# Infinite Sum Defined by $\int \frac{e^x}{x}dx$ vs. Exponential Function Taylor Series

Recently, when fiddling around with integration by parts, I noticed that it is possible to define infinite series that led to an integral. My calculus teacher noticed this, and told me to find

$$\int \frac{e^x}{x}dx$$

which I already knew didn't have a elementary function definition. After integrating by parts a few times, I found that this was leading to the summation

$$e^x\sum_{k=1}^{\infty}\frac{(k-1)!}{x^k} + C$$

or, rather

$$\frac{e^x}{x}\sum_{k=0}^{\infty}\frac{k!}{x^{k}} + C$$

Which, to me, looked very similar to the Taylor series definition of e^x:

$$\sum_{k=0}^{\infty}\frac{x^k}{k!}$$

In that

$$\frac{k!}{x^{k}}^{-1} = \frac{x^k}{k!}$$

Is there some form of between what I have found and the exponential function's Taylor series that I don't yet understand?

• To make what you're doing rigorous, try doing it with a (convergent) definite integral, such as $\int_1^y \frac{e^x}{x} dx$. I think your approach leads to a non-convergent asymptotic expansion in this case. This is still a meaningful object (but not in the way you've seen in ordinary calculus). – Ian Feb 6 '16 at 16:56
• @Ian Alright, I'll see what I can get from there. – Addison Crump Feb 6 '16 at 16:59
• @Ian Also, what did you mean by "This is still a meaningful object"? – Addison Crump Feb 6 '16 at 17:15
• Non-convergent asymptotic expansions have a meaning, but not in the sense of $n$ tending to infinity. Instead they have a meaning in the sense of the parameter $x$ tending to some fixed value when $n$ is held finite. en.wikipedia.org/wiki/Asymptotic_expansion – Ian Feb 6 '16 at 17:17
• @Ian I solved the integral you gave me - it seems that $\int_{a}^{b}\frac{e^x}{x}dx$ is only convergent for $a,b>1$ or $a,b<-1$. – Addison Crump Feb 6 '16 at 17:40

I believe however that the following expansion is more widely used, due to simplicity: $$\int \frac{e^x}{x}dx=\int\frac{1}{x}\sum_{k=0}^{\infty}\frac{x^k}{k!}=\sum_{k=0}^{\infty}\int\frac{x^{k-1}}{k!}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!\cdot k}$$
• Well, the link is in that it is the integral of the Taylor series divided by $x$. Does not seem like there is a lot more to it XD – Jsevillamol Feb 6 '16 at 17:15