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

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SUMMARY

The integral \(\int_{-\infty}^{\infty} \exp(-(z-ia)^2)dz\) evaluates to \(\sqrt{\pi}\) for all real values of \(a\). By substituting \(u = z - ia\), the limits of integration change to \(-\infty - ia\) and \(\infty + ia\). The integrand remains analytic, allowing the evaluation of the integral along a modified contour that simplifies to the standard Gaussian integral, confirming the result of \(\sqrt{\pi}\).

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Kate2010
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Homework Statement



Prove that [itex]\int^{∞}_{-∞}[/itex] 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?
 
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Kate2010 said:

Homework Statement



Prove that [itex]\int^{∞}_{-∞}[/itex] 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 [itex]u=z-ia[/itex], just change the limits on the integral:

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

Now, that integrand is analytic so independent of path so that I can go from [itex]-\infty-ia[/itex] up to the point [itex]-\infty[/itex], go down the real axis to [itex]\infty[/itex] then up to the point [itex]\infty+ia[/itex]. The two vertical legs are zero because of the negative exponent so that we're left with just the ordinary gaussian integral which is [itex]\sqrt{\pi}[/itex]
 
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