Prove $\lim_{n \to \infty} ( \frac{\psi(-1/2+n i )- \psi(-1/2-n i)}{2i}- \frac{n}{n^2+1/4} ) = \frac{\pi}{2}$

Question

Consider the sequence:

$$A_n= \frac{\psi(-1/2+n i )- \psi(-1/2-n i)}{2i}- \frac{n}{n^2+1/4} $$

How would you prove that:

$$\lim_{n \to \infty} A_n= \frac{\pi}{2}$$

This sequence converges extremely fast, in fact for $n=10$ we have:

$$\left( \frac{\psi(-1/2+10 i )- \psi(-1/2-10 i)}{2i}- \frac{10}{100+1/4} \right)= \\ = 1.57079632679489661923132169 \ldots$$

Where all the digits shown are the same as for $\pi/2$.

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I have found this limit by considering the usual integral $\int_0^1 \frac{dx}{1+x^2}$, using the midpoint rule to derive a Riemann sum, then finding its explicit hypergeometric form, which simplified to the difference of digammas.

I was quite surprised by the accuracy of this method, because it only takes $10$ points to get $25$ correct digits, so my questions are:

• What other proofs can you offer for the limit?

• Why does it converge so fast? Is it because the midpoint rule is very accurate and the integrated function is very nice?

The motivation for this question is of course, generalization of this result, because any integral of the form:

$$\int_a^b \frac{P(x)}{Q(x)} c^x dx$$

where $P,Q$ are polynomials, can be approximated by a hypergeometric Riemann sum.

Answer 1

$$\begin{align} \frac{\psi(-1/2+n i )- \psi(-1/2-n i)}{2i} &=\frac{\psi(-1/2+n i )- \left(\psi(1/2-n i)-\frac1{-1/2-ni}\right)}{2i}\\ &=\frac{\psi(-1/2+n i )- \psi(1/2-n i)}{2i}+\frac{2n+i}{4n^2+1}\\ &=\frac{\psi(-1/2+n i )- \left(\psi(3/2-n i)-\frac1{1/2-ni}\right)}{2i}+\frac{2n+i}{4n^2+1}\\ &=\frac{\psi(-1/2+n i )- \psi(3/2-n i)}{2i}+\frac{4n}{4n^2+1}\\ &=\frac{\pi\cot{(\pi(3/2-ni))}}{2i}+\frac{4n}{4n^2+1}\\ &=\frac{\pi\cot{(3\pi/2-\pi ni)}}{2i}+\frac{4n}{4n^2+1}\\ &=\frac{\pi\tan{(\pi ni)}}{2i}+\frac{4n}{4n^2+1}\\ &=\frac{\pi i\tanh{(\pi n)}}{2i}+\frac{4n}{4n^2+1}\\ &=\frac{\pi \tanh{(\pi n)}}{2}+\frac{4n}{4n^2+1}\\ \end{align}$$ where I have used the recurrence formula and reflection formula for the digamma function shown here. Then our limit is simply $$\lim_{n\to\infty}\frac{\pi \tanh{(\pi n)}}{2}=\frac{\pi}2$$ as we have $$\lim_{n\to\infty}\tanh{(\pi n)}=1$$

Answer 2

We shall express the polygamma function $\psi^{(0)}(z)$ in terns of the $\Gamma(z)$ function and continue the derivation using the latter.

Let

$$f(n) = \frac{1}{2 i}\left(\psi ^{(0)}\left(-\frac{1}{2}+ n i\right)-\psi^{(0)} \left(-\frac{1}{2}- n i\right)\right)$$

This is a simplified version of the expression of the OP which neglects the term

$$- \frac{n}{n^2+1/4}$$

which is irrelevant for the limit.

Remembering the definition of the digamma function

$$\psi^{(0)}(z)=\frac{\partial\log (\Gamma (z))}{\partial z}$$

we are led to define the function

$$g(n) = -\frac{1}{2}\left(\log\left(\Gamma(-\frac{1}{2} + n i)\right) + \log\left(\Gamma(-\frac{1}{2} - n i)\right) \right)$$

so that

$$\frac{\partial g(n)}{ \partial n} = f(n)\tag{*}$$

Now $g$ can be written as

$$g(n) =-\frac{1}{2}\log\left(\Gamma(-\frac{1}{2} + n i) \Gamma(-\frac{1}{2} - n i) \right) = -\frac{1}{2}\log\left(\frac{4 \pi }{(1+4 n^2) \cosh(n \pi)}\right) $$

Here we have used the properties of $\log$ and the relation

$$\Gamma \left(-\frac{1}{2}+ n i\right) \Gamma \left(-\frac{1}{2}- n i\right)=\frac{4 \pi }{4 n^2+1}\frac{1}{\cosh(\pi n)}$$

which is a consequence of Euler's reflexion formula $\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(z)}$ for $z=\frac{3}{2}+i n$ and twofold application of $ \Gamma(x+1)= x\Gamma(x)$.

Now, using $(*)$ we get

$$f(n) = \frac{\pi \left(4 n^2+1\right) \tanh (\pi n)+8 n}{8 n^2+2}$$

from which the limit is easily obtained

$$\lim_{n\to \infty } \, f(n)=\frac{\pi }{2}$$

Discussion

The same limit is obtained for the more general expression

$$F(n) = \frac{1}{2 i}\left(\psi ^{(0)}\left(a + i b(n)\right)-\psi^{(0)} \left(a- i b(n)\right)\right)$$

where $a$ is real an $b(n)$ is a real function going to $\infty$ with $n$.

Using the asymptotic expansion

$$\psi ^{(0)}(z\to \infty)\simeq -\log \left(\frac{1}{z}\right)-\frac{1}{2 z}$$

we see that the limit boils down to

$$\frac{1}{2 i}\log(\frac{a-i b(n)}{a+i b(n)}) \simeq \frac{1}{2 i}\log(-1) = \frac{\pi}{2}$$