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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 Last edited by Greens; March 19th, 2019 at 07:57 AM. 
March 19th, 2019, 08:41 AM  #3  
Senior Member Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8  Quote:
 

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