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

[JasonPhysicist]

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.

 

[Someone2841]

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

 

[math771]

Hi there,

I understand your confusion with this expression. It may be helpful to think about the function F(z) in terms of its real and imaginary parts. We can rewrite F(z) as F(x+iy) = u(x,y) + iv(x,y), where u(x,y) = √(x^2 + y^2 + A^2) and v(x,y) = xy/√(x^2 + y^2 + A^2). Now, when we substitute z = it, we get F(it) = u(0,t) + iv(0,t) = iv(0,t) = 0 + it/√(t^2 + A^2). Similarly, when we substitute z = -it, we get F(-it) = u(0,-t) + iv(0,-t) = -iv(0,t) = 0 - it/√(t^2 + A^2). So, we can see that F(it) and F(-it) are complex conjugates of each other, which means that F(-it) = -F(it). This is why we get the result of 2i√(t^2 - A^2).

I hope this helps clear things up for you. Let me know if you have any further questions.