Can the integral $\int e^{dx}$ be solved?

I took Calculus I course, so I can compute easy integrals, but not observe and develop new things, like this particular integral. As far as I know, the $\frac{dx}{dy}$ notation is from the times when Leibniz thought about it as a fraction of infinitesimally small amounts of change, and this notation has many useful properties. For example, we can neatly express integral in that form, but we are not really multiplying by $dx$. It just happens that it is correct (but it isn't correct), so I suppose, that $\int e^{dx}$ has not sense. (I hope I am right till now.)

So, I have 2 hypothesis about the solution of this integral:

  1. This has no sense: the Leibniz notation just happens to have useful properties for multiplying, but it can't be used like that.
  2. You can somehow solve this. I have seen this case many times: the problem that seemed impossible to someone was so easy to other. I believe in mathematicians and I believe that they can figure out anything : )

I personally thing that the case 2 is correct. But is it really?

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    $\begingroup$ Can $x^!$ be solved? $\sqrt{{\rm True}}$? $\sum\limits_{x=!}^{\sqrt{ +}}$? $\endgroup$ May 20, 2021 at 21:14
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    $\begingroup$ @David To be fair, sense can arguably be made of $\int x^{dx}-1$ (as seen in youtu.be/VVF2nZ0WOY4 ) and a whole lot of mathematical discoveries have been made from unjustified formal manipulations, including the original solution to the Basel problem and solving the cubic, etc. $\endgroup$
    – Mark S.
    May 20, 2021 at 21:25
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    $\begingroup$ It has happened rather often in mathematics that one has extended formulas for example $e^x$ with $x$ real to $exp(A)$ where $A$ is a square matrix. But in the case you propose, it is not an extension ; beware of chimeras... $\endgroup$
    – Jean Marie
    May 20, 2021 at 21:26
  • $\begingroup$ Being totally whimsical and informal, you could Taylor expand, $e^{dx}=1+dx+\frac{(dx)^2}{2}+\cdots$, and then use the standard idea (en.m.wikipedia.org/wiki/Smooth_infinitesimal_analysis) that $(dx)^n=0$ for $n\geq 2$ to say $e^{dx}=1+dx$... but this doesn’t really lead anywhere interesting, so far as I can see. $\endgroup$ May 20, 2021 at 21:35
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    $\begingroup$ @DavidG.Stork $\sqrt{True}$ is in programming $True$ : ) $\endgroup$
    – User123
    May 20, 2021 at 21:42

3 Answers 3


As far as I am aware, there is not currently a standardised way to deal with the notation ${\int e^{dx}}$. I think really your question should not be posed as "can it be solved?" but instead "can we sensibly define a way to deal with it?". Using the word "solution" or "solved" implies that it has already got some unambiguous, derivable answer.

Edit. You asked in the comments about how we could potentially go about assigning this integral some sort of "answer". Currently, I don't know if you could. But just to point out, $$ \int x^{dx}-1 $$ was "given" an answer by considering this the same as $$ \int \frac{x^{dx}-1}{dx}dx $$ and then because $$ \lim_{a\to 0}\frac{x^a-1}{a} = \ln(x) $$ you view this as $$ \int \ln(x)dx $$ the same sort of semantical trick would not work here, because $$ \lim_{a\to 0}\frac{e^a}{a} $$ does not exist.

  • $\begingroup$ Yes, I asked about its solvability, but then in the main section I asked about the sense of it. So, one mathematician for the first case. Any other idea? $\endgroup$
    – User123
    May 20, 2021 at 21:29
  • $\begingroup$ @User123 currently, I'm not sure. I mean, if you consider how ${\int x^{dx}-1}$ was dealt with, it essentially was rewritten as ${\int \frac{x^{dx}-1}{dx}dx}$, and then you use the fact ${\lim_{a\to 0}\frac{x^a-1}{a} = \ln(x)}$ to get this is ${\int \ln(x)dx}$. Again, this is not a "solution", but really just a semantic way to get something that could be considered an answer. The same extension would not work here, since ${\lim_{a\to 0}\frac{e^{a}}{a}}$ does not exist $\endgroup$ May 20, 2021 at 21:33
  • $\begingroup$ Ok, thank you for the answer. I just hoped to get the solution, but obviously it can't exist : ) $\endgroup$
    – User123
    May 20, 2021 at 21:37
  • $\begingroup$ @User123 I misread what you wrote, apologies. Yes you are right $\endgroup$ May 20, 2021 at 21:43
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    $\begingroup$ @Riemann'sPointyNose: The problem with your approach (and the linked video) is that it multiplies the integrand by "1" ($\frac{dx}{dx}$) and then magically takes the limit involving just the denominator. Splitting an equation and taking limits of parts independently like this is not always valid... and I'm pretty sure not valid here either. $\endgroup$ May 20, 2021 at 21:58

Just a thought. Maybe complete nonsense. $$ \int e^{\mathrm dx}=\sum_{k=0}^\infty \int\frac{\mathrm dx^k}{k!}=1+\int\mathrm dx+\underbrace{\sum_{k=2}^\infty \int\frac{\mathrm dx^k}{k!}}_{=0}. $$ Hence, $$ \int e^{\mathrm dx}=1+x. $$

Edit: As per @user76284 comments it may make more sense to instead write $$ \int e^{\mathrm dx}-1=\sum_{k=1}^\infty \int\frac{\mathrm dx^k}{k!}=\int\mathrm dx+\underbrace{\sum_{k=2}^\infty \int\frac{\mathrm dx^k}{k!}}_{=0}=x. $$

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    $\begingroup$ To be fair, this is probably the only sort of "sensible" thing to assign to it $\endgroup$ May 20, 2021 at 21:50
  • $\begingroup$ Nice, the result. Can someone explain how "the video method" (youtu.be/VVF2nZ0WOY4) given by @MarkS in the comments doesn't have a limit (thus no sense), but this is valid (probably)? $\endgroup$
    – User123
    May 20, 2021 at 21:53
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    $\begingroup$ How did you get $\int (\mathrm{d}x)^0 = \int \stackrel{?}{=} 1$? $\endgroup$
    – user76284
    May 20, 2021 at 22:10
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    $\begingroup$ But that requires adding $\int 1$ to both sides, not $1$. $\endgroup$
    – user76284
    May 20, 2021 at 22:16
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    $\begingroup$ @user76284 So it would seem that maybe the original integral does not make sense but we can make sense of $\int e^{\mathrm dx}-1$. $\endgroup$ May 20, 2021 at 22:18

Note: This isn't a real answer to my own question, but just a reply on \DavidG.Stork 's comment. There is high probability that this is false, so please comment to tell me the mistake. I have only finished Calculus I, so I don't know much about this subject.

About the @DavidG.Stork 's comment and joking about such semantic hocus pocus: he gave an exercise: $$\int \frac{f(x)}{dx}$$

So I tried to do it (again, this is almost certainly not correct. Where did it get wrong?)

$$\int f'(x)\cdot dx=\lim_{n->\infty}\sum_{i=0}^{n}f(\frac{ix}{n})\cdot \frac{x}{n}= \sum_{i=0}^\infty f(i\cdot dx)\cdot dx$$

As there is a sum on both sides, we can divide by $(dx)^2$:

$$\int \frac{f'(x)}{dx}=\sum_{i=0}^\infty \frac{f(i\cdot dx)}{dx}=\lim_{n->\infty}\sum_{i=0}^{n}f(\frac{ix}{n})\cdot \frac{n}{x}$$

So, this might be an answer.

For example, $f(x)=x$: $f'(x)=1$: $$\int \frac{1}{dx}=\sum_{i=0}^\infty \frac{i\cdot dx}{dx}=\sum_{i=0}^\infty i$$

It doesn't converge. But if we did this many non-rigorous mathematical stuff, let's give this sum a Ramanujan value: $-\frac{1}{12}$. Again, this may not be correct, and if it is, then this isn't a real solution, because we used Leibniz notation and I just "extended" the definition of the integral. It is just a value that can be connected to such integral.


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