A longer look at the Si(x) function

Sinx/x can’t be integrated into an elementary function – instead we define:

Screen Shot 2016-05-21 at 9.31.12 PM

Where Si(x) is a special function.  This may sound strange – but we already come across another similar case with the integral of 1/x.  In this case we define the integral of 1/x as ln(x).  ln(x) is a function with its own graph and I can use it to work out definite integrals of 1/x.  For example the integral of 1/x from 1 to 5 will be ln(5) – ln(1) = ln(5).

The graph of Si(x) looks like this:

Screen Shot 2016-05-21 at 9.45.57 PM

Or, on a larger scale:

Screen Shot 2016-05-21 at 9.46.04 PM

You can see that it is symmetrical about the y axis, has an oscillating motion and as x gets large approaches a limit.  In fact this limit is pi/2.

Because Si(0) = 0,  you can write the following integrals as:

Screen Shot 2016-05-21 at 9.30.25 PM

How to integrate sinx/x ?

It’s all very well to define a new function – and say that this is the integral of sinx/x – but how was this function generated in the first place?

Well, one way to integrate difficult functions is to use Taylor and Maclaurin expansions.  For example the Maclaurin expansion of sinx/x for values near x=0 is:

Screen Shot 2016-05-21 at 9.31.00 PM

This means that in the domain close to x = 0, the function sinx/x behaves in a similar way to the polynomial above.  The last part of this expression O( )  just means everything else in this expansion will be x^6 or greater.

Graph of sinx/x

Screen Shot 2016-05-21 at 9.28.31 PM

Graph of 1 – x^2/6 + x^4/120

Screen Shot 2016-05-22 at 6.38.28 AM

In the region close to x=0 these functions behave in a very similar manner (this would be easier to see with similar scales so let’s look on a GDC):

So for the region above (x between 0 and 2) the 2 graphs are virtually indistinguishable.

Therefore if we want to integrate sinx/x for values close to 0 we can just integrate our new function 1 – x^2/6 + x^4/120 and get a good approximation.

Let’s try how accurate this is.  We can use Wolfram Alpha to tell us that:

Screen Shot 2016-05-22 at 6.49.08 AM

and let’s use Wolfram to work out the integral as well:

Screen Shot 2016-05-22 at 6.49.59 AM

Our approximation is accurate to 3 dp, 1.371 in both cases.  If we wanted greater accuracy we would simply use more terms in the Maclaurin expansion.

So, by using the Maclaurin expansion for terms near x = 0 and the Taylor expansion for terms near x = a we can build up information as to the values of the Si(x) function.