Here is the theorem of differentiation under the integral sign for the gauge integral:
Theorem: Let ##f:[a,b]\times [c,d]\rightarow \mathbb{R}## (where ##b## can be infinity) such that for each ##t\in [c,d]##, the function ##x\rightarrow f(x,t)## is measurable on ##[a,b]##.
Suppose that:
1) There exists ##t\in [c,d]## such ##x\rightarrow f(x,t)## is gauge integrable.
2) The partial derivative ##\frac{\partial f}{\partial t}## exists on ##[a,b]\times [c,d]##.
3) There are gauge integrable functions ##\alpha## and ##\omega## such that
\alpha(x)\leq \frac{\partial f}{\partial t}(x,t)\leq \omega(x)
for all ##x\in [c,d]## and ##t\in [a,b]##.
1) Then ##x\rightarrow f(x,t)## is gauge integrable for each ##t\in [c,d]##.
2) The function ##x\rightarrow \frac{\partial f}{\partial t}## is gauge integrable for each ##t\in [c,d]##
3) We have
\frac{d}{dt}\int_a^b f(x,t)dx = \int_a^b \frac{\partial f}{\partial t}dx
In particular, if ##f(x,t) = e^{-tx}\frac{\sin(x)}{x}##, then this is clearly measurable in ##x## since it is continuous.
Setting ##t=1## gives us ##e^{-x}\frac{\sin(x)}{x}## which is gauge integrable by the following theorem.
Theorem: A measurable function ##g## is gauge integrable iff there are gauge integrable functions ##g_1##, ##g_2## such that ##g_1\leq g\leq g_2##.
We have ##\frac{\partial f}{\partial t} = -e^{-tx}\sin(x)## exists on ##[0,+\infty]\times [\varepsilon,1]## and is easily seen to be gauge integrable by using the above Theorem. So the theorem applies, at least for ##t\in [\varepsilon, 1]##. So for those ##t##, we can indeed et
\int_0^{+\infty} e^{-tx}\frac{\sin(x)}{x}dx = \frac{\pi}{2} - \text{arctan}(t)
Then we would need to switch limit and integral to conclude that ##\int_0^{+\infty}\frac{\sin(x)}{x}dx = \frac{\pi}{2}##. This is provided in the document I linked.
