Help please: Fourier Transform of a Gaussian function showing integral equals 1

tjphop
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Homework Statement



Hi, the question is from a piece of coursework and before hand we were asked to find the Fourier transform G(K) of the function g(x)= e^(-∏(x^2)) (where g(x)= ∫ G(K)e^2∏ikx dx (integral from -∞ to ∞)). We were told to find G(K) by forming a differential equation in H(K), where H(K) is the Fourier transform for h(x)=g'(x), and using the fact that H(K)=2∏iK*G(K).
We had to show that G(K) = e^(-∏(k^2)) ∫ e^(-∏(x^2)) dx (integral from -∞ to ∞).

The question I am stuck on is the following:

Use the above results to show that ∫ e^(-∏(x^2) dx = 1 (integral from -∞ to ∞)

( i.e ∫ g(x) dx =1 (integral from -∞ to ∞) )

Homework Equations



Another previous part involved showing that the Fourier transform for l(x)=xg(x) was given by L(K) = (i/2∏) * d/dk ( G(K) ) , so this may be relevant too.


The Attempt at a Solution



I don't know how to proceed at all and have tried manipulating all the different expressions but to no avail. Any help would be much appreciated! Thanks
 
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Try integrating your expression for G(k) from k = -infinity to infinity. What does this give you?
 
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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...
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