My Math Forum Complicated integral

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March 9th, 2017, 03:03 PM   #11
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Quote:
 Originally Posted by SDK I'm not sure I understand correctly. You are interested in computing $I'(z)$? If this is the case you don't need to integrate anything. Simply apply the fundamental theorem of calculus and the chain rule. Assuming that $F(\lambda) = \int_c^\lambda I'(\lambda(z)) d \z$ for some conveniently chosen lower limit and you have $I(z) = F(\lambda_1) - F(\lambda_2)$ and now taking a derivative you get $$I'(z) = \arctan(b(\lambda_1)-a)\lambda_1^{\gamma}\cdot \frac{d\lambda_1}{dz} - \arctan(b(\lambda_2)-a)\lambda_2^{\gamma}\cdot \frac{d\lambda_2}{dz}$$
I still dont get the properly chosen lower limit, that the point, in my integral both limits depend on z, can you explain this further please?

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