Normalizing a biexponential function so its time integral is always unity

Question

I'm reading a paper that uses the following function:

\begin{align} f(t) &= \frac{1}{\tau_1-\tau_2}(e^{-t/\tau_1}-e^{-t/\tau_2}). \end{align}

The authors state that

The normalization adopted here ensures that the time integral of $f(t)$ is always unity.

I figured I'd verify this by taking the integral of $f$.

This is what I get:

\begin{align} \frac{1}{\tau_1-\tau_2}\bigg(\int e^{-t/\tau_1}dt-\int e^{-t/\tau_2}dt\bigg)&=\frac{e^{-t/\tau_2}\tau_2-e^{-t/\tau_1}\tau_1}{\tau_1-\tau_2}.\\ \end{align}

I'm not seeing how to reduce this to 1. Can anyone help me out?

Answer 1

You're supposed to take the definite integral over the entire time domain $$ \int_0^\infty \left(e^{-t/\tau_1}-e^{-t/\tau_2}\right)\, dt = \left[\tau_1e^{-t/\tau_1} - \tau_2e^{-t/\tau_2}\right]_0^\infty = \tau_1 - \tau_2 $$