Closed form for $n$-th derivative of exponential: $\exp\left(-\frac{\pi^2a^2}{x}\right)$

Question

I need the closed-form for the $n$-th derivative ($n\geq0 $):

$$\frac{\partial^n}{\partial x^n}\exp\left(-\frac{\pi^2a^2}{x}\right)$$

Thanks!

By following the suggestion of Hermite polynomials:

$$H_n(x)=(-1)^ne^{x^2}\frac{\partial^n}{\partial x^n}e^{-x^2}$$

and doing the variable change $x=\pi a y^{-\frac{1}{2}}$, I obtain:

$$\frac{\partial^n}{\partial x^n}=-2\left(\frac{y^{\frac{3}{2}}}{\pi a}\right)^n\frac{\partial^n}{\partial y^n}$$

and therefore

$$H_n(\pi a y^{-\frac{1}{2}})=(-1)^{n+1}e^{\frac{\pi^2a^2}{y}}2\left(\frac{\pi a}{y^{\frac{3}{2}}}\right)^n\frac{\partial^n}{\partial y^n}e^{-\frac{\pi^2a^2}{y}}$$

Finally

$$\frac{\partial^n}{\partial y^n}e^{-\frac{\pi^2a^2}{y}}=\frac{1}{2}e^{-\frac{\pi^2a^2}{y}}(-1)^{n+1}H_n(\pi a y^{-\frac{1}{2}})\left(\frac{y^{\frac{3}{2}}}{\pi a}\right)^n$$

Is this correct?

Answer 1

I obtained the following result using the Fourier transform: $$\frac{\partial^n}{\partial x^n}\exp\left(-\frac{\pi^2a^2}{x}\right)=\\\frac12\pi^2\,a^2(-1)^nn!\ x^{-n-2}\left({_2F_3}\left(\frac{n}{2}+1,\frac{n}{2}+\frac{3}{2};\frac{3}{2},\frac{3}{2},2;\frac{a^4\,\pi^4}{4\,x^2}\right)\pi^2a^2(n+1)\\-{_2F_3}\left(\frac n2+\frac12,\frac n2+1;\frac12,1,\frac32;\frac{a^4\pi^4}{4\,x^2}\right)2\,x\right).$$

Probably, it can be simplified.

Answer 2

Related problem: (I). Here is a formula for the $n$th derivative of integer order of $e^{\frac{c}{x}}$

Formula 1: This formula is valid only for $n \in \mathbb{N} \cup \left\{ 0\right\}$.

$$\left( \rm e^{\frac{c}{x}}\right)^{(n)} = {{\rm e}^{{\frac{c}{x}}}}\sum _{s=0}^{n} \sum _{k=0}^{n} \left( -1 \right)^{-k-s}\left[\matrix{n\\k+s}\right] \left\{\matrix{k+s\\s}\right\}{c}^{s}{x}^{-s-n},\quad n\in \mathbb{N} \cup \left\{ 0\right\} $$

where $\left[\matrix{n\\k+s}\right]$ and $\left\{\matrix{k+s\\s}\right\}$ are the Stirling numbers of the first kind and the second kind respectively.

Formula 2: Here is a unified formula which gives a complete solution to the problem of differentiation and integration of real order of the function in terms of the MeijerG function

$$\left(\rm e^{\frac{c}{x}}\right)^{(n)} = \left( -1 \right)^{n+1}{a}^{-n} \left( -1 \right) ^{n}G^{1, 1}_{1, 2}\left(-{\frac {a}{x}}\, \Big\vert\,^{1}_{1+n, n}\right) ,\quad n\in \mathbb{R}.$$

The last formula gives

1) derivatives of real order if $n>0$,

2) anti-derivatives of real order if $n<0$.

Note: The $n$th derivative of integer order of the function $e^{\frac{1}{x}}$ has to do with Lah nembers.

Answer 3

By induction, the formula \begin{equation}\label{g(t)-derivative} \bigl(e^{-1/t}\bigr)^{(i)}=\frac{e^{-1/t}}{t^{2i}} \sum_{k=0}^{i-1}(-1)^k{k!}\binom{i}{k}\binom{i-1}{k}{t^{k}}, \quad i\in\mathbb{N},\quad t\ne0 \end{equation} was established in Theorem 2.2 of the paper [1] below. Consequently, the derivative \begin{equation} \biggl[\exp\biggl(-\frac{a^2\pi^2}{x}\biggr)\biggr]^{(i)} =\exp\biggl(-\frac{a^2\pi^2}{x}\biggr)\biggl(\frac{a\pi}{x}\biggr)^{2i} \sum_{k=0}^{i-1}(-1)^k{k!}\binom{i}{k}\binom{i-1}{k}\biggl(\frac{x}{a^2\pi^2}\biggr)^{k}, \quad i\ge1 \end{equation} can be deduced readily.

One can also compute the $n$th derivative of the function $\exp\bigl(-\frac{a^2\pi^2}{x}\bigr)$ alternatively. For more information on alternative computations, please refer to the sites from [2] to [6] listed below.

References

1. Xiao-Jing Zhang, Feng Qi, and Wen-Hui Li, Properties of three functions relating to the exponential function and the existence of partitions of unity, International Journal of Open Problems in Computer Science and Mathematics 5 (2012), no. 3, 122--127; available online at https://doi.org/10.12816/0006128.
2. https://math.stackexchange.com/a/4262657/945479.
3. https://math.stackexchange.com/a/4261764/945479.
4. https://mathoverflow.net/a/405003/147732.
5. https://math.stackexchange.com/a/4262516/945479.
6. https://math.stackexchange.com/a/4262498/945479.