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:

$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} \left( \infty \right) = \infty $$

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.

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.

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 $x=\infty$