Integral which needs absolute value of zero

Question

I'm trying to solve an integral $$\int_0^t dx \,e^{-\nu \,|x-t|}$$ The trouble I'm having is that in the final stages this seems to need me to have a value for $\frac{0}{|0|}$, which I don't think is defined. My solution looks like this: First, we sub in $u = x-t, \,\,du = dx$: \begin{align} \int_0^{t'} du \,e^{-\nu \,|u|}\\ \end{align} Where $t'$ indicates that the new limit. Then we re-write the integral as $$\frac{|u|}{u}\int_0^t\frac{u}{|u|}e^{-\nu|u|}$$ Make another substitution $v = |u|,\,\,dv = \frac{|u|}{u}du$ to give \begin{align} &\quad \frac{|u|}{u}\int_0^{t''}\frac{u}{|u|}\frac{|u|}{u}e^{-\nu\,v}dv\\ &=\frac{|u|}{u}\int_0^{t''}e^{-\nu\, v}dv\\ &=\frac{|u|}{u}\cdot \frac{-1}{\nu}e^{-\nu\,v}\bigg|_0^{t''}\\ & = -\frac{|x-t|}{x-t}\frac{1}{\nu}e^{-\nu\,|x-t|}\bigg|_0^t \end{align} Putting the limits in, we get \begin{align}I &= \frac{1}{\nu}\left(-\frac{|t-t|}{t-t}e^{-\nu\,|t-t|}- \frac{|0-t|}{0-t}e^{-\nu\,|0-t|} \right)\\ &= \frac{1}{\nu}\left( -\frac{|0|}{0}-e^{-\nu t}\right) \end{align} where in the final step I have assumed that $t>0$. What am I missing?