Problem when integrating $e^x / x$. I made up some integrals to do for fun, and I had a real problem with this one. I've since found out that there's no solution in terms of elementary functions, but when I attempt to integrate it, I end up with infinite values. Could somebody point out where I go wrong?
So, I'm trying to determine: $$ \int{\frac{e^x}{x}} \, dx $$
Integrate by parts, where $u = 1/x$, and $v \, ' = e^x$. Then $u \, ' = - 1/x^2$, and $v=e^x$. So,
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \int{\frac{e^x}{x^2}} \, dx$$
Integrate by parts again, $u = 1/x^2$, $v \, ' = e^x$, so that $u \, ' = -2/x^3$ and $v=e^x$. So,
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + 2\int{\frac{e^x}{x^3}} \, dx$$
Repeat this process ad infinitum to get,
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + 2 \left( \frac{e^x}{x^3} + 3 \left( \frac{e^x}{x^4} + 4 \left( \frac{e^x}{x^5} + \, \cdots \right) \right) \right) $$
Expanding this gives,
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + \frac{2e^x}{x^3} + \frac{6 e^x}{x^4} + \frac{24 e^x}{x^5} + \cdots $$
And factoring that gives,
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} \left(  1 + \frac{1}{x} + \frac{2}{x^2} + \frac{6}{x^3} + \frac{24}{x^4} + \cdots \right) $$
Now, considering the series itself, the ratio between the $n^{th}$ term and the $(n-1)^{th}$ term = $\Large \frac{n}{x}$. Eventually, $n$ will be larger than $x$, so the ratio between successive terms will be positive, so (assuming $x$ is positive), the series diverges, meaning (and I'm sure everybody will cringe upon seeing notation used like this), that:
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} \left( \infty \right) = \infty $$
 A: As André Nicolas said, you've expressed an antiderivative of $e^x/x$ in terms of an asymptotic series! The proof is a little hairy, but the underlying idea is relatively simple, and I hope at least a little of that simplicity shows through in this answer.

An asymptotic expansion of a function $f$ around the point $p$, with respect to the variable $y(x)$, is a formal power series
$$a_0 + a_1 y + a_2 y^2 + a_3 y^3 + \ldots$$
whose $n$th partial sum
$$\tilde{f}_n(x) = a_0 + a_1 y + \ldots + a_n y^n$$
matches $f$ at $p$ with $o(y^n)$ error. In other words,
$$\lim_{x \to p} \frac{f(x) - \tilde{f}_n(x)}{y(x)^n} = 0.$$

As amin discovered, you're dealing with a special function called the exponential integral, defined as the integral
$$\operatorname{Ei}(x) = \int_{-\infty}^x \frac{e^u}{u}\;du.$$
The exponential integral is an antiderivative of $e^x/x$. Specifically, it's the antiderivative that goes to zero at $-\infty$. To focus on the power series part of your expression, consider the function $f$ defined by
$$\frac{e^x}{x} f(x) = \int_{-\infty}^x \frac{e^u}{u}\;du.$$
Your calculation suggests that
$$1 + \frac{1}{x} + \frac{2!}{x^2} + \frac{3!}{x^3} + \ldots$$
is an asymptotic expansion of $f$ around $-\infty$, with respect to the variable $y(x) = 1/x$. Let's prove it.

As N. S. pointed out, the $n$th partial sum of your series differs from $f$ by
$$\begin{align*}
f(x) - \tilde{f}_n(x) & = \frac{x}{e^x} (n+1)! \int_{-\infty}^x \frac{e^u}{u^{n+1}}\;du \\
& = x(n+1)! \int_{-\infty}^x \frac{e^{u-x}}{u^{n+1}}\;du
\end{align*}$$
Since $u \le x$ over the whole range of the integral,
$$\begin{align*}
\left| \int_{-\infty}^x \frac{e^{u-x}}{u^{n+1}}\;du \right|
& \le \left| \int_{-\infty}^x \frac{1}{u^{n+1}}\;du \right| \\
& = \tfrac{1}{n+1} \left| \frac{1}{x^{n+2}} \right| \\
& = \tfrac{1}{n+1} \left| y(x)^{n+2} \right|.
\end{align*}$$
Therefore,
$$\left|f(x) - \tilde{f}_n(x)\right| \le \left|y(x)^{n+1}\right|\,n!.$$
Since $y(x)$ goes to zero as $x$ goes to $-\infty$, it folows that
$$\lim_{x \to -\infty} \frac{f(x) - \tilde{f}_n(x)}{y(x)^n} = 0.$$
That means your power series really is an asymptotic expansion of $f$ around $-\infty$, with respect to the variable $y(x)$.

Your asymptotic expression for $\operatorname{Ei}$ near $-\infty$ is cool, but it would be much better if $\operatorname{Ei}$ had a nice, convergent Taylor expansion around $-\infty$, right? As it turns out, $\operatorname{Ei}$ does have a convergent Taylor expansion around $-\infty$, with respect to the variable $y(x) = 1/x$, but this Taylor expansion has a bizarre problem that makes it useless.
The coefficients of the Taylor expansion are given by the derivatives of $\operatorname{Ei}(x)$ with respect to $y(x)$ at $-\infty$. The first derivative is
$$\frac{d\operatorname{Ei}}{dy} = \frac{d\operatorname{Ei}}{dx} \frac{dx}{dy}
= \frac{e^x}{x} \left(-\frac{1}{y^2}\right) \\
= -\frac{e^{1/y}}{y}.$$
This first derivative is zero at $x = -\infty$. In fact, every derivative $d^n \operatorname{Ei}/dy^n$ is zero at $x = -\infty$. That means the Taylor expansion of $\operatorname{Ei}(x)$ at $-\infty$, with respect to the variable $1/x$, is
$$0 + 0\frac{1}{x} + 0\frac{1}{x^2} + 0\frac{1}{x^3} + \ldots$$
This Taylor expansion converges everywhere, as promised. It even converges to $\operatorname{Ei}(x)$ at $-\infty$. Everywhere else, though, it converges to the wrong value: $\operatorname{Ei}(x)$ is strictly negative for $x \in (-\infty, 0)$. In technical terms, we've discovered that the exponential integral is "smooth but not analytic" at $-\infty$. An asymptotic expansion is the best power series expansion you can hope to get for a function like this.
A: This part looks right:
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + \frac{2e^x}{x^3} + \frac{6 e^x}{x^4} + \frac{24 e^x}{x^5} + \cdots+ \frac{n!e^x}{x^{n+1}}+(n+1)!\int \frac{e^x}{x^{n+1}}$$
When you say "repeating to infinity" you want to take the limit of that...in order for your equality to hold, you need
$$\lim_n (n+1)!\int \frac{e^x}{x^{n+1}}=0 \,.$$
because that is your error in you partial sum "approximation". But not only the above limit is not 0, it actually makes no sense (an integral is a family of functions, what happens with the constant???). 
That's why formally, whenever you use a process like this, you need to prove that the difference between your n-th term and the limit goes to 0...
Your idea is similar to the following
\begin{eqnarray}
1&=&1+1-1\\
&=&1+1+1-2 \\
&=&1+1+1+1-3\\
&=&....
\end{eqnarray}
Taking limit to infinity you get
$$1=1+1+1+...+1+...= \infty$$
On this example you can see immediately that the "errror" in our appoximations don't go to 0, so our approximations are not approximations.
A: There is no elementary antiderivative for you this function.
You can take a look at here: http://en.wikipedia.org/wiki/Exponential_integral
What you have calculated here:
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} \left(  1 + \frac{1}{x} + \frac{2}{x^2} + \frac{6}{x^3} + \frac{24}{x^4} + \cdots \right) $$
is even something like a taylorexpansion of the integral at $x=\infty$
