Let $f(t):=\int_0^{\pi/2} \arccos\frac{t-\tan^2x}{t+\tan^2x}\,dx$, for $0\leq t\leq 1$. I would like to differentiate $f$ with respect to $t$ by taking the partial of the integrand:
$$
f'(t)
= \int_0^{\pi/2}\frac{\partial}{\partial t}
\left(\arccos\frac{t-\tan^2x}{t+\tan^2x}\right)\,dx
= -\int_0^{\pi/2} \frac{1}{\sqrt{t}}
\frac{\tan x}{t+\tan^2x}\,dx.
$$
I am not sure to be able to fully justify this step, in particular because $x$ can approach $\frac{\pi}2$ (from the left), where $\tan x\rightarrow +\infty$, and $t$ can approach $0$ (from the right), where $\frac 1{\sqrt{t}}\rightarrow +\infty$. Am I allowed to do this differentiation under the integral sign?
I also would like to be pointed to some reference about differentiation under the integral sign, for the Riemann integral. Any help would be very appreciated.
Best Answer
Here is an indirect proof, using the fact that integrals are often easier to control than derivatives: Write
$$ k(t, x) = \arccos\left(\frac{t+\tan^2 x}{t-\tan^2 x}\right). $$
Then for each $t > 0$ and $0 < x < \frac{\pi}{2}$, we have the bound
$$ \left| \frac{\partial k}{\partial t} \right| = \frac{1}{t} \cdot \frac{\sqrt{t}\tan x}{t+\tan^2 x} \stackrel{\text{(AM-GM)}}{\leq} \frac{1}{2t}. $$
So if $0 < a < b$, then by the Fundamental Theorem of Calculus (FToC) and the Fubini's Theorem,
\begin{align*} f(b) - f(a) &= \int_{0}^{\frac{\pi}{2}} (k(b, x) - k(a, x)) \mathrm{d}x \\ &= \int_{0}^{\frac{\pi}{2}} \int_{a}^{b} \frac{\partial k}{\partial t} \, \mathrm{d}t\mathrm{d}x \\ &= \int_{a}^{b} \int_{0}^{\frac{\pi}{2}} \frac{\partial k}{\partial t} \, \mathrm{d}x\mathrm{d}t \\ &= \int_{a}^{b} g(t) \mathrm{d}t, \end{align*}
where $g : (0, \infty) \to \mathbb{R}$ is defined by
$$ g(t) = \int_{0}^{\frac{\pi}{2}} \frac{\partial k}{\partial t} \, \mathrm{d}x. $$
Since $g$ is continuous, FToC again tells that $f$ is an antiderivative of $g$, and therefore $f$ is differentiable with $f' = g$ as desired.