Proving the inverse relationship between exponential function and natural logarithm I am working to prove that $E(x) = \displaystyle\sum_{n=0}^\infty \frac{x^n}{n!}$ and $L(x) = \displaystyle\int_1^x\frac{1}{t}dt $ are inverse functions to one another.
I have already found that $(L \circ E)(x) = x$ for all $x \in \mathbb{R}$, however I am struggling to prove that $(E \circ L)(x)=x$ for all $x > 0$.
My work for $L \circ E$ involved finding $(L \circ E)'(x) = 1$ to imply my result.
Any tips how to proceed for this $(E \circ L)$ result?
Thanks!
 A: Consider $$\left(\frac{(E\circ L)(x)}{x}\right)'=\frac{x\,(E\circ L)'(x)-(E\circ L)(x)}{x^2}\\=\frac{x\,(E'\circ L)(x)L'(x)-(E\circ L)(x)}{x^2} = \frac{(E\circ L)(x)-(E\circ L)(x)}{x^2} = 0$$
Then $(E\circ L)(x) = cx$. Evaluate at $x=1$.
A: Here's a solution in the same vein as your other direction. Let $$y(x)=(E\circ L)(x)=\sum_{n=0}^{\infty}\frac{1}{n!}\left(\int_1^x\frac{1}{t}dt\right)^n\ .$$
Then we have
$$y(1) = \sum_{n=0}^{\infty}\frac{1}{n!}\left(\int_1^1\frac{1}{t}dt\right)^n= 1\ ,$$
as each integral is $0$ for $n\geq 1$. Then you can differentiate $y$ to see
$$\begin{align}y'&=\sum_{n=0}^{\infty}\frac{n}{n!}\left(\int_1^x\frac{1}{t}dt\right)^{n-1}\frac{1}{x}\\
&=\frac{1}{x}\sum_{n=1}^{\infty}\frac{1}{(n-1)!}\left(\int_1^x\frac{1}{t}dt\right)^{n-1}\\
&=\frac{1}{x}y\ . \end{align}$$
Then we have the initial value problem
$$\begin{align}y'=\frac{1}{x}y\\ y(1)=1\ , \end{align}$$
which has the unique solution $$(E\circ L)(x)=y(x)=x\ .$$
You may want to justify 1) the term by term differentiation and 2) the solution to the ODE.
