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given: P = ∫ e ^(–x^2) dx limit is from - ∞ to +∞
Answer the following questions:
a) What are the possible methods of solving this problem?
b) Choose one and solve this problem. Show youre solution.
c) What is your conclusion?
d) What is the essence of this problem?

2007-04-27 00:27:33 · 3 answers · asked by blue 2 in Science & Mathematics Mathematics

3 answers

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 · answer #1 · answered by Pascal 7 · 2 0

P = ∫ e ^(–x^2) dx (x from -∞ to ∞)

P^2 = ∫e^(-x^2) dx ∫e^(-y^2) dy (x,y from -∞ to ∞)

P^2 = ∫∫e^(-x^2) e^(-y^2) dydx (x,y from -∞ to ∞)

P^2 = ∫∫e^-(x^2+y^2) dydx (x,y from -∞ to ∞)

Now switch to polar coordinates

P^2 = ∫∫e^(-r^2) rdrdt (t from 0 to 2pi, r from 0 to ∞)

P^2 = ∫∫e^(-r^2) rdrdt (t from 0 to 2pi, r from 0 to ∞)

P^2 = 2pi ∫e^(-r^2) rdr (r from 0 to ∞)

P^2 = 2pi (-1/2) ∫e^(-r^2) (-2r) dr (r from 0 to ∞)

P^2 = 2pi (-1/2) ∫e^(-r^2) (-2r) dr (r from 0 to ∞)

P^2 = 2pi (-1/2) [e^-r^2]|(r from 0 to ∞)

P^2 = 2pi (-1/2) [0-1]

P^2=pi

P=sqrt(pi)

2007-04-27 08:23:20 · answer #2 · answered by Astral Walker 7 · 0 0

Dang, that problem is HARD!! Can't help you dude, good luck!

2007-04-27 07:48:37 · answer #3 · answered by Anonymous · 1 1

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