I'm going to quote the relevant passage in the textbook and then ask my question.
The assumption that the infectives leave the infective class at rate $\alpha I$ per unit time requires a fuller mathematical explanation, since the assumption of a recovery rate proprotional to the number of infectives has no clear epidemiological meaning. We consider the "cohort" of members who were all infected at one time and let $u(s)$ denote the number of those who are still infective $s$ time units after having been infected. If a fraction $\alpha$ of these leave the infective class in unit time, then
$$ u'=-\alpha u$$
and the solution of this elementary differential equation is
$$u(s)=u(0)e^{-\alpha s} .$$
Thus, the fraction of infectives remaining infective $s$ units are having become infective is $e^{-\alpha s}$, so that the lenth of the infective period is distributed exponentially with mean $${\color{red}{\int^{\infty}_{0} e^{- \alpha s} \ ds = \frac{1}{\alpha}}}.$$
Now if the fraction of infectives remaining infective s units are having become infective is $e^{-\alpha s}$, then the fraction no longer infective $s$ units after having become infective is $1-e^{-\alpha s}$ which is the cdf of an exponential distribution.
Then since the pdf of an exponential distribution is $\alpha e^{- \alpha s}$, isn't then the mean infective period $$E(S) = \alpha \int_{0}^{\infty} se^{-\alpha s} \ ds = \frac{1}{\alpha} \ ?$$
For a non-negative random variable with density function $f(x)$ and cumulative distribution function $F(x)$, one way to calculate the mean is $\displaystyle \int_0^\infty x f(x) \; dx$ but another is $\displaystyle \int_0^\infty (1-F(x)) \; dx$.
You used the former while the book used the latter to get to the same result.
Note that you could write $E[X]= \displaystyle \int_0^\infty S(x) \; dx$ where $S(x)=1-F(x)$ is the survival function.