Integral $\frac{\sin(x)}{x}$ finite domain I have seen a question asking to find the value of $\int_{-100}^{100} \frac{\sin{x}}{x} dx$.
I have to confess that I didn't think this was possible. 
If I expand the $\sin$ using Taylor series, then unless the endpoints of the domain lie inside $(-1,1)$, the result will diverge. And in fact, I think it might even diverge if they lie inside this domain - is that correct?
On the other hand, we can use complex contour integrals to show the are bounded by the graph and the entire $x$ axis is equal to $\pi$. Therefore, we should expect the area bounded by the graph and this finite part of the $x$ axis to be both finite and less than $\pi$.
I know the antiderivative is $\operatorname{Si}(x)$ but I thought this was only defined for $x>0$. Perhaps we could try evaluating $\operatorname{Si}(400)-\operatorname{Si}(0)$ and doubling it or something.
In short, I have no idea what to do here. 
Thanks for your help.
 A: Instead of using Taylor series which is incredibly inefficent for values as large as $x=100$ (test for yourself how many terms you need to get an precision of say $10^{-3}$) you should derive an asymptotic expansion for this integral. An asymptotic expansion is something like the evil and not well behaved but powerful brother of an usual series expansion. It doesn't converge in a usual sense but gives incredible accurate results for large arguments as long as one takes the right number of terms (which means terminate it at a  meaningful order). I will not go into any formal details here, but just show you how to apply this procedure for the example at hand. Write
$$
I(z)=\int_{-z}^z \frac{\sin(x)}{x}=\int_{-\infty}^{\infty} \frac{\sin(x)}{x}-2\int_{z}^{\infty} \frac{\sin(x)}{x}
$$
The first integral is a classic and yields $\pi$, for the second let's integrate by parts with $\sin(x)=u'$ and $1/x=v$
$$
\int_{-z}^z \frac{\sin(x)}{x}=\pi-2\frac{\cos(z)}{z}+2\int_{z}^{\infty} \frac{\sin(x)}{x^2}
$$
we see immediatly that the remaining integral is much smaller then the rest of the rhs. so just we are lazy we just stop here and neglect the integral. Now doing the numbers (putting $z=100$) we get
$$
I(100)\approx 3.124346276144
$$
which is incredibly close to the real value $3.124450933778$ keeping in mind that we did just one integration by parts (the error is $\approx 10^{-5}$).
Integrating by parts $N-$ times we could derive the coefficent of the resulting series go as $N!$ which means that it is badly divergent! This is what i meaned by 'bad brother': An divergent series (for any fixed $x$) which gives incredibly exact result if we truncate it at a senseful point!
Fascinating,right?
A: If you know the function $Si(x)$, then the solution should be fairly simple.
In fact, the function is bounded from above and continuous, so there should be no Problem evaluating the integral, and also the Taylor Series converge everywhere. Your guess that the integral should be approximately $\pi$ is also correct.
The function $Si(x)$ is used for positive x because you can "mirror" the function to only use with positive values.
Observe, that for all x: $\frac{Sin(x)}{x} = \frac{Sin(-x)}{-x}$, so your integral evaluates to:
$\int \limits_{-100}^{100} \frac{Sin(x)}{x} = 2 Si(100)$
This is already not far from $\pi$, and I don't know if there is a very nice term for it.
