Using the Fundamental Theorem of Calculus to differentiate a function of two variables defined by an integral

Question

I'm having problems differentiating the following integral in $t$ $$\psi(x,t)=\int^{t}_{0}f(x + b(s-t),s)ds$$ with $b\in \mathbb{R}$ and $f(x,t) \in C^1(\mathbb{R}^2)$.

I will appreciate any help!

Answer 1

I solve it using the definition, "a trick" and some analysis exercises at penultimate and last lines.

First, call $f(x-b(s+t),s) = \pi(s,t)$.

Then, \begin{align*} \frac{\partial}{\partial t}\left[ \int_{0}^{t}\pi(s,t)ds \right] &=\lim_{h \to 0} \frac{\int_{0}^{t+h}\pi(s,t+h)ds - \int_{0}^{t}\pi(s,t)ds }{h}\\ &= \lim_{h \to 0} \frac{\int_{0}^{t+h}\pi(s,t+h)ds}{h} - \frac{\int_{0}^{t}\pi(s,t)ds}{h}\\ &= \lim_{h \to 0} \frac{\int_{0}^{t+h}\pi(s,t+h)ds}{h} - \frac{\int_{0}^{t+h}\pi(s,t)ds}{h} + \frac{\int_{0}^{t+h}\pi(s,t)ds}{h} - \frac{\int_{0}^{t}\pi(s,t)ds}{h}\\ &= \lim_{h \to 0} \frac{\int_{0}^{t+h}\pi(s,t+h) - \pi(s,t) ds}{h} + \frac{\int_{0}^{t+h}\pi(s,t)ds}{h} + \frac{\int_{t}^{0}\pi(s,t)ds}{h}\\ &= \lim_{h \to 0} \frac{\int_{0}^{t+h}\pi(s,t+h) - \pi(s,t) ds}{h} + \frac{\int_{t}^{t+h}\pi(s,t)ds}{h}\\ &= \lim_{h \to 0} \frac{\int_{0}^{t}\pi(s,t+h) - \pi(s,t) ds}{h} + \frac{\int_{t}^{t+h}\pi(s,t+h) - \pi(s,t) ds}{h} + \frac{\int_{t}^{t+h}\pi(s,t)ds}{h}\\ &= \lim_{h \to 0} \frac{\int_{0}^{t}\pi(s,t+h) - \pi(s,t) ds}{h} + \frac{\int_{t}^{t+h}\pi(s,t+h)ds}{h} - \frac{\int_{t}^{t+h}\pi(s,t) ds}{h} + \frac{\int_{t}^{t+h}\pi(s,t)ds}{h}\\ &= \lim_{h \to 0} \frac{\int_{0}^{t}\pi(s,t+h) - \pi(s,t) ds}{h} + \frac{\int_{t}^{t+h}\pi(s,t+h)ds}{h}\\ &= \lim_{h \to 0} \frac{\int_{0}^{t}\pi(s,t+h) - \pi(s,t) ds}{h} + \lim_{h \to 0} \frac{\int_{t}^{t+h}\pi(s,t+h)ds}{h}\\ & = \int_{0}^{t}\lim_{h \to 0}\frac{\pi(s,t+h) - \pi(s,t) ds}{h} + \lim_{h \to 0} \frac{1}{h} \cdot \int_{t}^{t+h}\pi(s,t+h)ds\\ &= \int_{0}^{t}\frac{\partial\pi(s,t)}{\partial x}ds + \pi(t,t) \end{align*}

Thanks for the comments!