Integrating Exponentials from First Principles

The functional equation alone pins down the antiderivative.

Here’s a quite long calculation showing that it’s possible to integrate exponential functions directly from first principles. The length of this calculation alone is a good selling point for the fundamental theorem of calculus, which will make this trivial!

Exponential Functions

An exponential function is a continuous nonconstant function $E\colon\mathbb{R}\to\mathbb{R}$ satisfying the law of exponents $E(x+y)=E(x)E(y)$ for all $x,y\in\mathbb{R}$ .

This is a functional characterization of exponentials; we don’t specify a formula for how to compute the function but instead we specify how the function ought to behave. There are several facts about exponentials we will need, that follow directly from this definition:

Exponentials are always nonzero
Exponentials are strictly increasing, or strictly decreasing
Exponentials are differentiable everywhere

Our main theorem here is that we compute the Darboux integral of an exponential on an interval, directly from this functional definition. To do so, we make use of minimal facts about integration, so that this can be given early on. One thing we do take advantage of is the ability to compute Darboux integrals with a sequence of partitions (instead of directly from the definition, with infima and suprema)

If $P_n$ is a sequence of shrinking partitions of $J$ and $\lim U(f,P_n)=\lim L(f,P_n)$ then $f$ is integrable on $J$ and $\int_J f$ is this common value.

Without further ado, here’s the advertised calculation!

Integrating Exponentials

Let $E$ be an exponential function, and $[a,b]$ an interval. Then $E$ is Darboux integrable on $[a,b]$ and

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

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:

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

Recalling our definition of $\exp$ as the exponential function with $\exp^\prime(0)=1$ (in Part II on differentiation), we have

On any interval $[a,b]$ the natural exponential is integrable, and $\int_{[a,b]}\exp = \exp(b)-\exp(a)$