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I have ended up with the following ratio of two definite integrals

\begin{equation} \frac{\int_{x_1}^{x_2}\alpha T I_0 e^{-\alpha l} d \lambda}{\int_{x_1}^{x_2} T I_0 e^{-\alpha l} d \lambda} \end{equation}

where $\alpha$ $T$ and $I_0$ are functions of $\lambda$. $\alpha$ and $T$ are experimentally known (i.e. not mathematical functions) and $I_0$ is completely unknown. What I really need is to know if the whole thing cancels ( in a mathematically justified way) to leave an integral over $\alpha$ (this bit I can actually do)



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No, that is an unjust manipulation, unless $\alpha$ is constant. But in general, we have $$ - \frac{\partial}{\partial l} \log \left( \int_{x_1}^{x^2} TI_0 e^{-\alpha l} \; d\lambda \right) = \frac{\int_{x_1}^{x^2} \alpha TI_0 e^{-\alpha l} \; d\lambda}{\int_{x_1}^{x^2} TI_0 e^{-\alpha l} \; d\lambda}, $$ Though I'm not sure if it will be helpful at all. – Sangchul Lee Sep 6 '12 at 15:24
@sos440, doesn't this assume continuity of the integrand not present in the question? – Karolis Juodelė Sep 6 '12 at 15:35
@KarolisJuodelė, of course those functions have to be not so peculiar. But under quite general and admissible restriction on them, the resulting integral, as a function of the variable $l$, behaves nicely. Just observe that this is a variant of the famous Laplace transform. – Sangchul Lee Sep 6 '12 at 15:45

@sos440, if alpha isn't dependent on lambda, the quotient cancels to alpha, and if it is dependent on lambda your derivative is incorrect because it doesn't incorporate d-alpha/d-lambda. Assuming that l = lambda, which seems to be your assumption as well.

I'm also uncertain as to whether you are allowed to ignore the definite integral when finding the derivative of the inside of the logarithm, but I'd be interested to know if there is a reason it is permitted in this case.

share|cite|improve this answer Its use was sort-of popularised by Feynman. As for the log function, he's applying chain rule here. – Deepak Aug 25 '15 at 15:45

If $T I_0 e^{-\alpha l}$ is positive, then $$ \frac{\int_{x_1}^{x_2}\alpha T I_0 e^{-\alpha l} d \lambda}{\int_{x_1}^{x_2} T I_0 e^{-\alpha l} d \lambda} $$ is a "weighted average" of $\alpha$. So then at least you can say that it is between the minimum of $\alpha$ and the maximum of $\alpha$.

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