Let $f_1:\mathbb{R}\to \mathbb{R}$ be a locally integrable function (that is $f_1\in L^1_{loc}(\mathbb{R})$). Let us define $f_{n+1}:=\int_0^x f_n(t)\,dt$ for all $n\ge1$. We consider the series $S(x)=\sum_{n=1}^{+\infty} f_n(x)$.

The problem asks: prove that $S$ pointwise converges for all $x\in\mathbb{R}$ and find a closed form for the sum.

For the part about the pointwise convegence I think one can consider:

$|f_n(x)|=|\int_0^x\int_0^{t_1}...\int_0^{t_{n-2}} f_1(t_{n-1})\,dt_{n-1}\,dt_{n-2}\,...\,dt_1 |\le||f_1||_{L^1(0,x)}\frac{x^{n-2}}{(n-2)!}=C\frac{x^{n-2}}{(n-2)!}$

with $C\ge 0$ depending on $x$. By Stirling approximation $\frac{x^{n}}{(n)!}$ is asymptotic to $\frac{1}{\sqrt{2\pi n}}\big(\frac{xe}{n}\big)^n$ which gives a converging series.

But now what about the closed form for the series? How can we proceed? I am not even sure about what one means by "closed form" in this case. Thank you all!


2 Answers 2


They key is for $m \ge 0$, $f_{m+2}$ can be expressed as a single integral over $f_1$. More precisely,

$$f_{m+2}(x) = \int_0^x \frac{(x-y)^m}{m!} f_1(y) dy\tag{*1}$$

One can show this by induction.

The case $m = 0$ is trivial, it is essentially the definition of $f_2$.

Assume $(*1)$ is true for some $m$. By definition of $f_{m+3}$ and induction assumption, we have

$$\begin{align}f_{m+3}(x) &= \int_0^x f_{m+2}(y) dy = \int_0^x \int_0^y \frac{(y-z)^m}{m!} f_1(z) dzdy\\ &= \int_0^x \int_0^x \frac{(y-z)^m}{m!}\theta(y-z)f_1(z) dzdy \end{align} $$ where $\theta(t)$ is the Heaviside step function.

Notice $\displaystyle\;\frac{(y-z)^m}{m!}\theta(y-z)\;$ is $L^\infty$ on $[0,x]^2$. Since the product of a $L^\infty$ function with a $L^1$ function is $L^1$. Using Fubini's theorem, we can exchange order of integration and get

$$f_{m+3}(x) = \int_0^x \int_0^x \frac{(y-z)^m}{m!}\theta(y-z)f_1(z) dydz = \int_0^x \frac{(x-z)^{m+1}}{(m+1)!}f_1(z)dz $$ So $(*1)$ is also true for $m+1$. By principle of induction, $(*1)$ is true for all $m \ge 0$.

As a result, we find

$$\sum_{n=1}^\infty f_n(x) = f_1(x) + \sum_{m=0}^\infty f_{m+2}(x) = f_1(x) + \sum_{m=0}^\infty \int_0^x \frac{(x-z)^m}{m!}f_1(z) dz$$

Using DCT (which is essentially using Fubini in disguise ), one can exchange the order of summation and integration. At the end, we obtain: $$\sum_{n=1}^\infty f_n(x)= f_1(x) + \int_0^x \left(\sum_{m=0}^\infty \frac{(x-z)^m}{m!}\right) f_1(z) dz = f_1(x) + \int_0^x e^{x-z} f_1(z) dz $$


Here is an idea of what you should expect in the closed formula. This is kind of cheating, but again, it is just to give you an idea. Assume the series $f_n$ satisfies all conditions so that we can derive inside the sum(uniform convergence, $f_1$ is differenciable, etc)

We must have then

$$S'(x)- S(x)= f_1'(x)\;\forall\; x\in \mathbb{R}.$$ Solving this differential equation with $S(0)=f_1(0)$ we find

$$S(x)= f_1(x) + e^x \int_0^x f_1(t)e^{-t} dt $$


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