# College Math Teaching

## August 28, 2017

### Integration by parts: why the choice of “v” from “dv” might matter…

We all know the integration by parts formula: $\int u dv = uv - \int v du$ though, of course, there is some choice in what $v$ is; any anti-derivative will do. Well, sort of.

I thought about this as I’ve been roped into teaching an actuarial mathematics class (and no, I have zero training in this area…grrr…)

So here is the set up: let $F_x(t) = P(0 \leq T_x \leq t)$ where $T_x$ is the random variable that denotes the number of years longer a person aged $x$ will live. Of course, $F_x$ is a probability distribution function with density function $f$ and if we assume that $F$ is smooth and $T_x$ has a finite expected value we can do the following: $E(T_x) = \int^{\infty}_0 t f_x(t) dt$ and, in principle this integral can be done by parts….but…if we use $u = t, dv = f_x(t), du = dt, v = F_x$ we have:

\

$t(F_x(t))|^{\infty}_0 -\int^{\infty}_0 F_x(t) dt$ which is a big problem on many levels. For one, $lim_{t \rightarrow \infty}F_x(t) = 1$ and so the new integral does not converge..and the first term doesn’t either.

But if, for $v = -(1-F_x(t))$ we note that $(1-F_x(t)) = S_x(t)$ is the survival function whose limit does go to zero, and there is usually the assumption that $tS_x(t) \rightarrow 0$ as $t \rightarrow \infty$

So we now have: $-(S_x(t) t)|^{\infty}_0 + \int^{\infty}_0 S_x(t) dt = \int^{\infty}_0 S_x(t) dt = E(T_x)$ which is one of the more important formulas.