Closed form for $n$-th derivative of exponential: $\exp\left(-\frac{\pi^2a^2}{x}\right)$ 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?
 A: 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.
A: 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.
A: Here we show that OPs approach is regrettably not correct and provide an alternative to answer the problem.
This answer is based upon the $n$-th derivative of the composite of two functions. It is stated as identity (3.56) in H.W. Gould's Tables of Combinatorial Identities, Vol. I and called:

Hoppe Form of Generalized Chain Rule
Let $D_x$ represent differentiation with respect to $x$ and $z=z(x)$. Hence $D^n_x f(x)$ is the $n$-th derivative of $f$ with respect to $x$.  The following is valid for $n\geq 0$:
\begin{align*}
D_x^n f(z)=\sum_{k=0}^nD_z^kf(z)\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}z^{k-j}D_x^nz^j\tag{1}
\end{align*}

Let's denote for convenience only $c:=\pi^2a^2$.

The following is valid for $n\geq 1$
\begin{align*}
D_x^ne^{-\frac{c}{x}}=e^{-\frac{c}{x}}\sum_{k=0}^{n-1}(-1)^k\binom{n}{k}\binom{n-1}{k}k!c^{n-k}x^{-2n+k}\tag{2}
\end{align*}

Using Hoppes formula (1) we obtain by setting
\begin{align*}
f(z)=e^{z}\qquad\quad\text{and}\quad\qquad z=z(x)=-\frac{c}{x}
\end{align*}
the following for $n\geq 1$

\begin{align*}
D_x^ne^{-\frac{c}{x}}&=\sum_{k=0}^nD_z^ke^z\frac{(-1)^k}{k!}
\sum_{j=0}^k(-1)^j\binom{k}{j}\left(-\frac{c}{x}\right)^{k-j}D_x\left(-\frac{c}{x}\right)^j\tag{3}\\
&=e^{-\frac{c}{x}}\sum_{k=0}^n\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}
\left(-\frac{c}{x}\right)^{k-j}n!c^j\binom{n+j-1}{j-1}\frac{(-1)^{n+j}}{x^{n+j}}\tag{4}\\
&=e^{-\frac{c}{x}}\sum_{k=0}^n\frac{n!}{k!}(-1)^nc^k\sum_{j=0}^k(-1)^j\binom{k}{j}\binom{n+j-1}{j-1}x^{-n-k}\tag{5}\\
&=e^{-\frac{c}{x}}\sum_{k=0}^n\frac{n!}{k!}(-1)^nc^k\binom{n-1}{k-1}(-1)^kx^{-n-k}\tag{6}\\
&=e^{-\frac{c}{x}}\sum_{k=0}^n(-1)^{n+k}\binom{n}{k}\binom{n-1}{k-1}c^kx^{-n-k}\\
&=e^{-\frac{c}{x}}\sum_{k=0}^{n-1}(-1)^{k}\binom{n}{k}\binom{n-1}{k}c^{n-k}x^{-2n+k}\tag{7}\\
\end{align*}
and the claim follows.

Comment:

*

*In (3) we apply Hoppes formula (1)


*In (4) we use
\begin{align*}
D_z^ke^z=e^z=e^{-\frac{c}{x}}\quad\text{and}
\quad D_x\left(-\frac{c}{x}\right)^j=n!c^j\binom{n+j-1}{j-1}(-1)^{n+j}
\end{align*}


*In (5) we do some rearrangement


*In (6) we use the identity
\begin{align*}
\sum_{j=0}^k(-1)^j\binom{k}{j}\binom{n+j-1}{j-1}=\binom{n-1}{k-1}(-1)^k
\end{align*}

*

*In (7) we replace $k$ with $n-k$ and set the upper index limit to $n-1$

Small examples: $n=1,2,3$

Using formula (2) we obtain for $n=1,2,3$
\begin{align*}
D_x^1e^{-\frac{c}{x}}&=e^{-\frac{c}{x}}\frac{c}{x^2}\\
D_x^2e^{-\frac{c}{x}}&=e^{-\frac{c}{x}}\frac{c}{x^4}\left(c-2x\right)\tag{8}\\
D_x^3e^{-\frac{c}{x}}&=e^{-\frac{c}{x}}\frac{c}{x^6}\left(c^2-6cx+6x^2\right)\\
\end{align*}

Hermite polynomials:
If we consider Hermite polynomials in the form
\begin{align*}
H_n(x)=-e^{x^2}D_xe^{-x^2}
\end{align*}
and apply a substitution
\begin{align*}
x=x(y)=\sqrt{\frac{c}{y}}=\sqrt{c}y^{-\frac{1}{2}}
\end{align*}
in order to obtain
\begin{align*}
H_n(x(y))&=-e^{(x(y))^2}D_ye^{-(x(y))^2}\\
&=-e^{\frac{c}{y}}D_ye^{-\frac{c}{y}}
\end{align*}
we are in the same situation as above, since we have to calculate the $n$-th derivative of a composition of functions, which is not feasible with OPs approach. It could be done with Hoppes formula instead.

Let's look at a small example by taking $n=2$. In the following I omit most of the time  the argument $y$ and write $xx^\prime$ instead of $x(y)x^\prime(y)$ to ease reading.

\begin{align*}
H_2(x(y))&=e^{x^2}D_y^2\left(e^{-x^2}\right)\\
&=-e^{x^2}D_y\left(-2xx^\prime e^{-x^2}\right)\\
&=-e^{x^2}\left(D_y\left(-2xx^\prime\right)e^{-x^2}+(-2xx^\prime)D_y\left(e^{-x^2}\right)\right)\\
&=-e^{x^2}\left(\left(-2{x^\prime}^2-2xx^{\prime\prime}\right)e^{-x^2}+4x^2{x^\prime}^2e^{-x^2}\right)\\
&=-2{x^\prime}^2-2xx^{\prime\prime}+4x^2{x^\prime}^2
\end{align*}
Since
\begin{align*}
x&=x(y)=\sqrt{c}\,y^{-\frac{1}{2}}\\
x^\prime&=x^\prime(y)=-\frac{1}{2}\sqrt{c}\,y^{-\frac{3}{2}}\\
x^{\prime\prime}&=x^{\prime\prime}(y)=\frac{3}{4}\sqrt{c}\,y^{-\frac{5}{2}}\\
\end{align*}
we finally obtain
\begin{align*}
H_2(x(y))&-2{x^\prime}^2-2xx^{\prime\prime}+4x^2{x^\prime}^2\\
&=-2\frac{1}{4}c\,y^{-3}-2\sqrt{c}\,y^{-\frac{1}{2}}\frac{3}{4}\sqrt{c}\,y^{-\frac{5}{2}}
+4cy^{-1}\frac{1}{4}c\,y^{-3}\\
&=-\frac{1}{2}c\,y^{-3}-\frac{3}{2}c\,y^{-3}+c^2\,y^{-4}\\
&=-2c\,y^{-3}+c^2\,y^{-4}\\
&=\frac{c}{y^4}\left(c^2-2y\right)
\end{align*}
in accordance with (8). So, we can apply Hoppes formula or perform some equivalent calculation.

A: 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

*

*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.

*https://math.stackexchange.com/a/4262657/945479.

*https://math.stackexchange.com/a/4261764/945479.

*https://mathoverflow.net/a/405003/147732.

*https://math.stackexchange.com/a/4262516/945479.

*https://math.stackexchange.com/a/4262498/945479.

