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Formally using the inverse Mellin transform for x>0:

$$e^x f(x\tfrac{d}{dx})e^{-x}=e^x f(x\tfrac{d}{dx}) \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} \frac{x^{-s}}{(-s)!} ds$$

$$=e^x \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} f(-s) \frac{x^{-s}}{(-s)!} ds.$$

Let $$f(x)=\binom{x+\alpha+\beta}{\beta},$$

then

$$e^x \binom{x\tfrac{d}{dx}+\alpha+\beta}{\beta}e^{-x}=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x)$$

where $L_{\beta}^{\alpha}(x)$ is the generalized Laguerre function and $K(-\beta,\alpha+1,x),$ Kummer's confluent hypergeometric function.

For an elaboration, see the notes The Inverse Mellin Transform, Bell Polynomials, a Generalized Dobinski Relation, and the Confluent Hypergeometric Functions.

See also Rodriguez-like formula (attributed to Poole, 1936) on pg. 59 of Bateman's (et al.) Higher Transcendental Functions Vol. I:

$$(x\tfrac{d}{dx}+\alpha)_{n} h(x) = x^{1-\alpha}D^{n}[x^{n+\alpha-1}h(x)],$$

with $D=\tfrac{d}{dx}$, leading to

$$e^x \binom{x\tfrac{d}{dx}+\alpha+n}{n}e^{-x}=e^x x^{-\alpha}\tfrac{D^{n}}{n!}[x^{n+\alpha}e^{-x}]=L_{n}^{\alpha}(x).$$

This can be generalized by using the fractional integro-derivative representation of $K(a,b,x)$ (see Eqn. 13.2.1 on pg. 505 of Abramowitz and Stegun):

$$K(a,b,x)= e^x \tfrac{(b-1)!}{x^{b-1}}\int_{0}^{x} e^{-t}\tfrac{(x-t)^{a-1}}{(a-1)!} \tfrac{t^{b-a-1}}{(b-a-1)!} dt=e^x \tfrac{(b-1)!}{x^{b-1}}D^{-a}[e^{-x}\tfrac{x^{b-a-1}}{(b-a-1)!}],$$

leading to

$$e^x {x^{-\alpha}}\tfrac{D^{\beta}}{\beta!}[x^{\beta+\alpha}e^{-x}]=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x).$$

Edit (Sept. 2019):

Abstracting, (as Jeans said) "leaving the operators hungry for something to differentiate":

$$\binom{x\tfrac{d}{dx}+\alpha+\beta}{\beta}={x^{-\alpha}}\frac{:Dx:^{\beta}}{\beta!}x^{\alpha}=L_{\beta}^{\alpha}(-:xD:)$$

$$=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,-:xD:),$$

where, by definition, for any two operators $:AB:^{\beta}=A^{\beta}B^{\beta}$.

Formally using the inverse Mellin transform for x>0:

$$e^x f(x\tfrac{d}{dx})e^{-x}=e^x f(x\tfrac{d}{dx}) \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} \frac{x^{-s}}{(-s)!} ds$$

$$=e^x \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} f(-s) \frac{x^{-s}}{(-s)!} ds.$$

Let $$f(x)=\binom{x+\alpha+\beta}{\beta},$$

then

$$e^x \binom{x\tfrac{d}{dx}+\alpha+\beta}{\beta}e^{-x}=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x)$$

where $L_{\beta}^{\alpha}(x)$ is the generalized Laguerre function and $K(-\beta,\alpha+1,x),$ Kummer's confluent hypergeometric function.

For an elaboration, see the notes The Inverse Mellin Transform, Bell Polynomials, a Generalized Dobinski Relation, and the Confluent Hypergeometric Functions.

See also Rodrigues-like formula (attributed to Poole, 1936) on pg. 59 of Bateman's (et al.) Higher Transcendental Functions Vol. I:

$$(x\tfrac{d}{dx}+\alpha)_{n} h(x) = x^{1-\alpha}D^{n}[x^{n+\alpha-1}h(x)],$$

with $D=\tfrac{d}{dx}$, leading to

$$e^x \binom{x\tfrac{d}{dx}+\alpha+n}{n}e^{-x}=e^x x^{-\alpha}\tfrac{D^{n}}{n!}[x^{n+\alpha}e^{-x}]=L_{n}^{\alpha}(x).$$

This can be generalized by using the fractional integro-derivative representation of $K(a,b,x)$ (see Eqn. 13.2.1 on pg. 505 of Abramowitz and Stegun):

$$K(a,b,x)= e^x \tfrac{(b-1)!}{x^{b-1}}\int_{0}^{x} e^{-t}\tfrac{(x-t)^{a-1}}{(a-1)!} \tfrac{t^{b-a-1}}{(b-a-1)!} dt=e^x \tfrac{(b-1)!}{x^{b-1}}D^{-a}[e^{-x}\tfrac{x^{b-a-1}}{(b-a-1)!}],$$

leading to

$$e^x {x^{-\alpha}}\tfrac{D^{\beta}}{\beta!}[x^{\beta+\alpha}e^{-x}]=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x).$$

Edit (Sept. 2019):

Abstracting, (as Jeans said) "leaving the operators hungry for something to differentiate":

$$\binom{x\tfrac{d}{dx}+\alpha+\beta}{\beta}={x^{-\alpha}}\frac{:Dx:^{\beta}}{\beta!}x^{\alpha}=L_{\beta}^{\alpha}(-:xD:)$$

$$=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,-:xD:),$$

where, by definition, for any two operators $:AB:^{\beta}=A^{\beta}B^{\beta}$.

Formally using the inverse Mellin transform for x>0:

$$e^x f(x\tfrac{d}{dx})e^{-x}=e^x f(x\tfrac{d}{dx}) \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} \frac{x^{-s}}{(-s)!} ds$$

$$=e^x \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} f(-s) \frac{x^{-s}}{(-s)!} ds.$$

Let $$f(x)=\binom{x+\alpha+\beta}{\beta},$$

then

$$e^x \binom{x\tfrac{d}{dx}+\alpha+\beta}{\beta}e^{-x}=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x)$$

where $L_{\beta}^{\alpha}(x)$ is the generalized Laguerre function and $K(-\beta,\alpha+1,x),$ Kummer's confluent hypergeometric function.

For an elaboration, see the notes The Inverse Mellin Transform, Bell Polynomials, a Generalized Dobinski Relation, and the Confluent Hypergeometric Functions.

See also Rodriguez-like formula (attributed to Poole, 1936) on pg. 59 of Bateman's (et al.) Higher Transcendental Functions Vol. I:

$$(x\tfrac{d}{dx}+\alpha)_{n} h(x) = x^{1-\alpha}D^{n}[x^{n+\alpha-1}h(x)],$$

with $D=\tfrac{d}{dx}$, leading to

$$e^x \binom{x\tfrac{d}{dx}+\alpha+n}{n}e^{-x}=e^x x^{-\alpha}\tfrac{D^{n}}{n!}[x^{n+\alpha}e^{-x}]=L_{n}^{\alpha}(x).$$

This can be generalized by using the fractional integro-derivative representation of $K(a,b,x)$ (see Eqn. 13.2.1 on pg. 505 of Abramowitz and Stegun):

$$K(a,b,x)= e^x \tfrac{(b-1)!}{x^{b-1}}\int_{0}^{x} e^{-t}\tfrac{(x-t)^{a-1}}{(a-1)!} \tfrac{t^{b-a-1}}{(b-a-1)!} dt=e^x \tfrac{(b-1)!}{x^{b-1}}D^{-a}[e^{-x}\tfrac{x^{b-a-1}}{(b-a-1)!}],$$

leading to

$$e^x {x^{-\alpha}}\tfrac{D^{\beta}}{\beta!}[x^{\beta+\alpha}e^{-x}]=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x).$$

Formally using the inverse Mellin transform for x>0:

$$e^x f(x\tfrac{d}{dx})e^{-x}=e^x f(x\tfrac{d}{dx}) \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} \frac{x^{-s}}{(-s)!} ds$$

$$=e^x \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} f(-s) \frac{x^{-s}}{(-s)!} ds.$$

Let $$f(x)=\binom{x+\alpha+\beta}{\beta},$$

then

$$e^x \binom{x\tfrac{d}{dx}+\alpha+\beta}{\beta}e^{-x}=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x)$$

where $L_{\beta}^{\alpha}(x)$ is the generalized Laguerre function and $K(-\beta,\alpha+1,x),$ Kummer's confluent hypergeometric function.

For an elaboration, see the notes The Inverse Mellin Transform, Bell Polynomials, a Generalized Dobinski Relation, and the Confluent Hypergeometric Functions.

See also Rodriguez-like formula (attributed to Poole, 1936) on pg. 59 of Bateman's (et al.) Higher Transcendental Functions Vol. I:

$$(x\tfrac{d}{dx}+\alpha)_{n} h(x) = x^{1-\alpha}D^{n}[x^{n+\alpha-1}h(x)],$$

with $D=\tfrac{d}{dx}$, leading to

$$e^x \binom{x\tfrac{d}{dx}+\alpha+n}{n}e^{-x}=e^x x^{-\alpha}\tfrac{D^{n}}{n!}[x^{n+\alpha}e^{-x}]=L_{n}^{\alpha}(x).$$

This can be generalized by using the fractional integro-derivative representation of $K(a,b,x)$ (see Eqn. 13.2.1 on pg. 505 of Abramowitz and Stegun):

$$K(a,b,x)= e^x \tfrac{(b-1)!}{x^{b-1}}\int_{0}^{x} e^{-t}\tfrac{(x-t)^{a-1}}{(a-1)!} \tfrac{t^{b-a-1}}{(b-a-1)!} dt=e^x \tfrac{(b-1)!}{x^{b-1}}D^{-a}[e^{-x}\tfrac{x^{b-a-1}}{(b-a-1)!}],$$

leading to

$$e^x {x^{-\alpha}}\tfrac{D^{\beta}}{\beta!}[x^{\beta+\alpha}e^{-x}]=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x).$$

Edit (Sept. 2019):

Abstracting, (as Jeans said) "leaving the operators hungry for something to differentiate":

$$\binom{x\tfrac{d}{dx}+\alpha+\beta}{\beta}={x^{-\alpha}}\frac{:Dx:^{\beta}}{\beta!}x^{\alpha}=L_{\beta}^{\alpha}(-:xD:)$$

$$=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,-:xD:),$$

where, by definition, for any two operators $:AB:^{\beta}=A^{\beta}B^{\beta}$.

Formally using the inverse Mellin transform for x>0:

$$e^x f(x\tfrac{d}{dx})e^{-x}=e^x f(x\tfrac{d}{dx}) \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} \frac{x^{-s}}{(-s)!} ds$$

$$=e^x \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} f(-s) \frac{x^{-s}}{(-s)!} ds.$$

Let $$f(x)=\binom{x+\alpha+\beta}{\beta},$$

then

$$e^x \binom{x\tfrac{d}{dx}+\alpha+\beta}{\beta}e^{-x}=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x)$$

where $L_{\beta}^{\alpha}(x)$ is the generalized Laguerre function and $K(-\beta,\alpha+1,x),$ Kummer's confluent hypergeometric function.

For an elaboration, see the notes The Inverse Mellin Transform, Bell Polynomials, a Generalized Dobinski Relation, and the Confluent Hypergeometric Functions.

See also Rodriguez-like formula on pg. 59 of Bateman's (et al.) Higher Transcendental Functions Vol. I:

$$(x\tfrac{d}{dx}+\alpha)_{n} h(x) = x^{1-\alpha}D^{n}[x^{n+\alpha-1}h(x)],$$

with $D=\tfrac{d}{dx}$, leading to

$$e^x \binom{x\tfrac{d}{dx}+\alpha+n}{n}e^{-x}=e^x x^{-\alpha}\tfrac{D^{n}}{n!}[x^{n+\alpha}e^{-x}]=L_{n}^{\alpha}(x).$$

This can be generalized by using the fractional integro-derivative representation of $K(a,b,x)$ (see Eqn. 13.2.1 on pg. 505 of Abramowitz and Stegun):

$$K(a,b,x)= e^x \tfrac{(b-1)!}{x^{b-1}}\int_{0}^{x} e^{-t}\tfrac{(x-t)^{a-1}}{(a-1)!} \tfrac{t^{b-a-1}}{(b-a-1)!} dt=e^x \tfrac{(b-1)!}{x^{b-1}}D^{-a}[e^{-x}\tfrac{x^{b-a-1}}{(b-a-1)!}],$$

leading to

$$e^x {x^{-\alpha}}\tfrac{D^{\beta}}{\beta!}[x^{\beta+\alpha}e^{-x}]=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x).$$

Formally using the inverse Mellin transform for x>0:

$$e^x f(x\tfrac{d}{dx})e^{-x}=e^x f(x\tfrac{d}{dx}) \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} \frac{x^{-s}}{(-s)!} ds$$

$$=e^x \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} f(-s) \frac{x^{-s}}{(-s)!} ds.$$

Let $$f(x)=\binom{x+\alpha+\beta}{\beta},$$

then

$$e^x \binom{x\tfrac{d}{dx}+\alpha+\beta}{\beta}e^{-x}=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x)$$

where $L_{\beta}^{\alpha}(x)$ is the generalized Laguerre function and $K(-\beta,\alpha+1,x),$ Kummer's confluent hypergeometric function.

For an elaboration, see the notes The Inverse Mellin Transform, Bell Polynomials, a Generalized Dobinski Relation, and the Confluent Hypergeometric Functions.

See also Rodriguez-like formula (attributed to Poole, 1936) on pg. 59 of Bateman's (et al.) Higher Transcendental Functions Vol. I:

$$(x\tfrac{d}{dx}+\alpha)_{n} h(x) = x^{1-\alpha}D^{n}[x^{n+\alpha-1}h(x)],$$

with $D=\tfrac{d}{dx}$, leading to

$$e^x \binom{x\tfrac{d}{dx}+\alpha+n}{n}e^{-x}=e^x x^{-\alpha}\tfrac{D^{n}}{n!}[x^{n+\alpha}e^{-x}]=L_{n}^{\alpha}(x).$$

This can be generalized by using the fractional integro-derivative representation of $K(a,b,x)$ (see Eqn. 13.2.1 on pg. 505 of Abramowitz and Stegun):

$$K(a,b,x)= e^x \tfrac{(b-1)!}{x^{b-1}}\int_{0}^{x} e^{-t}\tfrac{(x-t)^{a-1}}{(a-1)!} \tfrac{t^{b-a-1}}{(b-a-1)!} dt=e^x \tfrac{(b-1)!}{x^{b-1}}D^{-a}[e^{-x}\tfrac{x^{b-a-1}}{(b-a-1)!}],$$

leading to

$$e^x {x^{-\alpha}}\tfrac{D^{\beta}}{\beta!}[x^{\beta+\alpha}e^{-x}]=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x).$$