How to show $\sum_{k=0}^{\infty} \frac{1}{k!} \left( \int_{1}^{x} \frac{1}{t} \ dt \right)^k =x$?

Question

There are a lot of ways to show that $e^x$ and $\ln(x)$ are inverse functions of each other depending on how you define them. I am trying to show that given the definitions

$$ e^x:= \sum_{k=0}^{\infty} \frac{x^k}{k!} \qquad \text{and} \qquad \ln(x) := \int_{1}^{x} \frac{1}{t} \ dt$$

then $$ e^{\ln(x)}=\sum_{k=0}^{\infty} \frac{1}{k!} \left( \int_{1}^{x} \frac{1}{t} \ dt \right)^k =x$$

---

My attempt:

My idea was to show that $\frac{d^2}{dx^2} e^{\ln(x)} =0$, and then use the initial conditions I can get by evaluating the definitions at specific values to figure out that the constant of integration must be $0$.

Doing this I get $$ \frac{d^2}{dx^2}\sum_{k=0}^{\infty} \frac{1}{k!} \left( \int_{1}^{x} \frac{1}{t} \ dt \right)^k = \sum_{k=0}^{\infty} \frac{1}{k!} \frac{d^2}{dx^2} \left( \int_{1}^{x} \frac{1}{t} \ dt \right)^k $$ From here I use the fact that $\frac{d^2}{dx^2} f(g(x)) = g'(x)^2 f''(g(x)) + f'(g(x))g''(x)$, which applied to this gives me \begin{align*} =& \sum_{k=0}^{\infty} \frac{1}{k!} \left[\left(\frac{1}{x}\right)^2 k(k-1)\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{k-2} + k\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{k-1} \left(-\frac{1}{x^2}\right) \right]\\ =&\frac{1}{x^2}\left[ \sum_{k=0}^{\infty} \frac{1}{(k-2)!}\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{k-2} - \sum_{k=0}^{\infty} \frac{1}{(k-1)!}\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{k-1} \right]= \frac{1}{x^2} \left(\frac{1}{(-2)!}\right) \left[\ln(x) \right]^{-2} \end{align*} which is the point where I noticed I may have made several mistakes in the process, since this last result didn't make much sense to me.

Could anyone tell me where my mistakes are? And also, does anyone know another way to rigorously show this result from the definitions in the beginning? Thank you!

Answer 1

It is best to avoid conventional symbols for these functions while dealing with such problems because doing so runs the risk of inadvertently using some of their properties without proof.

So let $$f(x) =\sum_{k=0}^{\infty} \frac{x^k} {k!}, x\in\mathbb{R}, g(x) =\int_{1}^{x}\frac{dt}{t},x>0$$ From these definitions we get $$f'(x) =f(x), g'(x) =\frac{1}{x}$$ and therefore if $h(x) =g(f(x))-x $ then $$h'(x) =g'(f(x)) f'(x) -1=\frac{1}{f(x)}\cdot f(x) - 1=0$$ It follows $h$ is constant with $$h(x) =h(0)=g(f(0))=g(1)=0$$

---

One can't apply similar technique to show that $f(g(x)) =x$, but that can be deduced from $g(f(x)) =x$ by observing that $f, g$ are strictly monotone and hence each is invertible.

Answer 2

By taking into account the comment by @Pythagoras, my attempt can be corrected to get the following:

\begin{align} \frac{d^2}{dx^2} e^{\ln(x)} & = \frac{d^2}{dx^2}\left(\underbrace{1}_{k=\color{blue}{0}} + \underbrace{\int_{1}^{x} \frac{1}{t} \ dt}_{k=\color{blue}{1}} +\sum_{k={\color{blue}{2}}}^{\infty} \frac{1}{k!} \left( \int_{1}^{x} \frac{1}{t} \ dt \right)^k \right)\\ & = -\frac{1}{x^2}+\sum_{k=2}^{\infty}\frac{1}{k!}\left[\left(\frac{1}{x}\right)^2 k(k-1)\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{k-2} + k\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{k-1} \left(-\frac{1}{x^2}\right) \right] \\ & \overset{\color{purple}{j =k-1}}{=} -\frac{1}{x^2}+\left[\frac{1}{x^2}\sum_{\color{purple}{j }=1}^{\infty}\frac{1}{(\color{purple}{j }-1)!}\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{\color{purple}{j }-1} - \frac{1}{x^2}\sum_{k=2}^{\infty}\frac{1}{(k-1)!}\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{k-1} \right] \\ & = -\frac{1}{x^2}+\left[\underbrace{\frac{1}{x^2}(1)}_{j=\color{green}{1}} + \frac{1}{x^2}\sum_{j=\color{green}{2}}^{\infty}\frac{1}{(j-1)!}\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{j-1} - \frac{1}{x^2}\sum_{k=2}^{\infty}\frac{1}{(k-1)!}\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{k-1} \right] \\ & = -\frac{1}{x^2}+\frac{1}{x^2}\\ & = 0 \end{align} where we apply $\frac{d^2}{dx^2} f(g(x)) = g'(x)^2 f''(g(x)) + f'(g(x))g''(x)$ to $f(x) = x^k$ and $g(x) = \int_{1}^{x} \frac{1}{t} \ dt $. The previous result can thus be combined with $$ e^{\ln(\color{blue}{1})} = 1 +\sum_{k=1}^{\infty} \frac{1}{k!} \left( \underbrace{\int_{1}^{\color{blue}{1}} \frac{1}{t} \ dt}_{0} \right)^k =1 $$ And $$ \frac{d}{dx}e^{\ln(x)}\Bigg\vert_{x=\color{blue}{1}} = \frac{d}{dx}\left(1 +\int_{1}^{x} \frac{1}{t} \ dt +\sum_{k=2}^{\infty} \frac{1}{k!} \left(\int_{1}^{x} \frac{1}{t} \ dt \right)^k\right)\Bigg\vert_{x=\color{blue}{1}} = \frac{1}{\color{blue}{1}}+\sum_{k=2}^{\infty} \frac{k}{k!} \left( \underbrace{\int_{1}^{\color{blue}{1}} \frac{1}{t} \ dt}_{0} \right)^{k-1}\left(\frac{1}{\color{blue}{1}}\right) =1 $$ to conclude that $$ \boxed{e^{\ln(x)}=x} $$