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[tex] h \left( x \right) = \frac{2}{\pi}\int^{\infty}_{0} \left( dk \right) \left[ \int^{\infty}_{0} h \left( \varsigma \right) sin \left( k \varsigma \right) sin \left( x \varsigma \right) d\varsigma \right] [/tex]

and the limits of integration can be changed to

[tex] h \left( x \right) = \frac{1}{2 \pi}\int^{\infty}_{-\infty} \left( dk \right) \left[ \int^{\infty}_{-\infty} h \left( \varsigma \right) sin \left( k \varsigma \right) sin \left( x \varsigma \right) d\varsigma \right] [/tex]

I'm trying to understand why this factor of 1/4 arises...

My understand is this

- in the first case, one integral gives us a fourier coefficent [tex] A_{k} [/tex] the other integration is effectively doing this; [tex]\sum^{\infty}_{k=0} A_{k} sin \left( kx \right)[/tex]

- in the second case, one integral gives [tex] 2 A_{k} [/tex], because it integrates across 2 periods, but it can't matter which period you integrate across to find the fourier coefficent. The other integral is effectively [tex]2 \sum^{\infty}_{k=0} A_{k} sin \left( kx \right)[/tex] since it is summing over 2 periods instead of 1.

Is this a correct way of looking at it? I'm more confident I know whats going on after writing this post and having to articulate it...