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
Derivative of improper integral
2
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calculus
integration
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0Yes, for $g(s) = \frac{1}{s^2}$ it works and equals $-\frac{\pi}{2r^2}$. – 2011-09-18
1 Answers
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$ \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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0@Nimza: If you found Michael's answer useful, don't forget to click the check mark to the left of his answer. – 2011-09-18