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 February 28th, 2008, 02:00 PM #1 Member   Joined: Jan 2008 Posts: 34 Thanks: 0 Sine/Cosine Integral values I don't know how to use cosine integrals or sine integrals - but I had this problem: [-((p*i)/2))Ci(p/2)] + [(p/2)Si(p/2)] + [((p*i)/2)Ci(-p/2)] + (p/2)Si(-p/2)]= ? Where p=pi, i=the imaginary unit, Ci=Cosine integral, Si=Sine integral Can anyone provide an answer to the above value? February 28th, 2008, 06:11 PM #2 Member   Joined: Feb 2008 Posts: 89 Thanks: 0 Greetings: The equation simplifies to zero. Note that cos(pi/2) = cos(-pi/2) = 0. Thus we are left with (p/2)Si(p/2) + (p/2)Si(-p/2) = pi/2 - pi/2 = 0. Regards, Rich B. rmath4u2@aol.com February 28th, 2008, 08:32 PM #3 Member   Joined: Jan 2008 Posts: 34 Thanks: 0 No, I don't mean cos(pi/2) or cos(-pi/2) - I know what that is. What I'm trying to do is a cosineintegral/sineintegral. But I don't know how to use them or evaluate them. Can anyone elaborate on how to find the cosineintegral of (pi/2) and (-pi/2) and the sineintegral of (pi/2) and (-pi/2)? February 29th, 2008, 09:16 AM #4 Senior Member   Joined: May 2007 Posts: 402 Thanks: 0 Well, it's true that Si(-x)=-Si(x). This cancels the sinintegral. The Ci(-x) for x>0 is, however, not defined. Still, Mathematica gives the value of -pi^2/2. Tags integral, sine or cosine, values Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post johnny Calculus 4 September 23rd, 2010 04:00 AM RMG46 Trigonometry 8 June 15th, 2010 12:36 PM achmeineye Trigonometry 6 March 12th, 2009 10:17 PM Vay Trigonometry 3 January 25th, 2009 05:22 AM karel747 Calculus 5 April 1st, 2008 11:20 PM

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