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 March 19th, 2019, 07:13 AM #1 Senior Member   Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8 I can't calculate a simple infinity integra!!! $\displaystyle x(t) = \begin{cases} 1 \quad ,0 \leq t \leq 1 \\ 0 \quad ,else \end{cases}$ The book says to calculate the integral below which must be 1, but I find infinity. $\displaystyle \lim_{T \to \infty} \int_{-T}^{T} |x(t)|^2 \cdot dt = 1$ My calculation $\displaystyle \lim_{T \to \infty} \int_{-T}^{T} |x(t)2|^2 \cdot dt = \lim_{T \to \infty} \int_{-T}^{T} |1|^2 \cdot dt = 1^2 \cdot \lim_{T \to \infty} \int_{-T}^{T} dt = 1^2 \cdot \lim_{T \to \infty} t |_{-T}^{T} = 1^2 \cdot \lim_{T \to \infty} (T+T) = \infty$ What am I doing wrong? March 19th, 2019, 07:52 AM #2 Member   Joined: Oct 2018 From: USA Posts: 88 Thanks: 61 Math Focus: Algebraic Geometry I would try splitting the integral into pieces: $\displaystyle \lim_{T \to \infty} \int_{-T}^{T} |x(t)|^2 \cdot dt =$ $\displaystyle \lim_{T \to \infty} \int_{-T}^{0} |x(t)|^2 \cdot dt + \int_{0}^{1} |x(t)|^2 \cdot dt + \int_{0}^{T} |x(t)|^2 \cdot dt$ Since we know the value of the function along the bounds we can substitute within the integral: $\displaystyle \lim_{T \to \infty} \int_{-T}^{0} 0 \cdot dt + \int_{0}^{1} |1|^2 \cdot dt + \int_{-T}^{0} 0 \cdot dt = 0 + t|^{1}_{0} + 0 = 1$ The issue you ran into is that you substituted one across the whole interval (-T,T) which would just be the area of an infinitely long rectangle, so that's why you got infinity Thanks from topsquark and babaliaris Last edited by Greens; March 19th, 2019 at 07:57 AM. March 19th, 2019, 08:41 AM   #3
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 Originally Posted by Greens I would try splitting the integral into pieces: $\displaystyle \lim_{T \to \infty} \int_{-T}^{T} |x(t)|^2 \cdot dt =$ $\displaystyle \lim_{T \to \infty} \int_{-T}^{0} |x(t)|^2 \cdot dt + \int_{0}^{1} |x(t)|^2 \cdot dt + \int_{0}^{T} |x(t)|^2 \cdot dt$ Since we know the value of the function along the bounds we can substitute within the integral: $\displaystyle \lim_{T \to \infty} \int_{-T}^{0} 0 \cdot dt + \int_{0}^{1} |1|^2 \cdot dt + \int_{-T}^{0} 0 \cdot dt = 0 + t|^{1}_{0} + 0 = 1$ The issue you ran into is that you substituted one across the whole interval (-T,T) which would just be the area of an infinitely long rectangle, so that's why you got infinity
Thanks I understand now! Tags calculate, infinity, integra, simple 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 mrluck Probability and Statistics 2 October 25th, 2017 09:59 PM mhhojati Complex Analysis 1 July 4th, 2013 04:16 AM Math Noob Algebra 2 May 16th, 2012 10:06 PM stephanos Advanced Statistics 4 July 22nd, 2010 05:54 AM -DQ- Algebra 5 September 14th, 2009 05:13 AM

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