Proof wanted: average area (Si(x)-Ci(x))

[Function]

Hello

 

I just disovered some new functions: sinIntegral(x) (Si(x)) and cosIntegral(x) (Ci(x)) and just played around with them on GeoGebra.

 

Now, I found something kind of 'beautiful': the average area between those functions, limited by [math]a-1[/math] and [math]a[/math] (so just basically any 'area-block' with width 1), equals [math]\frac{\pi}{2}[/math] when the number of elements of which the average is taken reaches [math]\infty[/math]:

 

[math]\lim_{n\to\infty}{\left[\frac{\int_{1}^{n}{Si(x) dx}-\int_{1}^{n}{Ci(x) dx}}{n-1}\right]}=\frac{\pi}{2}[/math].

 

In order to prove it, I worked out the left side, but I don't really know how to do it further:

 

[math]\lim_{n\to\infty}{\left[\frac{n\cdot Si(n)+\cos{n}-Si(1)-\cos{1}-n\cdot Ci(n)+\sin{n}+Ci(1)-\sin{1}}{n-1}\right]}[/math]

 

According to L'Hôpital:

 

[math]=\lim_{n\to\infty}{\left(n\cdot\frac{\sin{n}}{n}+Si(n)+\sin{n}-n\cdot\frac{\cos{n}}{n}-Ci(n)-\cos{n}\right)}[/math]

 

[math]=\lim_{n\to\infty}{\left(Si(n)-Ci(n)+2\sin{n}-2\cos{n}\right)}[/math]

 

[math]=\lim_{n\to\infty}{\left(Si(n)-Ci(n)\right)}+2\lim_{n\to\infty}{\left(\sin{n}-\cos{n}\right)}[/math]

 

[math]=\frac{\pi}{2}+2\lim_{n\to\infty}{\left(\sin{n}-\cos{n}\right)}[/math]

 

Now, I know that the first term is [math]\frac{\pi}{2}[/math], so the second term should be [math]0[/math], but how can one prove this? As far as I know, [math]\lim_{n\to\infty}{\left(\sin{n}-\cos{n}\right)}[/math] is not specified, for the values of [math]\cos{n}[/math] and [math]\sin{n}[/math] alternate.

 

Can someone help me?

 

Thanks

 

Function

Edited February 28, 2014 by Function

[Function]

Note: Si = sine-integral function and Ci = cosine-integral function

[overtone]

never mind, bad response

Edited June 4, 2014 by overtone