# Relationship between the modified Bessel function of the second kind and the Psi function?

While reading this paper this paper, I was stumble upon a strange relationship between the modified Bessel function of the second kind and the Psi function that is:

$$c{K_1}\left( c \right) \approx 1 + \frac{{{c^2}}}{2}\left( {\ln \left( {\frac{c}{2}} \right) + {{\text{C}}_0}} \right) + \frac{{{c^4}}}{{16}}{{\text{C}}_1}$$

Where $${C_0}$$= −φ(1) /2 − φ(2) /2 and C1 = −φ(2)/2 − φ(3)/2

The paper said that φ(•) denotes the Psi function and point a citation to

. I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals Series and Products, New York, NY, USA:Academic, 2007.

Which yield the following definition and integral representation

Also, is it possible to extent this approximation for a modified Bessel function of the second kind of order $$M$$ like this $${K_M}\left( {2\sqrt x } \right)$$ ?

• For reference purposes, $\psi(x)$ is known as the digamma function Jan 19, 2021 at 2:27

The DLMF (eq. 10.3.1) provides the following sum formula for $$K_n(x)$$ with nonnegative integer $$n$$:

\begin{align} K_n(x) &=\frac12 \left(\frac{x}{2}\right)^{-n}\sum_{k=0}^{n-1}\frac{(n-k-1)!}{k!}\left(-\frac{x^2}{4}\right)^k+(-1)^{n+1}\ln\left(\frac{x}{2}\right)I_n(x)\\ &\qquad\quad\quad +(-1)^n \frac12 \left(\frac{x}{2}\right)^n \sum_{k=0}^\infty\left[\psi(k+1)+\psi(n+k+1)\right]\frac{(x^2/4)^k}{k!(n+k)!} \end{align}

For the case of $$n=1$$, this reduces to $$K_1(x) =\frac1x+\ln\left(\frac{x}{2}\right)I_1(x) - \frac{x}{4} \sum_{k=0}^\infty\left[\psi(k+1)+\psi(n+k+1)\right]\frac{(x^2/4)^k}{k!(k+1)!}$$

Since $$I_1(x)\approx x/2$$ for small $$x$$, we find

\begin{align} x K_1(x) &\approx 1+\ln\left(\frac{x}{2}\right)\cdot \frac{x^2}{2}-\frac{1}{4}(\psi(1)+\psi(2))x^2-(\psi(2)+\psi(3))\frac{x^4}{32}\\ &=1+\frac{x^2}{2}\left(\ln\left(\frac{x}{2}\right)-\frac12 \psi(1)- \frac12 \psi(2)\right) -\left(\frac12 \psi(2)+\frac12 \psi(3)\right)\frac{x^4}{16} \end{align} which is the stated result.

• When x is small why does the infinite sum that contain the Psi function becomes finite ? Jan 19, 2021 at 4:54
• Because it's an expansion in powers of $x$, so I've only kept powers up to $x^4$. Jan 19, 2021 at 5:58
• Does this series converge faster than just using taylor expansion? Jan 19, 2021 at 7:57
• Probably not, but that's because the above isn't that far from a Taylor series. Indeed, if you rearrange the main sum formula as $K_n(x)+(-1)^n \ln(x/2)I_n(x)$, then what remains on the right-hand side is just the Taylor series. Jan 19, 2021 at 14:28