Standard Gaussian integral. The value is √π.
Proof:
Let P=[-∞, ∞]∫e^(-x²) dx. Then P² = [-∞, ∞]∫e^(-x²) dx * [-∞, ∞]∫e^(-x²) dx. Obviously, it doesn't matter which dummy variable is used, so we change the variable in the second integral to y. Then we have:
[-∞, ∞]∫e^(-x²) dx * [-∞, ∞]∫e^(-y²) dy
y is constant in terms of x, so:
[-∞, ∞]∫e^(-x²) [-∞, ∞]∫e^(-y²) dy dx
And x is constant in terms of y, so:
[-∞, ∞]∫[-∞, ∞]∫e^(-(x²+y²)) dy dx
This is the double integral of a function of two variable over the whole xy-plane. Therefore, we make a change of variables to polar coordinates. The range of integration will now be θ∈[0, 2π], r∈[0, ∞]:
[0, 2π]∫[0, ∞]∫e^(-r²) r dr dθ
Now, r e^(-r²) has an elementary antiderivative, namely -1/2 e^(-r²). So this is:
[0, 2π]∫ (-1/2 e^(-∞²)) - (-1/2 e^(-0²) dθ
[0, 2π]∫1/2 dθ
π
Therefore, P²=π, and since P is positive (since e^(-x²) is everywhere positive), this implies that P=√π. Q.E.D.
2007-04-27 01:28:08
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answer #1
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answered by Pascal 7
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P = â« e ^(–x^2) dx (x from -â to â)
P^2 = â«e^(-x^2) dx â«e^(-y^2) dy (x,y from -â to â)
P^2 = â«â«e^(-x^2) e^(-y^2) dydx (x,y from -â to â)
P^2 = â«â«e^-(x^2+y^2) dydx (x,y from -â to â)
Now switch to polar coordinates
P^2 = â«â«e^(-r^2) rdrdt (t from 0 to 2pi, r from 0 to â)
P^2 = â«â«e^(-r^2) rdrdt (t from 0 to 2pi, r from 0 to â)
P^2 = 2pi â«e^(-r^2) rdr (r from 0 to â)
P^2 = 2pi (-1/2) â«e^(-r^2) (-2r) dr (r from 0 to â)
P^2 = 2pi (-1/2) â«e^(-r^2) (-2r) dr (r from 0 to â)
P^2 = 2pi (-1/2) [e^-r^2]|(r from 0 to â)
P^2 = 2pi (-1/2) [0-1]
P^2=pi
P=sqrt(pi)
2007-04-27 08:23:20
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answer #2
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answered by Astral Walker 7
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Dang, that problem is HARD!! Can't help you dude, good luck!
2007-04-27 07:48:37
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answer #3
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answered by Anonymous
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