I've got a derivation of $\int _{0} ^\infty 2^x dx = -1 / \ln 2$ and am wondering if anybody has studied or derived this before.
Using the Borel summation, $\sum _{n=0} ^\infty 2^n = -1$.
What we will do to solve the integral is manually perform Riemann integration by splitting $[0, \infty)$ into lots of little rectangles and then use the infinite sum above to add them up.
The sum approximates the integral by splitting $[0, \infty)$ into rectangles of width $1$. If we want to split it into rectangles of with $1/2^a$ instead, then we can pre-multiply by the width of the rectangles, and do a double summation, first over the integers, and second over each rectangle in the range of that integer.
For integer $n$, the integral $\int _{n} ^{n+1} 2^x dx$ is approximated by splitting the interval $[n, n+1]$ into $2^a$ rectangles as $\int _{x=n} ^{n+1} 2^x dx \approx \frac{1}{2^a} \sum _{k=0} ^{2^a-1} 2^{n + k/2^a}$
The original integral is equal to the sum of these over all $n$, with the limit taken as $a \to \infty$.
$$ \begin{equation} \begin{split} \int _0 ^\infty 2^x dx &= \lim _{a \to \infty} \sum _{n=0} ^\infty \frac{1}{2^a} \sum _{k=0} ^{2^a-1} 2^{n + k/2^a} \end{split} \end{equation} $$
Pulling the $n$ summation all the way to the left
$$ \begin{equation} \begin{split} (1) &= \sum _{n=0} ^\infty 2^n \cdot \lim _{a \to \infty} \frac{1}{2^a} \sum _{k=0} ^{2^a-1} 2^{k/2^a} \end{split} \end{equation} $$
The summation over $n$ on the left is just equal to $-1$. The limit expression on the right is equal to the ordinary Riemann integral $\int _0 ^1 2^x dx = 1/\ln 2$. Therefore the original improper integral is equal to
$$ \begin{equation} \begin{split} (2) &= -\frac{1}{\ln 2} \end{split} \end{equation} $$
Remember this derivation lives in a world where $\sum _{n=0} ^\infty 2^n = -1$, so I'm wondering whether this integral has been studied and deemed interesting by mathematicians, or if there are other non-infinite solutions besides $-1/\ln 2$ that may be argued to be correct solutions.
Thanks!!!!!
Nonsensical "equalities" like yours appear quite frequently in fields which use generating functions heavily, such as combinatorics, number theory, and QFT. (Arguably, they are a part of complex analysis as well, but I would argue that complex analysis almost always works with series that converge.)
Unfortunately, I don't know of any practical uses for this particular equality.
The standard interpretation for such equalities is via analytic continuation or as the leading term in an asymptotic series. Terry Tao has a great blog post on this topic. In situations where one can apply the above techniques (and one can here — see below), there is a strong enough uniqueness property that there is rarely another "sensible" resummation.
In your case, one can write a simpler "proof": for $\alpha>0$, we have $$\int_0^{\infty}{e^{-\alpha x} dx}=\frac{1}{\alpha}$$ The latter is analytic on $\mathbb{C}\setminus\{0\}$, so let $\alpha=-\ln{(2)}$ (which, of course, is not positive).
(Another method, closer to the OP's: $$\int_0^{\infty}{2^x dx}=\sum_{k=0}^{\infty}{2^k\int_0^1{2^x dx}}=\left(\sum_{k=0}^{\infty}{2^k}\right)\int_0^1{2^x dx}$$ where the former is the monotone convergence theorem and the latter factoring out a constant.)