Beta function of the conjugate arguments

Question

How to simplify (integrate)

$$\text{B}\left(\dfrac{m}{2}+ix, \dfrac{m}{2}-ix\right)$$ when $m\in\mathbb{N}$?

eg. when $m=1$ WolframAlpha simplifies the expression to $\pi \ \text{sech} (\pi x)$

Answer 1

Recall that $$\mathrm{B}(x,y)\mathrm{B}(x+y, 1-y) = \dfrac{\pi}{x\sin(\pi y)} \ \ \ (1)$$

$$ \mathrm{B}(x+1,y) = \dfrac{x}{x+y}\mathrm{B}(x,y) \ \ \ (2)$$

Let $X=\dfrac{m}{2}+ix$. Then, $$\begin{align} \mathrm{B}\left(X + \bar{X}, 1-\bar{X}\right) & = \mathrm{B}\left(m, 1-\bar{X}\right)\\ &=\dfrac{m-1}{m-1+1-\bar{X}}\mathrm{B}\left(m-1, 1-\bar{X}\right) \ \ \ (\text{if} \ m>1)\\ &=\vdots \ \ (\text{By repeatedly applying (2)})\\ &=\dfrac{(m-1)!}{\displaystyle\Pi(m-k+1-\bar{X})}\mathrm{B}(1,1-\bar{X}) \ \ \ \ \ (\text{Where product is taken from} \ k=1,…,m-1)\\ \end{align}$$

So,

$$\begin{align} \mathrm{B}(X,\bar{X})&=\dfrac{\pi}{X\mathrm{B}(1,1-\bar{X})(m-1)!\sin \pi\bar{X}}\displaystyle\prod_{k=1}^{m-1}(X+1-k) \ \ (\text{by (1)})\\ &=\dfrac{\pi}{\mathrm{B}(1,1-\frac{m}{2}+ix)(m-1)!\sin\pi(\frac{m}{2}-ix)}\displaystyle\prod_{k=2}^{m-1}(1+\frac{m}{2}+ix-k) \ \ (\text{if}\ m>2)\\ &=\begin{cases} \dfrac{i\pi(1-n+ix)(-1)^{n}\text{cosech}(\pi x)}{(2n-1)!} \ \ \displaystyle\prod_{k=2}^{2n-1}(1+n+ix-k) & \text{if} \ m=2n\\ \dfrac{\pi(\frac{1}{2}-n+ix)(-1)^n\text{sech}(\pi x)}{(2n)!} \ \displaystyle\prod_{k=2}^{2n}(\frac{3}{2}+n+ix-k) & \text{if} \ m=2n+1 \end{cases} \end{align}$$

Dealing with the excluded cases below:

$$\mathrm{B}\left(\frac{1}{2}+ix, \frac{1}{2}-ix\right) = \dfrac{\pi(\frac{1}{2} + ix)}{(\frac{1}{2}+ix)\sin\pi (\frac{1}{2}-ix)} = \pi \text{sech} \pi x$$

$$\mathrm{B}\left(1+ix, 1-ix\right) = \dfrac{i\pi x}{(1+ix)\sin\pi (1-ix)} = \dfrac{x}{1+ix}\pi\text{cosech}\pi x$$