Completing the square for integration of e^( )

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The discussion revolves around solving an integral involving the exponential function e^(-a*x^2 + b*x) as part of a quantum mechanics problem. The method of "completing the square" is suggested to simplify the integral, transforming it into a more manageable form. The integral ultimately separates into a product of e^(-k^2y^2) and a constant, allowing for the extraction of the constant from the integral. However, the resulting integral cannot be expressed in elementary functions and is defined as the "Error function" or "erf(x)." The key takeaway is the importance of the change of variables in simplifying the integral for further evaluation.
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Homework Statement


I'm working on problem 2.22 from Griffith's Intro. to Quantum Mechanics (a free particle problem). I am stuck on the final integral from part b. Part a of the problem is normalizing:
A*e-a*x2 which I did. Part b wants the general, time-dependent wave function.

Homework Equations


Griffith says that integrals of the form
∫e-(a*x2+b*x) can be solved by "completing the square." Griffith gives the example of defining y to be (a)(1/2)*(x+(b/(2a)) and using that definition to convert (a*x2+b*x) to y2 - (b2/(4*a)) I presume we are supposed to substitute in the converted expression into the integral and work from there.


The Attempt at a Solution


I have the following integral:
∫e-k2/(4*a)+i*(k*x-(h*k2)/(2*m)*t)
I presume using the "completing the square" method is supposed to make this integral work out; but I don't see how this will help. If I do convert this into something like:
ey + (something about a, h, and m),
how do I integrate with y?

Thanks,
Vance
 
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The exponential, e^{-k^2(y^2- b^2/4a}) can be separated as e^{-k^2y^2}e^{k^2b^2/4} and then the constant, e^{k^2b^2/4a} can be taken out of the integral, leaving \int e^{-k^2y^2}dy. That integral cannot be done in terms of 'elementary' functions (polynomials, fractions, trig functions, exponential, logarithms). That integral is defined to be the "Error function" or "erf(x)".
 
Thank-you so much. The missing part was the change of variables.
Vance
 

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