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I want to know if the function sinx/x is integrable and if it is, then what's its integral?

My high school book says its a non-integrable function while WolframAlpha says its integral is Si(x) + constant

Please shed some light on this topic and explain from basic level like what is Si(x), who used it for the first time etc.

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The indefinite integral, you mean. Of course, $\mathrm{Si}(x)$ is simply defined to be a definite integral of $(\sin x)/x$, so of course you're going to get this plus a constant. –  anon Jun 26 '12 at 13:42
    
See this Wikipedia page for the definition of $\operatorname{Si}(x)$. –  Zev Chonoles Jun 26 '12 at 13:43
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2 Answers

up vote 9 down vote accepted

There is no elementary function whose derivative is $\frac{\sin x}{x}$. However, antiderivatives of this function come up moderately frequently in applications, for example in signal processing. So it has been convenient to give one of its antiderivatives, $\int_0^x\frac{\sin t}{t}\,dt$, a name. The standard name is $\text{Si}(x)$. In the old days, one could find tables of $\text{Si}(x)$. Nowadays, it is a built-in function in many programs, including Maple and Mathematica.

There are many other instances in mathematics where a non-elementary function has been given a special name. So there is an enormous variety of such special function, among them the error function of probability theory, and the Gamma function.

Remark: If you write down a medium-ugly elementary function, like $\frac{e^x}{x^2+1}$, or even $\sqrt{1+x+x^5}$, the function will usually not have an elementary antiderivative. this makes for real trouble when we are trying to invent new arclength problems. Recall that the arclength of $y=f(x)$ involves $\int_a^b \sqrt{1+(f'(x))^2}\,dx$. If we start with a quite nice function $f(x)$, the function $\sqrt{1+(f'(x))^2}$ can be pretty complicated, and ordinarily does not have an elementary antiderivative. That's why most of the calculus book examples in the arclength section are quite artificial.

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In general, a function $f:\mathbb{R} \longrightarrow \mathbb{R}$ is integrable if it is bounded and the set of discontinuities (i.e. $x=0$ in this case) have measure zero. Intuitively, this more or less amounts to the function being defined except at reasonably few exceptional points (i.e. a finite number of points as in this case is fine), so the function is integrable since it is clearly also bounded. As the others have mentioned and linked to, the integral is nontrivial and has a special function defined to be its antiderivative called $Si(x)$.

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