For a definite integral $$\int^b_af(x,h)dx,$$
can we in general show that $$\lim_{h\to \infty} \int^b_af(x,h)dx = \int^b_a \lim_{h\to \infty} f(x,h) dx.$$
Intuitively this seems to make sense, as an integral can be approximated by a Riemann sum, and we could use the limit sum rule iteratively. However, I did not found how to formalise this, can anyone help me with this.
It doesn't always work! For each $h \ge 1$, define
$$ f(x, h) = \begin{cases} h & \text{if } 0 \leq x \leq \frac{1}{h} \\ 0 & \text{if } \frac{1}{h} < x \leq 1 \end{cases} $$
Then, for all $h \ge 1$ $$ \int_0^1 f(x, h) \ dx = \int_0^{1/h} h \ dx = 1 $$
Therefore, $\displaystyle \lim_{h \to \infty} \int_0^1 f(x, h) \ dx = 1$. However, for all $0 < x \leq 1$, we have $\displaystyle \lim_{h \to \infty} f(x, h) = 0$. Thus, $\displaystyle \int_0^1 \lim_{h \to \infty} f(x, h) \ dx = 0$.
Let $q(x)=\lim_{h\to\infty}f(x,h)$ for all $x\in[a,b]$. We say that $f(x,h)$ converges pointwise to $q(x)$. This is not a sufficient condition for interchanging limits and integrals.
A necessary and sufficient condition for such a swappage is that $f(x,h)$ converge uniformly to $q(x)$. That is, for any $\epsilon>0$ there is an $H\ge0$ such that for all $h\ge H$ and $x\in[a,b]$, $$|f(x,h)-q(x)|<\epsilon.$$ If that condition is satisfied, the following holds $$\lim_{h\to\infty}\int_a^b f(x,h)dx=\int_a^b q(x)dx.$$
