Help with expression ##F(it)-F(-it)## in the Abel-Plana form

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The discussion centers on the expression ##F(it)-F(-it)## derived from the Abel-Plana formula, where $$F(z)=\sqrt{z^2 + A^2}$$ and ##A## is a positive real number. The correct evaluation shows that for ##t>A##, the result is ##F(it)-F(-it)=2i\sqrt{t^2 -A^2}##. The confusion arises from the behavior of the function around branch points ##z=\pm iA##, leading to the conclusion that ##F(it)=-F(-it)##, which is validated through the properties of even functions. The participant expresses uncertainty about proving this relationship.

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I´m having a problem with the value of the expression

##F(it)-F(-it)##, found on the Abel-Plana formula, where $$F(z)=\sqrt{z^2 + A^2}$$, with ##A## being a positive real number (F(z) is analytic in the right half-plane).

Well, I know the result is ##F(it)-F(-it)=2i\sqrt{t^2 -A^2}##, for ##t>A##

Starting from the fact that the function has branch points ##z=\pm iA## I´d have to go around around these points to obtain the above result. However, to obtain it, I should have

$$F(-it)=-F(it)$$, which means ##F(it)=i\sqrt{t^2 -A^2}## and ##F(-it)=-i\sqrt{t^2 -A^2}##.

I honestly can't see why that happens and I can't formulate a proof for it.

Any help would be appreciated.
 
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I think ##F(it)-F(-it)=0##, no? This is unsurprising given that the only occurrence of z in the formula is squared, and ##z^2## is an even function. In any case:

##F(it)=\sqrt{(it)^2+A^2}=\sqrt{-t^2+A^2}##
and
##F(-it)=\sqrt{(-it)^2+A^2}=\sqrt{-t^2+A^2}##
 
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