Derivative of improper integral Please help me to show next equality:
$$\frac{d}{dr} \int\limits_r^\infty \frac{sg(s)\;ds}{\left(s^2-r^2\right)^{1/2}} = 
 r \int\limits_r^\infty \frac{g'(s)\;ds}{\left(s^2-r^2\right)^{1/2}}$$
g(s) is considered to be smooth enough. I tried to calculate left statement using the definition of derivative. Left integral's increment is
$$
\begin{align}
& \int\limits_{r+\Delta r}^\infty \frac{sg(s)\;ds}{\left( s^2 - (r+\Delta r)^2 \right)^{1/2} } - \int\limits_r^\infty \frac{sg(s)\;ds}{\left(s^2 - r^2\right)^{1/2}} \\  \\
& = \int\limits_r^\infty \left[ \frac{(s+\Delta r)g(s+\Delta r) }{\left( (s+\Delta r)^2 - (r+\Delta r)^2 \right)^{1/2}} - \frac{sg(s)}{\left( s^2 - r^2 \right)^{1/2}} \right] \; ds
\end{align}
$$
Calculation implies
$$\frac{d}{dr} \int\limits_{r}^{\infty} \frac{sg(s)\;ds}{\left(s^2-r^2\right)^{1/2}} = \int\limits_r^\infty \frac{s(r-s)g(s)+sg'(s)+g(s)}{\left(s^2-r^2\right)^{1/2}}\;ds$$
So I think it's a wrong way. After, i've tried to do the change s = ru. I've reached
$$
\frac{d}{dr} \int\limits_r^\infty \frac{sg(s)\;ds}{\left(s^2-r^2\right)^{1/2}} = 
 \int\limits_r^\infty \frac{s^2 g'(s) + sg(s) }{r \left(s^2-r^2\right)^{1/2} } ds
$$

$$
\begin{align}
s & = r\sec\theta \\
ds & = r\sec\theta\tan\theta\;d\theta
\end{align}
$$
$$
\begin{align}
\int_r^\infty \frac{sg(s)\;ds}{\sqrt{s^2-r^2}}
& = r^2 \int_0^{\pi/2} (\sec^2\theta)\; g(r\sec\theta)\;d\theta.
\end{align}
$$
