Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

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

share|cite|improve this question
It looks as if you have discovered an asymptotic expansion. There are divergent series such that if one is sensible and cuts off early, one gets a good approximation, but if you go too far, there is trouble. Such things are even useful! –  André Nicolas Dec 5 '12 at 20:33

2 Answers 2

up vote 6 down vote accepted

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 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.

share|cite|improve this answer
Thanks, the example helped a lot. –  SiliconCelery Dec 5 '12 at 20:56

There is no elementary antiderivative for you this function.

You can take a look at here:

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 $x=\infty$

share|cite|improve this answer
"Like" a Taylor expansion. But divergent for every $x$ ... –  GEdgar Dec 5 '12 at 20:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.