# Debye function

In mathematics, the family of Debye functions is defined by $D_{n}(x)={\frac {n}{x^{n}}}\int _{0}^{x}{\frac {t^{n}}{e^{t}-1}}\,dt.$
The functions are named in honor of Peter Debye, who came across this function (with n = 3) in 1912 when he analytically computed the heat capacity of what is now called the Debye model.

## Mathematical properties
### Relation to other functions

The Debye functions are closely related to the polylogarithm.

### Series expansion

They have the series expansion $D_{n}(x)=1-{\frac {n}{2(n+1)}}x+n\sum _{k=1}^{\infty }{\frac {B_{2k}}{(2k+n)(2k)!}}x^{2k},\quad |x|<2\pi ,\ n\geq 1,$ where $B_{n}$ is the n-th Bernoulli number.

### Limiting values

$\lim _{x\to 0}D_{n}(x)=1.$ If $\Gamma$ is the gamma function and $\zeta$ is the Riemann zeta function, then, for $x\gg 0$, $D_{n}(x)={\frac {n}{x^{n}}}\int _{0}^{x}{\frac {t^{n}\,dt}{e^{t}-1}}\sim {\frac {n}{x^{n}}}\Gamma (n+1)\zeta (n+1),\qquad \operatorname {Re} n>0,$

### Derivative

The derivative obeys the relation $xD_{n}^{\prime }(x)=n\left(B(x)-D_{n}(x)\right),$ where $B(x)=x/(e^{x}-1)$ is the Bernoulli function.

## Applications in solid-state physics
### The Debye model

The Debye model has a density of vibrational states $g_{\text{D}}(\omega )={\frac {9\omega ^{2}}{\omega _{\text{D}}^{3}}}\,,\qquad 0\leq \omega \leq \omega _{\text{D}}$ with the Debye frequency ω$_D$.

### Internal energy and heat capacity

Inserting g into the internal energy $U=\int _{0}^{\infty }d\omega \,g(\omega )\,\hbar \omega \,n(\omega )$ with the Bose–Einstein distribution $n(\omega )={\frac {1}{\exp(\hbar \omega /k_{\text{B}}T)-1}}.$ one obtains $U=3k_{\text{B}}T\,D_{3}(\hbar \omega _{\text{D}}/k_{\text{B}}T).$ The heat capacity is the derivative thereof.

### Mean squared displacement

The intensity of X-ray diffraction or neutron diffraction at wavenumber q is given by the Debye-Waller factor or the Lamb-Mössbauer factor. For isotropic systems it takes the form $\exp(-2W(q))=\exp \left(-q^{2}\langle u_{x}^{2}\rangle \right).$ In this expression, the mean squared displacement refers to just once Cartesian component u$_x$ of the vector u that describes the displacement of atoms from their equilibrium positions. Assuming harmonicity and developing into normal modes, one obtains $2W(q)={\frac {\hbar ^{2}q^{2}}{6Mk_{\text{B}}T}}\int _{0}^{\infty }d\omega {\frac {k_{\text{B}}T}{\hbar \omega }}g(\omega )\coth {\frac {\hbar \omega }{2k_{\text{B}}T}}={\frac {\hbar ^{2}q^{2}}{6Mk_{\text{B}}T}}\int _{0}^{\infty }d\omega {\frac {k_{\text{B}}T}{\hbar \omega }}g(\omega )\left[{\frac {2}{\exp(\hbar \omega /k_{\text{B}}T)-1}}+1\right].$ Inserting the density of states from the Debye model, one obtains $2W(q)={\frac {3}{2}}{\frac {\hbar ^{2}q^{2}}{M\hbar \omega _{\text{D}}}}\left[2\left({\frac {k_{\text{B}}T}{\hbar \omega _{\text{D}}}}\right)D_{1}{\left({\frac {\hbar \omega _{\text{D}}}{k_{\text{B}}T}}\right)}+{\frac {1}{2}}\right].$ From the above power series expansion of $D_{1}$ follows that the mean square displacement at high temperatures is linear in temperature $2W(q)={\frac {3k_{\text{B}}Tq^{2}}{M\omega _{\text{D}}^{2}}}.$ The absence of $\hbar$ indicates that this is a classical result. Because $D_{1}(x)$ goes to zero for $x\to \infty$ it follows that for $T=0$ $2W(q)={\frac {3}{4}}{\frac {\hbar ^{2}q^{2}}{M\hbar \omega _{\text{D}}}}$ (zero-point motion).