Evaluating $\int{ \frac{x^n}{1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}}}dx$ using Pascal inversion

(Note: I apreciate very much who marked this as a duplicate but I would like an answer for why my proof is wrong)

This is my solution, I have no clue why it failed. Let's start:

define $$I_n(m) = \int_{0}^{x} \frac{t^m}{1 + t + t^2/2 + ... + t^n/n!}\ dt$$

so it should be true that

$$\sum_{m=0}^{n}\frac{I_n(m)}{m!} = x$$

Then I use Pascal inversion:

$$\sum_{m=0}^n \frac{n! I_n(m)}{m!} = n!x$$

$$\sum_{m=0}^n {n\choose m} B_n(m) = n!x$$

where $B_n(m) = (n-m)!I_n(m)$

by Pascal's formula:

$$I_n(n) = (-1)^nxn! \sum_{m=0}^{n} \frac{(-1)^m}{(n-m)!}$$

what did I do wrong ????

• I can't say specifically where you went wrong, but I think notation is dragging you down - for instance, you write $I_n(m)$, but there's a dependence on $x$ there that you don't capture at all. Commented Jun 29, 2015 at 0:37

You may observe that $$\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)'=1 + x + \frac{x^2}{2} + \cdots + \frac{x^{n-1}}{(n-1)!}$$ giving $$\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)-\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)'=\frac{x^n}{n!}$$ and \begin{align} \int \frac{x^n}{1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}}dx&=n!\int\frac{\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)-\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)'}{1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}}\:dx\\\\ &=n!\int dx-n!\int\frac{\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)'}{\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)}\:dx. \end{align} Thus
$$\int \frac{x^n}{1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}}dx=n!\:x-n!\ln \left| 1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right|+C.$$
• This is a very, very good answer. I never would have imagined the solution would be so nice. $+1$! Commented Jun 29, 2015 at 0:07