Equation:

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)

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?

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.