# How can I prove this identity involving the digamma function?

I'm trying to prove an identity involving the digamma function $\psi(z)$, but I can't seem to figure out a way to do it. Can anyone help me out? The identity is

$$\psi\left(\frac{m}{2} + iy\right) + \psi\left(\frac{m}{2} - iy\right) = \psi\left(\frac{n}{2} + iy\right) + \psi\left(\frac{n}{2} - iy\right)$$

where $m$ and $n$ are integers satisfying $m + n = 2$, and $y$ is any nonzero real number.

I've tried looking at a couple of the integral representations of $\psi(z)$ listed on the Wikipedia page, but I haven't been able to figure it out from those. I think there's probably some simple integral identity I'm forgetting that would make the whole thing work out - for instance, if I could show that

$$\int_0^\infty \frac{e^{nt/2}\cos(yt)}{\sinh(t/2)}\mathrm{d}t = \int_0^\infty \frac{e^{-nt/2}\cos(yt)}{\sinh(t/2)}\mathrm{d}t$$

for $n \in \mathbb{Z}, y \neq 0$, I think I'd be set. So what I'm hoping to get is a pointer to some relation like that which I could use. Of course I'd be happy with a full proof if you prefer ;-)

P.S. I know this sounds kind of "homework-y," but it's not for a homework assignment; I'm trying to verify a calculation in a physics paper.

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The first relation can be rewritten as

$$\psi\left(\frac{m}{2} + iy\right) + \psi\left(\frac{m}{2} - iy\right) = \psi\left(\frac{2-m}{2} + iy\right) + \psi\left(\frac{2-m}{2} - iy\right)$$

or

$$\psi\left(\frac{m}{2} + iy\right) + \psi\left(\frac{m}{2} - iy\right) = \psi\left(1-\left(\frac{m}{2} + iy\right)\right) + \psi\left(1-\left(\frac{m}{2} - iy\right)\right)$$

If we use the reflection formula for the digamma function (which can be derived from the reflection formula for the gamma function)

$$\psi(1-z)=\psi(z)+\pi\cot\pi z$$

$$\cot\left(\frac{m\pi}{2}+i\pi y\right)+\cot\left(\frac{m\pi}{2}-i\pi y\right)=0$$

or since the cotangent is an odd function,

$$\cot\left(\frac{m\pi}{2}+i\pi y\right)=\cot\left(-\frac{m\pi}{2}+i\pi y\right)$$

which can be rewritten as

$$\cot\left(\frac{m\pi}{2}+i\pi y\right)=\cot\left(\frac{m\pi}{2}+i\pi y-m\pi\right)$$

and it is clear that both sides of the equation will agree only if $m\in\mathbb Z$ due to the periodicity of the cotangent.

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