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

Question

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?

Answer 1

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.

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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.

Answer 2

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 $$ \int e^{\mathrm dx}-1=\sum_{k=1}^\infty \int\frac{\mathrm dx^k}{k!}=\int\mathrm dx+\underbrace{\sum_{k=2}^\infty \int\frac{\mathrm dx^k}{k!}}_{=0}=x. $$

Answer 3

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.