I have the following:
$$P = \sum_{j=1}^{6} \sum_{k=1}^{6} A(k,j) \int_0^h e^{(\alpha^{(k)} -\overline\alpha^{(j)})x} dx$$
Where A is a constant that depends on $j$ and $k$. Taking the integral yields:
$$P = \sum_{j=1}^{6} \sum_{k=1}^{6} A(k,j) \frac{1}{(\alpha^{(k)} -\overline\alpha^{(j)})} e^{(\alpha^{(k)} -\overline\alpha^{(j)})x} \bigg\vert_0^h$$
Where $\alpha$ is either pure real or pure imaginary and $\alpha_{2n} = -\alpha_{2n-1}$ $n=1,2,3$. For certain combinations of $j$ and $k$ the denominator $(\alpha^{(k)} -\overline\alpha^{(j)})$ and numerator $e^{(\alpha^{(k)} -\overline\alpha^{(j)})x} \bigg\vert_0^h$ will equal to zero, resulting in $\frac{0}{0}$.
The question is how can I justify that $P$ will equal to zero (is it even possible?), or do I need to apply l'Hopital's rule?
Thanks in advance