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March 19th, 2019, 07:13 AM   #1
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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?
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March 19th, 2019, 07:52 AM   #2
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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.
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March 19th, 2019, 08:41 AM   #3
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Quote:
Originally Posted by Greens View Post
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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