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Susanne217
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Homework Statement
Hi I am working on a solution for the integral the integral
[tex]\int_{-\infty}^{\infty} (\frac{sin(x)}{x})^2 dx[/tex]
Homework Equations
The Attempt at a Solution
I know from theory that
[tex]\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{a} f(x) dx + \int_{-\infty}^{a} f(x) dx[/tex]
So that [tex]\int_{-\infty}^{1} (\frac{sin(x)}{x})^2 dx + \int_{1}^{\infty} (\frac{sin(x)}{x})^2 dx [/tex]
since [tex](\frac{sin(x)}{x})^2 = \frac{sin^{2} x}{x^2}[/tex]
and by trig identity
[tex]\frac{sin^{2} x}{x^2} = \frac{1-cos(2x)}{2} \cdot x^{-2} [/tex]
for clarity that gives me an integral that I need to solve where
[tex]\int_{-t}^{t} \frac{1-cos(2x)}{2x^2} dx = \int^{1}_{-t}\frac{1-cos(2x)}{2x^2} dx + \int^{t}_{1}\frac{1-cos(2x)}{2x^2} dx[/tex]
What I am simply am burning to know is that the correct approach?
I have read in another thread in this forum that the purpose of the task is to first find a corresponding taylor series expansion
so anyway since [tex]\frac{sin^2 x}{x^2} = \frac{1-cos(2x)}{2x^2} = \frac{-(cos(2))-1}{2}\cdot (x-1) - \frac{cos(2) - 4 \cdot sin(2) + 3}{2} \cdot (x-1)^2 + 2 \cdot (7 sin(2) -6) \cdot (x-1)^3 + \cdot + ?[/tex]
I used [tex]x_0 = 1[/tex] since there is hole x = 0
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