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Working on three-body dispersion forces I got the following quantity: $$\frac{\partial } {{\partial \lambda }}\int\limits_\lambda ^{\pi - \lambda } {d\theta } \int\limits_\lambda ^{\pi - \lambda } {d\phi f\left( {\theta ,\phi } \right)} $$ where f is independent of $\lambda$. My question is: how can i take the derivative under the double integral sign? In one-dimensional case we have: $$\frac{d} {{dx}}\int\limits_{a\left( x \right)}^{b\left( x \right)} {f\left( {x,y} \right)} dy = f\left( {x,b\left( x \right)} \right)b'\left( x \right) - f\left( {x,a\left( x \right)} \right)a'\left( x \right) + \int\limits_{a\left( x \right)}^{b\left( x \right)} {f_x } \left( {x,y} \right)dy $$ Is there an analogous formula for the two-dimensional case?

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1 Answer 1

It seems we can implement the Leibniz integral rule twice. If we define

$$ F(\theta,\lambda) = \int_\lambda^{\pi-\lambda} d\phi\,f(\theta,\phi) $$


$$\begin{align}\frac{\partial } {{\partial\lambda }}\int_\lambda^{\pi-\lambda }d\theta\int_\lambda^{\pi -\lambda }{d\phi\,f\left( {\theta ,\phi } \right)} &= \frac{\partial}{\partial\lambda}\int_{\lambda}^{\pi-\lambda}d\theta \,F(\theta,\lambda) \\ &= F(\pi-\lambda,\lambda)(-1)-F(\lambda,\lambda)(1)+\int_\lambda^{\pi-\lambda}\frac{\partial F}{\partial\lambda}d\theta \end{align}$$ where $\partial F/\partial\lambda $ is again given by the Leibniz integral rule.

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