# Derivative of an Integral with max in the integrand [closed]

Can anyone explain why? $$\frac{d}{dx}\int_0^{\infty}\max[R-(x+1)B;-C]dF(R,\theta) = B\int_{(1+x)B-C}^{\infty}dF(R,\theta)$$

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## closed as off-topic by Dan Rust, TMM, user1337, Matthew Pressland, Davide GiraudoSep 17 '13 at 18:54

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I assume $\theta$ is just a parameter here; since it is not used, I will drop it. Note that the integral on the left is $$\int_0^{(1+x)B-C}(-C)\,dF(R)+\int_{(1+x)B-C}^\infty (R-(x+1)B)\,dF(R).$$ The two integrands agree at the common endpoint, therefore the two contributions to the derivative arising from the $x$-es in the limits cancel each other out (you have to be a bit careful in case $F$ has a discontinuity there, though). Then just differentiate the second integral under the integral sign.