Integrate exp(-(z-ia)^2) from z = - infinity to z = infinity

Kate2010
Messages
134
Reaction score
0

Homework Statement



Prove that \int^{∞}_{-∞} exp(-(z-ia)2)dz = √∏ for all real a.

Homework Equations


The Attempt at a Solution



If I use the substitution x = z-ia then dz = dx and if I use the limits x = -∞ to x = ∞ I get the correct answer. However, I do not know how to justify leaving the limits the same or if it is even ok?
 
Physics news on Phys.org
Kate2010 said:

Homework Statement



Prove that \int^{∞}_{-∞} exp(-(z-ia)2)dz = √∏ for all real a.

Homework Equations


The Attempt at a Solution



If I use the substitution x = z-ia then dz = dx and if I use the limits x = -∞ to x = ∞ I get the correct answer. However, I do not know how to justify leaving the limits the same or if it is even ok?

No it's not ok but you can still solve it by making that substitution u=z-ia, just change the limits on the integral:

\int_{-\infty}^{\infty}e^{-(z-ia)^2}dz=\int_{-\infty-ia}^{\infty+ia} e^{-u^2}du

Now, that integrand is analytic so independent of path so that I can go from -\infty-ia up to the point -\infty, go down the real axis to \infty then up to the point \infty+ia. The two vertical legs are zero because of the negative exponent so that we're left with just the ordinary gaussian integral which is \sqrt{\pi}
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top