Here is something I've found on the internet $$\begin{aligned} f-\int f&=1\\ \left(1-\int\right)f&=1\\ f&=\left(\frac1{1-\int}\right)1\\ &=\left(1+\int+\int\int+\dots\right)1\\ &=1+\int1+\int\int1+\dots\\ &= 1+x+\frac{x^2}2+\dots\\ &= e^x \end{aligned}$$ At first I interpreted this as a joke, but on closer inspection I'm not so sure...

The first thing I checked was the solution, and fairly enough, $e^x$ satisfies the initial equation. It is not the only solution, though: $\lambda e^x$ works jut as well for any $\lambda\in\mathbb R$. I'm not really familiar with integral equations; I don't know if the solution space is a vector space. Maybe it is and our goal is to find a basis for it. If that is the case, then $e^x$ may actually be the expected result.

That being settled, I started looking at the development.

  • Writing $\displaystyle f-\int f$ as $\displaystyle \left(1-\int\right)f$ is completely fine, its just operational calculus.
  • The "division of both sides" by $\displaystyle\left(1-\int\right)$ must actually mean applying its inverse $\displaystyle\left(\frac1{1-\int}\right)$ on both sides, right? This seems alright, but its not obvious at all to me why should this operator be invertible. In fact, the existence of multiple solutions suggests otherwise...
  • The most ridicule step surely is writing $\displaystyle \frac1{1-\int}=1+\int+\int\int+\dots$. If this is true, it settles my problem with the previous step. Is it, though? I can see how the infinite integral summation may be well-defined, but no further than that. I know it is referencing the formal power series equality $\displaystyle\frac1{1-X} = 1+X+X^2+\dots$, but I don't know if the same proof applies since it requires distributiveness.
  • Finally, there is an infinite amount of problematic constants being vanished into nonexistence at $\displaystyle 1+\int1+\int\int1+\dots$, a single of which would be enough to destroy the solution: $e^x+1$ does not satisfy the equation.

My guess is that this is really a joke, but I want to know just how much of it is true. Where does the problems arise and how big are they? Is there anything salvageable here?

  • 4
    $\begingroup$ This sort of thing is more commonly seen with a $\frac{d}{dx}$ operator. The indefinite integrals are a problem, but maybe one settled by taking the operator to mean $(\int f)(x) = \int_0^x f(t)\, dt$. Still a number of questions here I don't have the answers to. $\endgroup$
    – aschepler
    Commented Feb 28 at 0:27
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    $\begingroup$ It's cute. One issue is writing $f - \int f =1$, whereas the buried $+C$ that's not written means the $1$ is really irrelevant. $\endgroup$
    – Randall
    Commented Feb 28 at 0:27
  • $\begingroup$ Another (real) issue is: what does it mean to sum powers of a linear operator against itself an infinite number of times? $\endgroup$
    – Randall
    Commented Feb 28 at 0:29
  • 2
    $\begingroup$ This seems very much amenable to Banach’s fixed point theorem (if I’m not mistaken it’s called the Volterra integral operator), in order to prove the existence of the inverse operator. The rest is in my mind just cute symbolic manipulation (but really, an iterative solution is how Banach’s fixed point theorem is proved/ how the solution is constructed) so maybe it’s not so bad in hindsight. I’ll think/work it out more carefully later (unless someone else does it in the meantime). $\endgroup$
    – peek-a-boo
    Commented Feb 28 at 0:34
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    $\begingroup$ @Randall The $1$ is not irrelevant if we interpret $\int$ as a definite integral as ashepler mentions. $e^x$ is indeed the unique solution to $f - \int_0^x f(t) \, dt = 1$ (and not any other $\lambda e^x$ as OP mentions.) $\endgroup$ Commented Feb 28 at 1:00

3 Answers 3


In general if $T$ is a bounded linear operator on a Banach space $B$ with norm $||T|| < 1$ then $1-T$ is invertible, $1 + T + T^2 + \cdots$ converges in the Banach space topology, and

$$(1-T)^{-1} = 1 + T + T^2 + \cdots$$

just as in the case of real numbers. See Neumann series.

In this case, interpreting $\int$ as the operator $\mathcal{C}[0, 1-\epsilon] \to \mathcal{C}[0, 1-\epsilon]$ given by $(\int f)(x) = \int_0^x f(t) \, dt$, where $\mathcal{C}[0, 1-\epsilon]$ is the Banach space of continuous real-valued functions on $[0, 1- \epsilon]$ for some fixed $\epsilon > 0$, we note that $\int$ is a bounded linear operator and has norm $||\int|| = 1-\epsilon$ (see Integral as a linear operator is bounded). Then we know that $(1 - \int)$ is invertible and $(1-\int)^{-1}= 1 + \int + \int\int + \int\int\int + \cdots$, so these calculations are indeed justified.

  • 1
    $\begingroup$ I added an ‘alternative’ answer, mainly to satisfy my own curiosity, but do you have a quick argument for whether the domain of definition can be extended beyond $[0,1-\epsilon]$? $\endgroup$
    – peek-a-boo
    Commented Feb 28 at 3:34
  • $\begingroup$ No, it's not obvious to me how to extend it. (+1 to your answer!) $\endgroup$ Commented Feb 28 at 3:51
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    $\begingroup$ @peek-a-boo It is possible to extend the result, because we only need the spectral radius of $T$ to be $<1$ to apply the geometric series. For example on $C[0,1]$ it is well known that the spectrum of $\int$ is $\{0\}$ and so we can at least extend to this case (and probably others too). $\endgroup$
    – jd27
    Commented Feb 28 at 9:34
  • 1
    $\begingroup$ @jd27 ahh spectral radius $< 1$, of course. Ty. And indeed, for any $R>0$, on $C[-R,R]$, one can prove the bound $\|T^n\|\leq R^n/n!$, where $T=\int_0^{\cdot}$, so the spectral radius is $0$ in these cases too, and hence the Neumann series applies here as well :) $\endgroup$
    – peek-a-boo
    Commented Feb 28 at 11:26
  • 2
    $\begingroup$ This is amazing! Mathematics never fails to surprise me. $\endgroup$ Commented Feb 28 at 13:18

The answer by @Jair Taylor addresses the case of bounded operators of norm strictly less than $1$ (hence the restriction to the interval $[0,1-\epsilon]$), and justifies rigorously why the series works. I’m not sure if it’s possible to generalize that argument directly to work for larger domains. Hence, what follows is really more of an excuse to illustrate/apply some of the ideas in functional analysis, particularly around Banach’s fixed point theorem.

Your problem really is a fixed point problem, because $f-\int f=1$ rearranged gives $f=1+\int f$, i.e the function $f$ equals some other operator applied to $f$, namely $\Phi(f)=1+\int f$, i.e we have to solve the equation $f=\Phi(f)$. This is actually a very common theme when solving (nonlinear) integral equations, or even differential equations, or simply to prove many types of existence results: reformulate your original problem as a fixed-point problem (or whatever the case may be), and try to solve that new problem using the ‘fancy’ tools of functional analysis.

Some generalities.

First, a simple lemma.

Lemma 1.

Let $X$ be any set, suppose $\Phi:X\to X$ is a map such that for some $n\in\Bbb{N}$, the $n$-fold composition $\Phi^n$ has a unique fixed point $x$. Then, $x$ is also the unique fixed point of $\Phi$.

To prove this, note first that $\Phi^n(\Phi(x))=\Phi(\Phi^n(x))=\Phi(x)$, meaning that $\Phi(x)$ is a fixed point of $\Phi^n$, so by uniqueness of fixed points of $\Phi^n$, we must have $\Phi(x)=x$, i.e $x$ is a fixed point of $\Phi$. To show uniqueness, suppose $y$ is a fixed-point of $\Phi$. Then, $\Phi^n(y)=\Phi^{n-1}(y)=\dots=\Phi(y)=y$, so $y$ is a fixed point of $\Phi^n$, hence by uniqueness, $y=x$.

Next, we have a simple result about calculating fixed points in complete metric spaces; it’s just a mild (but useful) variant of Banach’s fixed point theorem.


Let $X$ be a non-empty complete metric space, and $\Phi:X\to X$ a map with the following property: for each $n\in\Bbb{N}$, $\Phi^n$ is a Lipschitz map, with constant $C_n$ such that $\sum_{n=0}^{\infty}C_n<\infty$. Then, $\Phi$ has a unique fixed point $\xi$; furthermore, for every $x\in X$, the sequence $\{\Phi^n(x)\}_{n=0}^{\infty}$ converges in $X$ to $\xi$.

The utility of this theorem is that it first tells us that a fixed point exists and is unique, and secondly it tells us how to calculate it: start with any arbitrary guess $x\in X$, and then successively refine your guess by calculating $\Phi(x),\Phi^2(x),\Phi^3(x),\dots$, and this sequence has a limit, and this is limit equals the sought after $\xi$.

To prove the theorem, note that since $\sum_{n=0}^{\infty}C_n<\infty$, we must have $C_n\to 0$ as $n\to\infty$, so there exists an $n_0$ such that $C_{n_0}<1$, and so $\Psi:=\Phi^{n_0}$ is a contraction in $X$, hence by Banach’s fixed point theorem, $\Psi$ has a unique fixed point, call it $\xi$; and furthermore, Banach’s theorem tells us that to calculate this fixed point, we can start with any $x\in X$, and then $\{\Psi^n(x)\}_{n=0}^{\infty}$ will have a limit and the limit will equal $\xi$.

We’re almost done, but not quite, because the theorem talks about $\Phi$, not $\Psi$. But now first applying lemma 1, we see that $\xi$ is also the unique fixed point of $\Phi$. Next, note that $\{\Phi^n(x)\}_{n=0}^{\infty}$ is a Cauchy sequence, because for any $n>m$, we have \begin{align} d(\Phi^n(x),\Phi^m(x))\leq\sum_{j=m}^{n-1}d(\Phi^{j+1}(x),\Phi^j(x))\leq\sum_{j=m}^{n-1}C_jd(\Phi(x),x). \end{align} Since $\sum_{k=0}^{\infty}C_k<\infty$, by making $n,m$ large enough, the RHS can be made as small as desired. Lastly, keep in mind that $\{\Psi^n(x)\}_{n=0}^{\infty}$ is actually a subsequence of $\{\Phi^n(x)\}_{n=0}^{\infty}$, which in fact converges to $\xi$. Recall that if we have a Cauchy sequence, and some subsequence converges, then the whole sequence converges to the same limit. This proves that $\Phi^n(x)\to \xi$ as $n\to\infty$, and hence completes the proof of the theorem.

Applying to the question at hand.

We fix an $R>0$, and consider the (complex) Banach space $X=C([-R,R];\Bbb{C})$, and fix any function $h\in X$, and consider the map $\Phi_h:X\to X$ defined by $(\Phi_h f)(x):= h(x)+\int_0^xf(t)\,dt$. Then, a simple exercise in induction shows that for all $f,g\in X$, all $n\in\Bbb{N}$ and all $x\in [-R,R]$, we have \begin{align} |(\Phi_h^nf)(x)-(\Phi_h^ng)(x)|&\leq \|f-g\|\cdot\frac{|x|^n}{n!}, \end{align} and so $\|\Phi_h^nf-\Phi_h^ng\|\leq \|f-g\|\cdot\frac{R^n}{n!}$, meaning that $\Phi_h^n$ is Lipschitz continuous with Lipschitz constant at most $\frac{R^n}{n!}$; the ratio test now implies that $\sum_{n=0}^{\infty}\frac{R^n}{n!}<\infty$, and so by the theorem, $\Phi_h$ has a unique fixed point. Having a unique fixed point means that for every function $h\in X$, there exists a unique function $f\in X$ such that for all $x\in [-R,R]$, \begin{align} f(x)=h(x)+\int_0^xf(t)\,dt, \end{align} or in other words, the operator $I-\int_0^x:X\to X$ is bijective.

In your particular case, fix $h(x)\equiv \lambda$, and let us start off with the function $f_0\equiv 1$. Then, (calling it $\Phi$ instead of $\Phi_{\lambda}$ for simplicity)

  • $(\Phi f_0)(x):=\lambda+\int_0^x1\,dt=\lambda+x$.
  • $(\Phi^2f_0)(x):=\lambda +\int_0^x(\Phi f)(t)\,dt=\lambda+\int_0^x(\lambda +t)\,dt=\lambda+\lambda x+\frac{x^2}{2}$.
  • We can keep going by induction to prove that for all $n\in\Bbb{N}$, $(\Phi^nf_0)(x)=\lambda\sum_{k=0}^{n-1}\frac{x^k}{k!}+\frac{x^n}{n!}$.

Note that we already know $\{\Phi^n(f_0)\}_{n=0}^{\infty}$ converges in $X$ (i.e converges uniformly) to some unique function $f$, which is a fixed point of $\Phi$, i.e satisfies the following equation for all $x\in [-R,R]$: \begin{align} f(x)&=\lambda+\int_0^xf(t)\,dt. \end{align} On the other hand, from the explicit formula for $\Phi^n(f_0)$, we can immediately see that it converges pointwise to $\lambda e^x$. Therefore, we see that $f(x)=\lambda e^x$ is the unique continuous function on $[-R,R]$ which satisfies the above integral equation. And finally, since $R>0$ was chosen arbitrarily, the above equality holds for all $x\in\Bbb{R}$.

In fact, if you think about it, solving this integral equation (via Banach’s fixed point method) is precisely one of the ways of defining the exponential function in the first place: one of the definitions is that $\exp$ is the unique function satisfying $f’=f$ with $f(0)=1$. If you integrate both sides of the ODE, you get the above integral equation with $\lambda=1$ (and indeed, the existence and uniqueness theorems for ODEs go through this route of recasting the ODE into an integral equation which is solvable by such approximating arguments… the underlying principle here is that, as in much of analysis, integrals are better behaved than derivatives). Thus, even if you didn’t know that the pointwise limit of $\Phi^n(f_0)(x)$ was $\lambda e^x$, you could have simply defined $e^x$ to be the limit (which we already proved exists).

Edit: More in line with the question.

The comment by @jd27 made me realize that we can definitely apply the geometric series in general. Again, fix the Banach space $X=C[-R,R]$ of continuous function $[-R,R]\to\Bbb{C}$, and consider the linear operator $T:X\to X$ given by $(Tf)(x):=\int_0^xf(t)\,dt$. Then, by induction, one can show that for all $n\in\Bbb{N}_0$, and all $x\in [-R,R]$, \begin{align} |(T^nf)(x)|\leq\|f\|\cdot\frac{|x|^n}{n!}, \end{align} and hence $\|T^n\|\leq \frac{R^n}{n!}$. So, the series $\sum_{n=0}^{\infty}\|T^n\|$ is finite. Since in any Banach space (completeness is key), absolute convergence implies convergence in norm, it follows that there is a well-defined operator $S=\sum_{n=0}^{\infty}T^n$. Finally, simple manipulations with convergent power series shows that $S$ is the inverse operator of $I-T$. Therefore, $(I-T)^{-1}:X\to X$ exists and can be calculated as $\sum_{n=0}^{\infty}T^n$. Evaluating this series on the constant function with value $\lambda$ shows that $(I-T)^{-1}(\lambda)=\lambda\cdot \exp|_{[-R,R]}$.

Finally, since $R>0$ was arbitrary, we have shown that $f(x)=\lambda e^x$ is the unique solution of the integral equation $f(x)-\int_0^xf(t)\,dt=\lambda$ on all of $\Bbb{R}$.

So, tldr is there’s lots of cool math behind the scenes.


It's actually the main idea behind the Cauchy-Lipschitz theorem. In general, say you want to solve the ODE: $$ y'(x) = f(x,y(x)) $$ with initial condition $y(x_0) = y_0$. In your case, $$ x_0 = 0 \quad y_0 = 1\quad f(x,y) = y $$ The idea is to reformulate it as a fixed point equation. Indeed, if you have a general equation to solve: $$ X = \Phi(X) $$ you can construct a solution by iterating $\Phi$. Formally, applying $\Phi$ an infinite number of times gives a fixed point as one more iteration won't change things (classic use of infinity as a mise en abyme). To make it rigorous, you need to prove that these iterates will converge and if $\Phi$ is continuous, the limit will be a fixed point. In particular, for linear equations, this amounts to the Taylor series of the inverse if you start at $0$: $$ X = AX+B \\ (1-A)X = B \\ X = \left(\sum_{n=0}^\infty A^n\right)B \\ X = A(A(...(0+B)...)+B)+B $$

Back to the ODE, the trick is to rather reformulate it into an integral equation: $$ y(x) = y_0+\int_{x_0}^xf(\xi,y(\xi))d\xi $$ This is an improvement on many levels. First off, it is a fixed point problem. Secondly, you do not need to specify the initial condition, it is included in the equation. Finally, it now makes sense to solve the equation in a more general class of functions, which is useful for topological considerations (like solve equations in $\mathbb R$ is easier than in $\mathbb Q$). The operator is the Picard mapping, and in your case, it's: $$ y(x)\to 1+\int_0^xy(\xi)d\xi $$

The technical part of the proof of Cauchy-Lipschitz is to demonstrate the convergence of the iterates of the Picard mapping, which involves regularity assumptions of $f$ (Lipschitz continuity) and the Banach fixed point theorem. You can check out for example V. Arnold's Ordinary Differential Equations for a more detailed and technical discussion.

Hope this helps.


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