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


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Posted (edited)

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 a-1 and a (so just basically any 'area-block' with width 1), equals \frac{\pi}{2} when the number of elements of which the average is taken reaches \infty:

 

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

 

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

 

\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]}

 

According to L'Hôpital:

 

=\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)}

 

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

 

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

 

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

 

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

 

Can someone help me?

 

Thanks :)

 

Function

Edited by Function
  • 3 months later...

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