The following computation allows to test whether $e(x)$ is at all well defined for $x>0$. We start with $$e(x):=\lim_{\epsilon\to0+}\int_\epsilon^1 h\left({x\over y}\right)\>dy\qquad(x>0)\ .$$ Substituting $$y:={x\over u}\qquad(x\leq u<x/\epsilon)$$we obtain $$e(x)=x\>\lim_{\epsilon\to0+}\int_x^{x/\epsilon}h(u)\>{du\over u^2}=x\int_x^\infty h(u)\>{du\over u^2}\qquad(x>0)\ .$$ If the improper integral on the RHS exists when $x>0$ your function $x\mapsto e(x)$ is defined for $x>0$. In order to arrive at a reasonable value for $e(0)$ you have to investigate (for the given function $h$) how the integral behaves when $x\to{0+}\>$.