TITLE: Exchanging derivative and integral
QUESTION [7 upvotes]: Define the function $g:[1,2]\rightarrow \mathbb{R}$ to be
$$
g(t):= \int_0^{\frac{\pi}4} \arctan\frac{\sin 2x}{t-2\sin^2 x}\,dx.
$$
I would like to differentiate obtaining
\begin{align*}
g'(t) &= \frac{\rm d}{{\rm d}t}\int_0^{\frac{\pi}4} \arctan\frac{\sin 2x}{t-2\sin^2 x}\,dx 
=\int_0^{\frac{\pi}4} \frac{\partial}{\partial t}\left(\arctan\frac{\sin 2x}{t-2\sin^2 x}\right)\,dx=\\
&= -\int_0^{\frac{\pi}4} \frac{\sin 2x}{2(t-1)\cos 2x + t(t-2)+2} \,dx.
\end{align*}
and I would like to be sure to be allowed to do that. 
Call $f(x,t)$ the integrand function, defined over $\left[0,\frac{\pi}4\right]\times [1,2]$. For all $t\in [1,2]$ we have $|f(x,t)|\leq \frac{\pi}2$, which is integrable over $\left[0,\frac{\pi}4\right]$, and for all $x\in \left[0,\frac{\pi}4\right]$ we have that $\frac{\partial f}{\partial t}(x,t)$ exists and satisfies $\left|\frac{\partial f}{\partial t}(x,t)\right|\leq \sin 2x$, which again is summable over $\left[0,\frac{\pi}4\right]$. 

Are this domination conditions enough to be allowed to derive under
  the integral sign?

REPLY [4 votes]: The conditions are sufficient to reach the conclusion. 
With $\displaystyle f(x,t) = \arctan\frac{\sin 2x}{t-2\sin^2 x}$, we have $\displaystyle \left|\frac{\partial f}{\partial t}(x,t)\right| \leqslant \sin 2x$ for $x \in [0,\pi/4]$.
Thus, for any sequence $t_n \to t \in [1,2]$, it follows from the mean value theorem that there exists $\xi_{n,x}$ between $t$ and $t_n$ such that
$$\frac{g(t_n) - g(t)}{t_n - t} = \int_0^{\pi/4}\frac{f(x,t_n) - f(x,t)}{t_n - t} \, dx = \int_0^{\pi/4}\frac{\partial f}{\partial t}(x,\xi_{n,x}) \, dx $$
Since $\displaystyle \left|\frac{\partial f}{\partial t}(x,\xi_{n,x})\right| \leqslant \sin 2x$ (which is integrable), it follows by the dominated convergence theorem that
$$g'(t) = \lim_{n \to \infty}\int_0^{\pi/4}\frac{f(x,t_n) - f(x,t)}{t_n - t} \, dx = \int_0^{\pi/4}\lim_{n \to \infty}\frac{f(x,t_n) - f(x,t)}{t_n - t} \, dx =\\ \int_0^{\pi/4}\frac{\partial f}{\partial t}(x,t) \, dx$$