# Ratios of similar definite integrals

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

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

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)

Thanks

Ali

-
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

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 $\alpha$.