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


Our imagination is stretched to the utmost, not, as in fiction, to imagine things which are not really there, but just to comprehend those things which are there.

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:


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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