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

In summary, to prove \int^{∞}_{-∞} exp(-(z-ia)2)dz = √∏ for all real a, one can make the substitution u=z-ia and change the limits of integration to \int_{-\infty-ia}^{\infty+ia} e^{-u^2}du, which simplifies to the gaussian integral \sqrt{\pi}. This is possible because the integrand is analytic and the vertical legs of the path of integration are zero due to the negative exponent.
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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]
 
Last edited:

1. What does the function exp(-(z-ia)^2) represent?

The function exp(-(z-ia)^2) represents a Gaussian distribution with mean ia and standard deviation 1.

2. What is the purpose of integrating this function from z = - infinity to z = infinity?

Integrating this function from z = - infinity to z = infinity allows us to calculate the total area under the Gaussian curve, which represents the probability of a random variable falling within a certain range of values.

3. Why is the integration of this function important in science?

The integration of this function is important in science because it is used in a variety of fields, such as physics, chemistry, and statistics, to model and analyze various phenomena and data sets.

4. Can this function be solved analytically or does it require numerical methods?

This function cannot be solved analytically and requires numerical methods, such as the Gaussian quadrature or Monte Carlo methods, to approximate the integral.

5. How does changing the value of a affect the shape of the Gaussian curve?

Changing the value of a shifts the mean of the Gaussian distribution, resulting in a different peak or center of the curve. A larger value of a will result in a wider and flatter curve, while a smaller value of a will result in a narrower and taller curve.

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