Nested integral question with integral as upper bound of integration. (Recursive Integral? Nested Integral?) In form: $\int_0^I f(x) dx = I$ NOTE: Even though this question has been marked "Answered" there are still some unanswered questions I would still appreciate answers for. Thank you!
Disclaimer: I'm not sure about the proper term for this type of problem, since I couldn't find it anywhere online except for this youtube video*, in which it is called a nested integral. However, once I looked the term up here to see if the question had been asked before, completely different things popped up, so I apologize for any confusion that may have been created.
*https://www.youtube.com/watch?v=xnFCncV-288&t=1s
When you compute the integral $\int_0^I 2x dx = I$, you get two solutions: 1 and 0. I have a few questions here:

*

*Is this a sort of definite integral? (If not, is there a proper term for this?) If so, how can it equal two things at once?

*Are both answers correct? If so, which answer would be more valid, if either?

*When this is extended to $\int_0^I 4x^3 dx = I$ (And past. i.e $\int_0^I 17x^{16} dx = I$), you start to get complex solutions. Are these valid answers? If so, how do you take an integral with a complex number as a bound of integration? (is there an intuitive graphical representation of this that could help me understand?)

*Given $\int_0^I f(x) dx = I$, and $F(I)-I-F(0)=0$, are there functions $f(x)$ that have infinitely many solutions*? Are there equations with an infinite amount of real solutions and a finite number of complex? Functions with a finite number of real and an infinite number of complex?$^{I have figured out a function that fits these criteria.}$ Are there functions with zero solutions of both, or functions that only have complex solutions? (I would also be interested in more functions with purely real solutions** outside of trivial ones like, $2x dx$, and $3x^2 dx$. Anything in the form $\int_0^I nx dx = I$, $n \in \Bbb R$, $n \neq 0$, is trivial and the solution is given by $I = \frac{2}{n}$, $0$.)

*Outside of $\int_0^I 1dx = I$, which is valid for everything.
**Another function I have found is $\int_0^I \ln (x)dx = I$, which gives the result $e^2$. In addition, if you take $\int_0^I \ln (nx) dx = I$, you get a solution in the form $\frac {e^2}{n}$ when $n \neq 0, e$. If $n = e^c$, where $c \in \Bbb N$, you get an answer in the form $\frac {e^2}{e^c}$, or $\frac {1}{e^{c-2}}$. If $n=e^{-c}$, $c \geq 1$, the rule appears to be $I = e^{c+2}$. The rule for when $c \lt 1$ is more complex, and I don't believe I fully understand it. If $c$ is in the form $\frac {1}{r}$, where $r \in \Bbb Z$, the solution is $e^{\frac {2r-1}{r}}$, and when in the form $\frac {k}{r}$, with $k \in \Bbb Z$ and $k \neq 1$, as long as it doesn't reduce into the previous form $\frac {1}{r_1}$, the solution is instead given by $e^{\frac {2r+k}{r}}$. I would like help understanding why this is. In the case $n^{-c}$, $c \in \Bbb Z$, the rule is simply $ce^2$. For the cases where $c$ is a fraction, the rule is again different. I will update this soon once I have more time.
Representation of what one of these integrals look like for anyone who might be confused: $$\int_0^{\int_0^{\int_0^I f(x) dx} f(x) dx} f(x) dx = I$$
Some things I've noticed: I am fairly sure you can use some sort of trigonometry to find something with infinite solutions, as it's periodic, but my math is not advanced enough to prove anything here. All I know is that if you want $n$ number of solutions it is very easy to generate it by taking $\int_0^I nx^{n-1} dx = I$ when $n \in \Bbb N$.
Also, bonus question: Is there a way to write this as an infinite series, or continued fraction, or something in that vein?
Bonus bonus question I thought of just as I was about to post this: What would the solution be to $\int_0^I \frac 1x dx = I$? Is there a solution?

*

*Note for this question: When plugged into Symbolab, it says that this integral "diverges." What does it mean in this context? Is it accurate?

Bonus bonus bonus question: Can every $f(x)$ that does not have an undefined section (as in $\int_0^I \frac 1x dx = I$) be solved for a numerical value? If not, why? Same for if so.

*

*Note for this question: For example, would this be solvable if, say, $f(x)$ was $e^{-x^2}$, as in the Gaussian Integral? Would the integral only be solvable if it had an elementary integral?

I couldn't find any tags I thought matched this well enough, so I just went with improper-integrals, integration, and calculus.
Thank you! I would appreciate answers to any of these questions, but I'll only mark the question "answered" if all of my main questions have been answered.
Note: I have slightly edited this to make some aspects of the question clearer and more precise.
Note 2: I have again edited this to update with some new findings and interesting tidbits, as well as slightly clarify some things.
Note 3: Same as Note 2.
 A: Consider a function $f$ that is continuous on an interval $J$ and $a\in J$. Then, for every $x$ in $J$, the definite integral
$$A(x)=\int_a^x f(t)\; dt$$
exists in $\mathbb R$ (as the limit of Riemann Sums). This defines a function $A$ from $J$ into $\mathbb R$. The problem in the linked video is to find the fixed points $I$ of the function $A$, that is the numbers $I$ satisfying $A(I)=I$.
In the case where $f$ has an elementary antiderivative, the Fundamental Theorem of Calculus implies that the equation written using an integral can be written as a "classic equation". In the given example where $f(x)=2x$ and $a=0$, we have
$$
\int^I_0 2x\; dx = I \iff I^2 = I
$$
To answer your questions,


*

*Is this a sort of definite integral? (If not, is there a proper term for this?) If so, how can it equal two things at once?


The quantity $\int_0^I 2x\; dx$ is a definite integral. Note that definite integrals are just a way of expressing numbers (for example, $\int_0^{2\pi} \sin(x)\; dx$ is just another way of writing $0$). Numbers can be expressed in a lot of ways.



*Are both answers correct? If so, which answer would be more valid, if either?


$\int_0^I f(x)\; dx = I$ is an equation. It can have $0$, $1$, finitely or infinitely many solutions. None of them is more valid (except if more conditions are given, such as we are looking for positive solutions.)



*When this is extended to $\int_0^I 4x^3 dx = I$ (And past. i.e $\int_0^I 17x^{16} dx = I$), you start to get complex solutions. Are these valid answers? If so, how do you take an integral with a complex number as a bound of integration? (is there an intuitive graphical representation of this that could help me understand?)


That should probably be an independent question, as it is a vast subject. The construction of definite integrals can be extended to line segments $[a,b]$ where $a$ and $b$ are complex numbers. Remember that complex numbers are points in the plane, so line segment between them is well-defined. Alternatively, you can look at line integrals.



*Given $\int_0^I f(x) dx = I$, and $F(I)-I-F(0)=0$, are there functions $f(x)$ that have infinitely many solutions*?


I will answer the first part only. You can always choose an antiderivative whose value at $0$ is $0$ (if defined at $0$), so the equation becomes $F(I)=I$, and you are looking for a function $F$ with infinitely many fixed points. For example, you can consider $F(x)=x+\sin(x)$ (so $f(x)=1+\cos(x)$). So your intuition about trigonometric functions was right!

Bonus bonus bonus question: Can every $f(x)$ that does not have an undefined section (as in $\int_0^I \frac 1x dx = I$) be solved for a numerical value? If not, why? Same for if so.

Not every equation has solutions that can be expressed using elementary functions. As for numerical approximations, if $f$ is continuous, then $A$ is differentiable, so you can use standard techniques (Intermediate Value Theorem or Newton's Method for the solutions of $A(I)-I=0$.
