how to differentiate the integral with singularity $${d\over dx} \, \int_{x}^{\infty} \frac{e^{-t}}{t \, \sqrt{t^2-x^2}} \, dt $$
How to move the derivative into the integral or simplify?
 A: Using the Leibniz integral rule,
$$ \frac{\partial}{\partial \, x} \int_{a(x)}^{b(x)} f(x,t) \, dt = f(x, b(x)) \, \frac{d b}{d x} - f(x, a(x)) \, \frac{d a}{d x} + \int_{a(x)}^{b(x)} \frac{d f(x, t)}{dx} \, dt, $$
then
\begin{align}
\frac{d I}{d x} &= 
{d\over dx} \, \int_{x}^{\infty} \frac{e^{-t}}{t \, \sqrt{t^2-x^2}} \, dt \\
&= - \lim_{t \to x} \frac{e^{-t}}{t \, \sqrt{t^2 - x^2}} + x \, \int_{x}^{\infty} \frac{e^{-t}}{t} \, (t^2 - x^2)^{-3/2} \, dt. 
\end{align}
The derivative then becomes a problem with the limit as $t \to x$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{{\displaystyle #1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
& \color{#44f}{\totald{}{x}\int_{x}^{\infty}
{\expo{-t} \over t\root{t^{2} - x^{2}}}\,\dd t} =
\totald{}{x}\int_{0}^{\infty}
{\expo{-t -\ x} \over
\pars{t + x}\root{t^{2} + 2xt}}\,\dd t
\\[5mm] \stackrel{\LARGE\color{red}{\S}}{=} &
\int_{0}^{\infty}\partiald{}{x}
\bracks{{\expo{-x -\ t} \over
\pars{x + t}\root{t^{2} + 2xt}}}\,\dd t
\end{align}
$\ds{\large\color{red}{\S}}$: Can you make this step ?.
