Question about interchanging the order of limit and integral I am trying to show that $\lim_{\theta\to0} \int^\frac{1}{2} _0 e^{-i\theta t \cos 2 \pi t} = 1$. Clearly, I need to show that it is valid to interchange the order of a limit and an integral under this situation. So I am trying to simplify it to the following:
Suppose that I have a function $f(x,y):\mathbb R^2 \to \mathbb R$, and $f$ continuous, does $\lim_{x \to c}\int^a _bf(x,y) dy = \int^a _b \lim_{x\to c} f(x,y)dy= \int^a _b f(c,y)dy$ hold? I think it should hold but I might need some sort of "uniform continuity" condition.
 A: Suppose $f(x,y)$ is defined for $a\le y\le b$ and all points $x$ in a set $A$ for which $x=c$ is a limit point.

If for each $x$ of $A$, $f(x,y)$ is integrable on $[a,b]$, and if $f(x,y)$ converges uniformly to $g(y)$ on $[a,b]$ as $x\to c$, then $g(y)$ is integrable on $[a,b]$ and 
$$\lim_{x\to c}\int_a^b f(x,y)\,dy=\int_a^b g(y)\,dy$$

PROOF:
Let $x_n$ be a sequence of points in $A$, none equal to $c$, that converges to $c$.  Then, $f(x_n,y)$ converges uniformly to $g(y)$.  
We first show that $g(y)$ is integrable on $[a,b]$.  Proceeding, for any given $\epsilon>0$, there exist a number $N(\epsilon)$ such that $|f(x_N,y)-g(y)|<\epsilon/4(b-a)$ for every $y\in [a,b]$.
Since $f(x_N,y)$ is integrable, there exists step functions $\sigma_1(y)$ and $\tau_1(y)$ such that  $\sigma_1(y)\le f(x_N,y)\le \tau_1(y)$ on $[a,b]$ and 
$$\int_a^b (\tau_1(y)-\sigma_1(y))\,dy<\frac{\epsilon}{2}$$
Now, define new step functions $\sigma(y) = \sigma_1(y)-\frac{\epsilon}{4(b-a)}$ and $\tau(y)=\tau_1(y)+\frac{\epsilon}{4(b-a)}$.  Then, for $y\in[a,b]$,
$$\sigma(y)<\sigma_1(x)+[g(y)-f(x_N,y)]\le g(y)\\\\
\tau(y)>\tau_1(y)+[g(y)-f(x_N,y)]\ge g(y)$$
and
$$\begin{align}
\int_a^b (\tau(y)-\sigma(y))\,dy&=\int_a^b \left(\tau_1(y)-\sigma_1(y)+\frac{\epsilon}{2(b-a)}\right)\\\\
&<\epsilon
\end{align}$$
whence we conclude $g(y)$ is integrable.
Finally, we choose $n$ so large that $|f(x_n,y)-g(y)|<\frac{\epsilon}{(b-a)}$.  Then, 
$$\begin{align}
\left|\int_a^b (f(x_n,y)-g(y))\,dy\right|&\le \int_a^b \left|f(x_n,y)-g(y)\right|\,dy\\\\
&\le \epsilon
\end{align}$$
and we are done!
