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)
Thanks
Ali
Best Answer
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 $\alpha$.