Note: Exponential Integration

We will show the argument for $E$ an increasing exponential (its base $E(1)>1$): an identical argument applies to decreasing exponentials (only switching $U$ and $L$ in the computations below).

To show $E(x)$ is integrable, we use @thm-compute-integral-sequence, which assures us it is enough to find a sequence $P_n$ of shrinking partitions where $\lim L(f,P_n)=\lim U(f,P_n)$. Indeed - for each $n$, let $P_n$ denote the evenly spaced partition of $[a,b]$ with widths $\Delta_n = (b-a)/n$

$$P_n=\{a,a+\Delta_n , a+2\Delta_n,\cdots, a+n\Delta_n=b\}$$

We will begin by computing the lower sum. Because $E$ is continuous, it achieves a maximum and minimum value on each interval $P_i=[t_i,t_{i+1}]$. And, since $E$ is monotone increasing, this value occurs at the leftmost endpoint. Thus,

$$ \begin{align*} L(E,P_n)&=\sum_{0\leq i < n} \inf_{P_i}\{E(x)\}|P_i|\\ &= \sum_{0\leq i < n} E(t_i)\Delta_n\\ &= \sum_{0\leq i < n} E(a+i\Delta_n)\Delta_n \end{align*} $$

Using the law of exponents for $E$ we can simplify this expression somewhat:

$$ \begin{align*} E(a+i\Delta_n)&=E(a)E(i\Delta_n)\\ &=E(a)E(\Delta_n+\Delta_n+\cdots+\Delta_n)\\ &= E(a)E(\Delta_n)E(\Delta_n)\cdots E(\Delta_n)\\ &= E(a)E(\Delta_n)^i \end{align*} $$

Plugging this back in and factoring out the constants, we see that the summation is actually a partial sum of a geometric series:

$$ \begin{align*} \sum_{0\leq i < n} E(a+i\Delta_n)\Delta_n&= \sum_{0\leq i < n}E(a)E(\Delta_n)^i \Delta_n\\ &= E(a) \Delta_n\sum_{0\leq i < n}E(\Delta_n)^i \end{align*} $$

Having previously derived the formula for the partial sums of a geometric series, we can write this in closed form:

$$\sum_{0\leq i < n}E(\Delta_n)^i=\frac{1-E(\Delta_n)^n}{1-E(\Delta_n)}$$

But, we can simplify even further! Using again the laws of exponents we see that $E(\Delta_n)^n$ is the same as $E(n\Delta_n)$, and $n\Delta_n$ is nothing other than the width of our entire interval, so $b-a$. Thus the numerator becomes $1-E(b-a)$, and putting it all together yields a simple expression for $L(E,P_n)$:

$$L(E,P_n)=E(a)\Delta_n \frac{1-E(b-a)}{1-E(\Delta_n)}$$

Some algebraic re-arrangement is beneficial: first, note that by the laws of exponents we have

$$ \begin{align*} E(a)(1-E(b-a))&=E(a)-E(b-a)E(a)\\ &=E(a)-E(b) \end{align*} $$

Thus for every $n$ we have

$$L(E,P_n)=\left(E(a)-E(b)\right)\frac{\Delta_n}{1-E(\Delta_n)}$$

We are interested in the limit as $n\to\infty$: by the limit laws we can pull the constant $E(a)-E(b)$ out front, and only concern ourselves with the fraction involving $\Delta_n$. There’s one final trick: look at the negative reciprocal of this fraction:

$$\frac{-1}{\frac{\Delta_n}{1-E(\Delta_n)}}=\frac{E(\Delta_n)-1}{\Delta_n}$$

Because we know $E(0)=1$ for all exponentials, this latter term is none other than the difference quotient defining the derivative for $E$! Since we have proven $E$ to be differentiable, we know that evaluating this along any sequence converging to zero yields the derivative at zero. And as $\Delta_n\to 0$ this implies

$$\lim \frac{E(\Delta_n)-E(0)}{\Delta_n}= E^\prime(0)$$

Thus, our original limit $\Delta_n/(1-E(\Delta_n))$ is the negative reciprocal of this, and

$$ \begin{align*}\lim L(E,P_n)&=\lim \left(E(a)-E(b)\right)\frac{\Delta_n}{1-E(\Delta_n)}\\ &= \left(E(a)-E(b)\right)\lim \frac{\Delta_n}{1-E(\Delta_n)}\\ &=\left(E(a)-E(b\right)) \frac{-1}{E^\prime(0)}\\ &=\frac{E(b)-E(a)}{E^\prime(0)} \end{align*} $$

Phew! That was a lot of work! Now we have to tackle the upper sum. But luckily this will not be nearly as bad: we can reuse most of what we’ve done! Since $E$ is monotone increasing, we know that the maximum on any interval occurs at the rightmost endpoint, so

$$ \begin{align*} U(E,P_n)&=\sum_{0\leq i < n} \sup_{P_i}\{E(x)\}|P_i|\\ &= \sum_{0\leq i < n} E(t_{i+1})\Delta_n\\ &= \sum_{0\leq i < n} E(a+(i+1)\Delta_n)\Delta_n \end{align*} $$

Comparing this with our previous expression for $L(E,P_n)$, we see (unsurprisingly) its identical except for a shift of $i\mapsto i+1$. The law of exponents turns this additive shift into a multiplicative one:

$$ \begin{align*} U(E,P_n) &= \sum_{0\leq i < n} E(a+(i+1)\Delta_n)\Delta_n\\ &= \sum_{0\leq i < n} E(\Delta_n)E(a+i\Delta_n)\Delta_n\\ &=E(\Delta_n) \sum_{0\leq i < n}E(a+i\Delta_n)\Delta_n\\ &= E(\Delta_n)L(E,P_n) \end{align*} $$

Thus, $U(E,P_n)=E(\Delta_n)L(E,P_n)$ for every $n$. Since $E$ is continuous,

$$\lim E(\Delta_n)=E(\lim \Delta_n)=E(0)=1$$

And, as $L(E,P_n)$ converges (as we proved above) we can apply the limit theorem for products to get

$$ \begin{align*}\lim U(E,P_n) &=\lim (E(\Delta_n)L(E,P_n))\\ &=\left(\lim E(\Delta_n)\right)\left(\lim L(E,P_n)\right)\\ &= \lim L(E,P_n)\\ &= \frac{E(b)-E(a)}{E^\prime(0)} \end{align*} $$

Thus, the limits of our sequence of upper and lower bounds are equal! And, by the argument at the beginning of this proof, that squeezes $L(E)$ and $U(E)$ to be equal as well. Thus, $E$ is integrable on $[a,b]$ and its value is what we have squeezed:

$$\int_{[a,b]}E = \frac{E(b)-E(a)}{E^\prime(0)}$$