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 $limL(f,P_n)=limU(f,P_n)$. Indeed - for each $n$, let $P_n$ denote the evenly spaced partition of $[a,b]$ with widths ${\mathrm{\Delta }}_n=(b-a)/n$ $$P_n=a,a+{\mathrm{\Delta }}_n,a+2{\mathrm{\Delta }}_n,\cdots ,a+n{\mathrm{\Delta }}_n=b$$

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,

$$\begin{array}{rlrl}L(E,P_n)& =\sum _{0\le i<n}\underset{P_i}{inf}E(x)|P_i|& =\sum _{0\le i<n}E(t_i){\mathrm{\Delta }}_n& =\sum _{0\le i<n}E(a+i{\mathrm{\Delta }}_n){\mathrm{\Delta }}_n\end{array}$$

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

$$\begin{array}{rlrlr}E(a+i{\mathrm{\Delta }}_n)& =E(a)E(i{\mathrm{\Delta }}_n)& =E(a)E({\mathrm{\Delta }}_n+{\mathrm{\Delta }}_n+\cdots +{\mathrm{\Delta }}_n)& =E(a)E({\mathrm{\Delta }}_n)E({\mathrm{\Delta }}_n)\cdots E({\mathrm{\Delta }}_n)& =E(a)E({\mathrm{\Delta }}_n{)}^i\end{array}$$

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

$$\begin{array}{rlr}\sum _{0\le i<n}E(a+i{\mathrm{\Delta }}_n){\mathrm{\Delta }}_n& =\sum _{0\le i<n}E(a)E({\mathrm{\Delta }}_n{)}^i{\mathrm{\Delta }}_n& =E(a){\mathrm{\Delta }}_n\sum _{0\le i<n}E({\mathrm{\Delta }}_n{)}^i\end{array}$$

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

$$\sum _{0\le i<n}E({\mathrm{\Delta }}_n{)}^i=\frac{1-E({\mathrm{\Delta }}_n{)}^n}{1-E({\mathrm{\Delta }}_n)}$$

But, we can simplify even further! Using again the laws of exponents we see that $E({\mathrm{\Delta }}_n{)}^n$ is the same as $E(n{\mathrm{\Delta }}_n)$, and $n{\mathrm{\Delta }}_n$ 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 $L(E,P_n)$:

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

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

$$\begin{array}{rlr}E(a)(1-E(b-a))& =E(a)-E(b-a)E(a)& =E(a)-E(b)\end{array}$$

Thus for every $n$ we have

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

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

$$\frac{-1}{\frac{{\mathrm{\Delta }}_n}{1-E({\mathrm{\Delta }}_n)}}=\frac{E({\mathrm{\Delta }}_n)-1}{{\mathrm{\Delta }}_n}$$

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 ${\mathrm{\Delta }}_n\to 0$ this implies

$$lim\frac{E({\mathrm{\Delta }}_n)-E(0)}{{\mathrm{\Delta }}_n}=E^{\prime }(0)$$

Thus, our original limit ${\mathrm{\Delta }}_n/(1-E({\mathrm{\Delta }}_n))$ is the negative reciprocal of this, and

$$\begin{array}{rlrlr}limL(E,P_n)& =lim(E(a)-E(b))\frac{{\mathrm{\Delta }}_n}{1-E({\mathrm{\Delta }}_n)}& =(E(a)-E(b))lim\frac{{\mathrm{\Delta }}_n}{1-E({\mathrm{\Delta }}_n)}& =(E(a)-E(b))\frac{-1}{E^{\prime }(0)}& =\frac{E(b)-E(a)}{E^{\prime }(0)}\end{array}$$

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

$$\begin{array}{rlrl}U(E,P_n)& =\sum _{0\le i<n}\underset{P_i}{sup}E(x)|P_i|& =\sum _{0\le i<n}E(t_{i+1}){\mathrm{\Delta }}_n& =\sum _{0\le i<n}E(a+(i+1){\mathrm{\Delta }}_n){\mathrm{\Delta }}_n\end{array}$$

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:

$$\begin{array}{rlrlr}U(E,P_n)& =\sum _{0\le i<n}E(a+(i+1){\mathrm{\Delta }}_n){\mathrm{\Delta }}_n& =\sum _{0\le i<n}E({\mathrm{\Delta }}_n)E(a+i{\mathrm{\Delta }}_n){\mathrm{\Delta }}_n& =E({\mathrm{\Delta }}_n)\sum _{0\le i<n}E(a+i{\mathrm{\Delta }}_n){\mathrm{\Delta }}_n& =E({\mathrm{\Delta }}_n)L(E,P_n)\end{array}$$

Thus, $U(E,P_n)=E({\mathrm{\Delta }}_n)L(E,P_n)$ for every $n$. Since $E$ is continuous, $$limE({\mathrm{\Delta }}_n)=E(lim{\mathrm{\Delta }}_n)=E(0)=1$$

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

$$\begin{array}{rlrlr}limU(E,P_n)& =lim(E({\mathrm{\Delta }}_n)L(E,P_n))& =(limE({\mathrm{\Delta }}_n))(limL(E,P_n))& =limL(E,P_n)& =\frac{E(b)-E(a)}{E^{\prime }(0)}\end{array}$$

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:

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