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How do I integrate $$\int\frac{\sin(x)}xdx$$? I tried using integration by parts, but it led me to nowhere. Please help.

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You have to use the (nonelementary) sine integral in the general case; however, if you're interested in the limit from 0 to $\infty$, there is an m.SE question devoted to that topic... – J. M. May 5 '11 at 3:03

The function $f(x)=\sin(x)/x$ does not admit an elementary antiderivative, i.e., there is no formula for its integral (using quotients of polynomials, trig. functions, logartithms, exponentials, i.e., the usual functions you study in calculus).

Symbolic integration is the part of calculus that deals with finding antiderivatives. There is a fairly sophisticated algorithm due to Risch and implementing it shows that there is no nice formula for $\displaystyle \int\frac{\sin(x)}x dx$. The algorithm is sufficiently elaborate that apparently no software package can currently find antiderivatives for all functions for which it is possible. The Wikipedia page I linked to has references to the original (and nice) paper.

A few years ago, Matthew Wiener posted a fairly readable account of the algorithm on sci.math; here is a pdf of the post.

For a nice full length exposition of the mathematics involved, I highly recommend the book by Manuel Bronstein,"Symbolic Integration 1 (transcendental functions)" (2 ed.), 1997, Springer-Verlag.

Now, not all is bad news here: One can integrate term by term the power series for $\sin(x)/x$ expression and obtain the power series of its antiderivative, (that converges everywhere), and there are numerical methods to approximate very decently this function. Finally, one can compute explicitly (for example, using methods of complex analysis) that $$ \int_0^\infty\frac{\sin(x)}x dx=\frac{\pi}2. $$

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The pdf was typed by Apollo Hogan. I'm afraid I lost the link to the original post. – Andrés E. Caicedo May 5 '11 at 3:36
here is the original sci.math post. – J. M. May 7 '11 at 3:22
@J.M. : Thanks! – Andrés E. Caicedo May 12 '11 at 16:37
See also – Andrés E. Caicedo Apr 12 '13 at 2:50

If you ask Wolfram Alpha, it will tell you that the integral is $\text{Si}(x)+C$. If you ask it what is $\text{Si}(x)$, it will tell you, among other things, that $$\text{Si}(x)=\int_0^x \frac{\sin t}{t}dt$$ Not very helpful! But one can get some information out of all this unhelpfulness.

If there were an expression for your integral in terms of elementary functions, Wolfram Alpha, which is really pretty good, would very likely produce such an expression. And indeed it can be proved that there is no such expression.

But the integral that you want shows up naturally in a number of applications, for example in optics. So it is convenient to have a name for it, and $\text{Si}(x)$ is as far as I know the only one in common use.

Some definite integrals involving $\sin(x)/x$ can be evaluated explicitly, but of course not by the usual technique of finding an indefinite integral and then substituting.

There is nothing particularly mysterious about a function given by a simple formula not having an indefinite integral given by a combination of elementary functions. In fact "most" elementary functions do not have an elementary antiderivative.

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To add: one should consider him/herself incredibly lucky if an integral s/he encounters in applications has a (simple) closed form. – J. M. May 5 '11 at 5:49

There is no indefinite integral that can be written in elementary functions. However, as sometimes happens, the definite integral on certain endpoints is known; see:

Solving the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$?

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