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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 $$ \frac{d}{dr} \int\limits_r^\infty \frac{sg(s)\;ds}{\left(s^2-r^2\right)^{1/2}} = \int\limits_r^\infty \frac{s^2 g'(s) + sg(s) }{r \left(s^2-r^2\right)^{1/2} } ds $$

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Are there some hypothesis about $g$? Be careful during the calculation, the low bound is not constant. Maybe you should do the change $ru=s$ to make this bound constant. – Davide Giraudo Sep 18 '11 at 18:25
It looks like there is some missing information about $g$ that will imply that these quantities are even defined in the first place. Once you have that information about $g$, this looks like an application of integration by parts. – alex.jordan Sep 18 '11 at 18:42
g(x) is smooth enough and it's all what I know about it. – Nimza Sep 18 '11 at 18:58
Smooth doesn't seem to be enough, for example if we take $g=1$, the LHS doesn't exist. Did you try to show the result for example with $g(s)=\frac 1{s^2}$? – Davide Giraudo Sep 18 '11 at 19:11
Yes, for $g(s) = \frac{1}{s^2}$ it works and equals $-\frac{\pi}{2r^2}$. – Nimza Sep 18 '11 at 19:56
up vote 1 down vote accepted

$$ \begin{align} s & = r\sec\theta \\ ds & = r\sec\theta\tan\theta\;d\theta \end{align} $$

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

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Thank you, it reduces my equality to a new one, that looks better – Nimza Sep 18 '11 at 19:59
Problem is solved! – Nimza Sep 18 '11 at 20:47
@Nimza: If you found Michael's answer useful, don't forget to click the check mark to the left of his answer. – J. M. Sep 18 '11 at 21:50

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