For $\kappa>0$, $\alpha>\kappa$, $x>0$, and for $0\leq z \leq \frac{\pi\sqrt{\kappa }}{\sqrt{\alpha -\kappa }}$, I want to show that the function $$g(z)=\sqrt{\frac{\alpha }{\kappa }-1} \cosh \left(\frac{1}{2} (z-\alpha x)\right) \tan \left(\frac{1}{2} z \sqrt{\frac{\alpha }{\kappa }-1}\right)$$
is increasing in $z$.
I believe this is true for any $z$ value, but showing it for $0\leq z \leq \frac{\pi \sqrt{\kappa }}{\sqrt{\alpha -\kappa }}$ would suffice. I'm not quite sure how to go about it. I've calculated the derivative w.r.t $z$ and shown that the derivative is positive when $z=0$ and explodes to $\infty$ as $z$ approaches $\frac{\pi \sqrt{\kappa }}{\sqrt{\alpha -\kappa }}$. But I have not shown that the derivative is positive anywhere in this interval.
Any help would be greatly appreciated.
Let's write $$f=\sqrt {\frac{\alpha }{\kappa }-1}$$ Then we have:
For $f>0$, $\alpha>0$, $x>0$, and for $0\leq z \leq \frac{\pi}{f}$, show that the function $$g(z)=f \cosh \left(\tfrac{1}{2} (z-\alpha x)\right) \tan \left(\tfrac{1}{2} z f\right)$$
is increasing in $z$. $$g'(z)=\tfrac12 f^2 \cosh \left(\tfrac{1}{2} (z-\alpha x)\right) \sec^2 \left(\tfrac{1}{2} z f\right) + \tfrac12 f \sinh \left(\tfrac{1}{2} (z-\alpha x)\right) \tan \left(\tfrac{1}{2} z f\right)$$
For $g'(z) \geqslant 0$ to be true we require:
$$f \cosh \left(\tfrac{1}{2} (z-\alpha x)\right) \sec^2 \left(\tfrac{1}{2} z f\right) >- \sinh \left(\tfrac{1}{2} (z-\alpha x)\right) \tan \left(\tfrac{1}{2} z f\right)$$
If $z \geqslant \alpha x$ then the $\sinh$ term is positive and we have:
$$f \coth \left(\tfrac{1}{2} (z-\alpha x)\right) >- \sin \left(\tfrac{1}{2} z f\right) \cos \left(\tfrac{1}{2} z f\right) $$
$$2f \coth \left(\tfrac{1}{2} (z-\alpha x)\right) >- \sin zf $$
But the LHS is positive and the RHS is negative so we are ok.
Otherwise, $z<\alpha x$ then the $\sinh$ term is negative and we require:
$$f \cosh \left(\tfrac{1}{2} (\alpha x - z)\right) \sec^2 \left(\tfrac{1}{2} z f\right) > \sinh \left(\tfrac{1}{2} (\alpha x - z)\right) \tan \left(\tfrac{1}{2} z f\right)$$
$$2f \coth \left(\tfrac{1}{2} (\alpha x - z)\right) > \sin zf $$
Unfortunately, this isn't always true. For example $f=0.1, \alpha x=100, z=10$ breaks it. So I don't think your result is true. E&OE