My Math Forum I can't calculate a simple infinity integra!!!

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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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Joined: Oct 2015
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Quote:
 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!

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