# What does one mean when $\int \frac{\sin x}x$ doesn't exist?

Well I say that by taylor's expansion: $$\int\frac{\sin x}x=\int\frac{x-x^3/6+x^5/120+...}x=x-x^3/18+x^5/480+...+\mathbb{C}$$ It's another thing that there doesn't exists a closed form for the sum/difference.But it does exists.So I am now confused about:

• Does integration to every function exists?
• [I partly understand something told about elementary functions etc.]
• If it does, does there exists a closed form always?
• [I think no, but can't support my contradiction]
• Can taylor series be always used like this, atleast for approximation?
• [I think it always can be]

and similar ones, can somebody help?

• en.wikipedia.org/wiki/Sine_integral#Sine_integral Not elementary, but useful for enough people that it was given a name. – Will Jagy Aug 22 '14 at 20:56
• There may not be a elementary closed-form expression for the antiderivative, but that is not the same as saying that it does not exist. Indeed, the integral $\int_a^b (\sin(x)/x) dx$ exists for all finite $a,b$ since the integrand is continuous (after setting the value at $x=0$ to $1$). In addition, $\int_{-\infty}^\infty (\sin(x)/x) dx$ exists as an improper integral and equals $\pi$. – Bungo Aug 22 '14 at 20:59
• There is an antiderivative of $\frac{\sin x}{x}$. The antiderivative cannot be expressed in closed form using elementary functions. Every continuous function has an antiderivative. So do some non-continuous functions. Taylor series apply only to nicely behaved functions, which luckily include many of the useful ones. – André Nicolas Aug 22 '14 at 20:59
• there is nothing such as integral does not exist , How ever if the integral is bounded it might be not integrable at that interval only or it might have branch cut . who told you anyway it does not exist ?!. – Bswan Aug 22 '14 at 21:21
• You may be interested in this. It might be too advanced for you now, but you may want to try it later after you have taken more mathematics. It's interesting anyway. www2.maths.ox.ac.uk/cmi/library/academy/LectureNotes05/… – MPW Aug 22 '14 at 21:39

## 3 Answers

• Does integral always exist?

No. There are non-integrable functions. Integrability was a major reason to switch from Riemann integral to Lebesgue. However, continuous functions always have a primitive, so it's rather a theoretical concern, though a very important one.

• Does there always exist a closed form?

No. And you just pointed an example. Some such "bad" integrals get a name, when they are generally useful, and this give rise to a whole zoo of special functions. Yours above is called the sine integral.

• Can Taylor series always be used?

Only if the integrand can be expressed (easily) as a Taylor series. Often the series has no simple form, so it's just replacing a difficult problem with another as difficult. However, special function may often be defined as series expansion. And some functions are not analytic at all ($\exp (-1/x)$ is a classical example, around 0, and with Borel's lemma it's easy to get other ones, but see also here).

• The Gamma function has a simple pole at $0$, so if you want an antiderivative in a series around $0$ you'll need a logarithmic term to take care of that. Away from the poles, Taylor series do exist and can be used for integration. – Robert Israel Aug 22 '14 at 21:02
• @RobertIsrael Your're right of course. I don't know where I got this. :-) Maybe I had in mind that Gamma function does not satisfy an easy differential equation. Anyway, sorry for this! – Jean-Claude Arbaut Aug 22 '14 at 21:07

I don't know who told you $\int \dfrac{\sin x}{x}\; dx$ doesn't exist. It does, and it even has a name: ${\rm Si}(x)$. But it is not an elementary function.

1. Every continuous function has an antiderivative.
2. Not every elementary function has an elementary antiderivative. There is a well-developed theory behind this: see Risch algorithm. "Closed form", on the other hand, is a nebulous concept: it may depend on whether somebody cared enough about this integral to give it a name.
3. The Taylor series of any analytic function can be integrated term-by-term to give the Taylor series for an antiderivative.
1. The Lebesgue Integrability Condition says that any bounded, almost-everywhere continuous function on a compact interval is Riemann integrable. In particular, this means that, say, $$\int_1^x \frac{\sin t}{t} \, dt$$ is a well-defined function. In fact, in this particular case the integral from zero exists also, because the function $$f(t) := \begin{cases} \displaystyle\frac{sin(t)}{t} & \text{ if } t \neq 0 \\ 1 & \text{otherwise} \end{cases}$$ is continuous everywhere. This function is sometimes called $Si(t)$ (for Sine integral), and you can see a graph here: http://www.wolframalpha.com/input/?i=plot+integral+sin%28t%29%2Ft+dt

2. There is no closed form for this integral in terms of elementary functions. In fact that's the case for most integrals.

3. Yes, you can commute integrals and sums, but there are certain convergence issues to worry about.

• In regards to #2, alas, "almost all" integrals don't even have nice ways to write the "antiderivative". One tries not to become depressed about it... – colormegone Aug 22 '14 at 21:07
• I said "most" because I wanted to be vague. I mean, obviously if you start with something that's just an arbitrary continuous function that has no closed-form expression then you wouldn't expect its integral to have a closed form. So the relevant statement would be what proportion of elementary functions have integrals that are elementary. Unfortunately the set of elementary functions is countable up to the choice of finitely many real coefficients, so there's no natural measure on it that lets us talk about "almost all." It's a fundamentally discrete-ish object. – Daniel McLaury Aug 22 '14 at 21:14
• I wasn't making my comment as a criticism: I was merely remarking that the careful analytical techniques we learn and teach for "exact" integration only work on a minute (infinitesimal?) proportion of all the integrands we encounter (or can think up), something I warn students about in courses. It's part of what keeps numerical analysts and supercomputers employed... – colormegone Aug 22 '14 at 22:00
• Sure, it's just that what you said is a bit misleading. You can say that almost all integrals of continuous functions aren't elementary, but that's just because almost all continuous functions aren't elementary -- it's not relevant to the issue at hand, which is the integrals of elementary functions. You can't say that almost all integrals of elementary functions aren't elementary, because there's no natural probability measure on the set of elementary functions. – Daniel McLaury Aug 23 '14 at 4:29
• The original post asked about "every function", and was what I thought you were referring to in the phrase "most integrals". I guess my attempt at humor about what we face in dealing with integration misfired. – colormegone Aug 23 '14 at 4:43