How to compute the integral below?
$$ I(y) = \int_{0}^{y} \frac{\sum_{i=1}^{k} e^{-a_i \cdot x - b_i}}{\sum_{i=1}^{k} e^{-a_i \cdot x - b_i} + \sum_{i=k+1}^{m} e^{-a_i \cdot x - b_i}} \; dx $$
where $a_i \geq 0$, $b_i \geq 0$, $1\leq k < m$.
Background:
I tried solving it using Taylor series expansion. $$ I(y) = \int_{0}^{y} \frac{\sum_{i=1}^{k} e^{-a_i \cdot x - b_i}}{\sum_{i=1}^{m} e^{-a_i \cdot x - b_i}} \; dx \\ I(y) = \int_{0}^{y} \frac{\sum_{i=1}^{k} e^{-a_i \cdot x - b_i}}{e^{-a_j \cdot x -b_j}(1 + \sum_{i=1}^{j-1} e^{-(a_i - a_j) \cdot x - (b_i -b_j)} + \sum_{i=j+1}^{m} e^{-(a_i - a_j) \cdot x - (b_i -b_j)})} \; dx $$
I have assumed, there exists $a_j$ and $b_j$ such that $(a_i - a_j) \geq 0$, and $(b_i - b_j) \geq 0$. Substituting $a_l = a_i - a_j$, and $b_l = b_i - b_j$, I get $$ I(y) = \int_{0}^{y} \frac{(\sum_{i=1}^{k} e^{-a_i \cdot x - b_i})e^{a_j \cdot x +b_j}}{1 + \sum_{l=1}^{m-1} e^{-a_l \cdot x - b_l}} \; dx $$ $$ I(y) = \int_{0}^{y} \frac{(\sum_{i=1}^{k} e^{-a_i \cdot x - b_i})e^{a_j \cdot x +b_j}}{1 - (-\sum_{l=1}^{m-1} e^{-a_l \cdot x - b_l})} \; dx $$ Using Taylor Series Expansion ,I get $$ I(y) = \int_{0}^{y} (\sum_{i=1}^{k} e^{-a_i \cdot x - b_i})e^{a_j \cdot x +b_j} \cdot \left( \sum_{n=0}^{\infty} (-1)^{n} (\sum_{l=1}^{m-1} e^{-a_l \cdot x - b_l})^{n} \right) \; dx $$
Is there any method to complete this computation of the integral?
Any help will be appreciated.