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 Test 
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Joined: 25 Feb 2008, 13:32
Posts: 106
Location: Cape Town, South Africa
Post Test
Equation:

Image

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05 Mar 2008, 17:55

Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Re: Test
I assume that 'psi' is an explicit and differentiable function of 'x'.For my convenience I choose to replace 'psi' with 'y'.Then the given integral can be written as:

I = intgrl{exp(-yx)} wrt x from 0 to infinity.

An elementary use of the chain rule leads to the following expression:

I = {-1/(y+xy')}{exp(-yx)} wrt x from 0 to infinity(y' denotes the first derivative of y wrt x).To calculate the value of the integral explicitly we require an expression y(x) in terms of x.Then the value is simply:

I(infinity)-I(0)

As a special case if y is a constant function we have y'= 0 and exp{-infinity}=0.We also have exp(0)=1.Hence,

I=0-(-1/y)=1/y (y is not 0)


12 Mar 2008, 12:13

Joined: 19 Mar 2008, 13:44
Posts: 9
Location: Beirut, Lebanon
Post Re: Test
OK Sameed! Could you please show us how you apply your expression of "I"
to find the value of the original integral when y = x ?We should get 1/2*root (pi), shouldn't we?


19 Mar 2008, 14:27

Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Re: Test
I is defined only when y+xy' is not zero.Therefore, I(0) is not defined when y=x.However,we can evaluate this integral by switching to polar coordinates which yields the familiar sqrt{pi}/2.Thanks for pointing it out.


01 Apr 2008, 05:55
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