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I was trying to find the integral of $\frac{\sin x}{x}$ recently, and nothing I tried appeared to get me any closer to a solution, so I looked it up and apparently it's a Non-integrable function.

What makes a function Non-integrable and is there a way to solve it? Why can't it be solved with the "normal" methods.

Forgive my error-ridden terminology, I'm new.

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  • $\begingroup$ It pretty much means there's no good antiderivative for it based on the function itself. Try graphing it and you'll see. $\endgroup$ – pie314271 Jan 24 '17 at 4:15
  • $\begingroup$ Sorry, but what do you mean by no "good" antiderivative? $\endgroup$ – smaude Jan 24 '17 at 4:16
  • $\begingroup$ Basically, you simply can't integrate it no matter what. The antiderivative doesn't exist. $\endgroup$ – pie314271 Jan 24 '17 at 4:17
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    $\begingroup$ The term you're looking for is a function that doesn't have an elementary antiderivative. These are functions who have an antiderivative but you can't express it in terms of finite amounts of addition, multiplication, exponentation and trig function. $\endgroup$ – lordoftheshadows Jan 24 '17 at 4:17
  • $\begingroup$ A non integrable function is one where the definite integral can't be assigned a value. For example the Dirichlet function isn't integrable. You just can't assign that integral a number. $\endgroup$ – lordoftheshadows Jan 24 '17 at 4:20
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"Integrability" (more specifically, Lebesgue integrability) is a technical condition similar to absolute convergence of a sequence where a function is only considered integrable on $(a,b)$ if $$ \int_a^b|f(x)|dx < \infty.$$ By comparison with $1/x,$ $\sin(x)/x$ fails this criterion on $(-\infty, \infty)$ and is thus considered a non-integrable function on $(-\infty, \infty).$ This is likely the sense that was meant when you read that $\sin(x)/x$ is not integrable.

However, just like series can be convergent but not absolutely convergent, it turns out that since $\sin(x)/x$ oscillates as $|x|\rightarrow\infty$, the 'area under the curve' cancels out over the oscillations and is finite. The integral $$ \int_{-\infty}^\infty \frac{\sin(x)}{x}dx$$ can be defined as an improper Riemann integral and happens to equal $\pi.$ This is just a warning that 'not integrable' (in the Lebesgue sense) doesn't mean that the definite integral doesn't have a value.

All of this has little to do with your concern of finding an antiderivative for $\sin(x)/x$ and thus computing the integral over an interval using the fundamental theorem of calculus. As others have mentioned, it turns out that the antiderivative cannot be expressed as an elementary function, so it's no surprise you haven't been successful in applying various integration tricks.

Even though you cannot compute the integral in the usual way by taking an antiderivative, remember that, since $\sin(x)/x$ is a nice bounded, continuous function, the definite integral $\int_a^b\frac{\sin(x)}{x}dx$ exists a well-defined number (area under the curve) for any $a$ and $b$. Likewise, as others have mentioned, an antiderivative function $\mathrm{Si}(x) = \int_0^x\frac{\sin(t)}{t}dt$ exists and we have $$\int_a^b\frac{\sin(x)}{x}dx = \mathrm{Si}(b)-\mathrm{Si}(a).$$ It's just that $\mathrm{Si}(x)$ does not have a nice expression in terms of elementary functions like $\sin$, $\cos,$ $\ln$, $e^x,$ etc like you would usually compute with integration techniques.... it's its 'own thing' (called the sine-integral function).

Functions without an elementary antiderivative are usually called 'functions without an elementary antiderivative' and not called 'not integrable', although the latter would be perfectly reasonable terminology if it weren't already reserved for the technical condition I mentioned above.

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    $\begingroup$ This is a great answer. $\endgroup$ – WillO Jan 26 '17 at 4:04
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$\frac{\sin x}{x}$ is integrable (since it is continuous), it is just that its indefinite integral cannot be expressed as an elementary function. So its integral defines a special function Sine integral.

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  • $\begingroup$ Just to clarify. $\frac{sin(x)}{x}$ can't be expressed in finitely many combinations of elementary functions. $\endgroup$ – lordoftheshadows Jan 24 '17 at 4:23
  • $\begingroup$ The correct way to say is that it cannot be expressed as an elementary function. The elementary function itself is a finite combination of... My statement was not clear either, so I corrected it. $\endgroup$ – Momo Jan 24 '17 at 4:31
  • $\begingroup$ Yea. That's a lot nicer. $\endgroup$ – lordoftheshadows Jan 24 '17 at 4:32
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Whether the anti-derivative of a function can be expressed in terms of elementary functions is solved by Risch algorithm. $(\sin x)/x$ is one of those functions without a elementary anti-derivative.

With that being said, $(\sin x)/x$ still has an anti-derivative, namely $$ g(x) = \int_0^x \frac{\sin t}{t} \mathrm dt $$ Then $g'(x) = (\sin x)/x$.

In some sense $(\sin x)/x$ is not integrable, since $$ \int_{-\infty}^\infty \left| \frac{\sin x}{x} \right| \mathrm dx = +\infty $$ so the Lebesgue integral of $(\sin x)/x$ does not exist.

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