# Antiderivative of $\frac{1}{\ln(x)}$?

I was looking on wikipedia, and found that the following expression cannot be expressed in terms of elementary functions:

$$\int\frac{1}{\ln(x)}\text{d}x$$

Although the function looks simple, why is it impossible to express it in terms of elementary functions?

• There is no relation between the conciseness of a symbolic expression and its integrability. Even $1/x$ cannot be antiderived if you haven't defined the logarithm.
– user65203
Dec 16, 2014 at 17:34
• See Liouville's theorem and the Risch algorithm. Dec 16, 2014 at 22:11

You can prove that this antiderivative is non-elementary using differential algebra (in particular the Rothstein-Trager theorem). The fact is that, contrary to what you see in most first-year calculus courses, "most" elementary functions do not have elementary antiderivatives.

No, this is a well-known (to those who know it!) special function, called the log integral function, $\text{li}(x)$.

Sometimes this happens with functions that we are interested in and that's what makes special functions, well, special. Here's a list of other popular/useful integrals that give rise to special functions.

• Good point, if you could give some explanation as to why, I think the OP would appreciate it. Dec 16, 2014 at 17:35
• All tautologies are true. ;-) Dec 16, 2014 at 17:39
• You do not need the li function, as most values are just Ei(ln(x))=li(x) as seen in the “list” link. Sep 15, 2021 at 13:48

First of all every when mathematicians say that a function is integrable, they mean only that the integral is well defined — that is, that the integral makes mathematical sense.

In practical terms, integrability hinges on continuity: If a function is continuous on a given interval, it’s integrable on that interval. Additionally, if a function has only a finite number of some kinds of discontinuities on an interval, it’s also integrable on that interval.

Loosely speaking, integral is the limit of the Riemann sums of a function as the partitions get finer. If the limit exists then the function is said to be integrable (or more specifically Riemann-integrable) Riemann sum is an approximation that takes the form Σ f(x)*Δx.The sum is calculated by dividing the region up into shapes (rectangles), trapezoids), (parabolas), or (cubicss) that together form a region that is similar to the region being measured, then calculating the area for each of these shapes, and finally adding all of these small areas together. This approach can be used to find a numerical approximation for a definite integral even if the fundamental theorem of calculus does not make it easy to find a closed-form solution.(its pretty much simple application of definite integrals in finding area under curve) Actually 1/lnx is not closed form here is the wonderful article about closed formclosed form description Since I/lnx is not closed form so it is not elementary function so naively you can say that it has no elementary antiderivative

• "Actually 1/lnx is not closed form here is the wonderful article about closed formclosed form description Since I/lnx is not closed form so it is not integrable (Riemann...)" Utterly false! $e^{-x^2}$ is Riemann integrable in any bounded interval but $\int e^{-x^2}dx$ isn't elementary. $1/x$ isn't Riemann integrable in any interval that contains zero, but $\int(1/x)dx = \ln|x|$ is elementary. Jun 21, 2015 at 10:01