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?

  • $\begingroup$ 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). $\endgroup$ – Ian Feb 6 '16 at 16:56
  • $\begingroup$ @Ian Alright, I'll see what I can get from there. $\endgroup$ – Addison Crump Feb 6 '16 at 16:59
  • $\begingroup$ @Ian Also, what did you mean by "This is still a meaningful object"? $\endgroup$ – Addison Crump Feb 6 '16 at 17:15
  • $\begingroup$ 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 $\endgroup$ – Ian Feb 6 '16 at 17:17
  • $\begingroup$ @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 $a,b<-1$. $\endgroup$ – Addison Crump Feb 6 '16 at 17:40

You have just stumbled upon the Exponential integral (Ei) function, which is a non-elementary function. Without getting into too much detail, that means that the function cannot be fully simplified, and the best we can do is express it as an series, as you did.

I believe however that the following expansion is more widely used, due to simplicity: $$ \int \frac{e^x}{x}dx=\int\frac{1}{x}\sum_{k=0}^{\infty}\frac{x^k}{k!}=\sum_{k=0}^{\infty}\int\frac{x^{k-1}}{k!}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!\cdot k} $$

  • $\begingroup$ I know what it is, I'm trying to find the link between it and the taylor series definition of the exponential function. $\endgroup$ – Addison Crump Feb 6 '16 at 16:49
  • $\begingroup$ 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 $\endgroup$ – Jsevillamol Feb 6 '16 at 17:15
  • $\begingroup$ Yes, but the summation factors out the e^x, yet the summation still retains the multiplicative inverse of the summand in the Taylor series representation of e^x. $\endgroup$ – Addison Crump Feb 6 '16 at 17:18

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