(1)

\[ \Gamma\left(a+1,x\right)=a\Gamma\left(a,x\right)+x^{a}e^{-x} \] より、
\begin{align*} \frac{x^{a}}{a!} & =\frac{e^{x}}{a!}\left(\Gamma\left(a+1,x\right)-a\Gamma\left(a,x\right)\right)\\ & =e^{x}\left(\frac{\Gamma\left(a+1,x\right)}{\Gamma\left(a+1\right)}-\frac{\Gamma\left(a,x\right)}{\Gamma\left(a\right)}\right) \end{align*}

(1)-2

\begin{align*} \frac{x^{a}}{a!} & =\frac{d}{dx}\int\frac{x^{a}}{a!}dx\\ & =\frac{d}{dx}\frac{x^{a+1}}{(a+1)!}\\ & =\frac{d}{dx}\left(\frac{x^{a}}{a!}+\left[\frac{x^{k}}{k!}\right]_{k=a}^{k=a+1}\right)\\ & =\frac{d}{dx}\LHS+\left[\frac{x^{k-1}}{\left(k-1\right)!}\right]_{k=a}^{k=a+1}\\ & =e^{x}\left(1+\frac{d}{dx}\right)e^{-x}\LHS+\left[\frac{x^{k-1}}{\left(k-1\right)!}\right]_{k=a}^{k=a+1}\\ & =-e^{x}\int_{y}^{x}e^{-x}\left[\frac{x^{k-1}}{\left(k-1\right)!}\right]_{k=a}^{k=a+1}dx+\LHS\left(x\rightarrow y\right)\cnd{y=\begin{cases} 0 & a\geq0\\ \infty & a<0 \end{cases}}\\ & =-e^{x}\left[\left(\int_{\infty}^{x}+\int_{y}^{\infty}\right)e^{-x}\frac{x^{k-1}}{\left(k-1\right)!}dx\right]_{k=a}^{k=a+1}\\ & =-e^{x}\left[\frac{1}{\Gamma\left(k\right)}\left(\int_{\infty}^{x}+\int_{y}^{\infty}\right)e^{-x}x^{k-1}dx\right]_{k=a}^{k=a+1}\\ & =-e^{x}\left[\frac{1}{\Gamma\left(k\right)}\left(-\Gamma\left(k,x\right)+\Gamma\left(k\right)\theta\left(-y+1.0\right)\right)\right]_{k=a}^{k=a+1}\\ & =e^{x}\left(\frac{\Gamma\left(a+1,x\right)}{\Gamma\left(a+1\right)}-\frac{\Gamma\left(a,x\right)}{\Gamma\left(a\right)}\right) \end{align*}

(2)

\(\Re\left(a\right)>0\)のときの証明
\begin{align*} \sum_{k=0}^{n}\frac{x^{k+a}}{\left(k+a\right)!} & =\frac{d}{dx}\sum_{k=0}^{n}\int\frac{x^{k+a}}{\left(k+a\right)!}dx\\ & =\frac{d}{dx}\sum_{k=0}^{n}\frac{x^{k+a+1}}{\left(k+a+1\right)!}\\ & =\frac{d}{dx}\sum_{k=1}^{n+1}\frac{x^{k+a}}{\left(k+a\right)!}\\ & =\frac{d}{dx}\sum_{k=0}^{n}\frac{x^{k+a}}{\left(k+a\right)!}+\frac{d}{dx}\left[\frac{x^{k+a}}{\left(k+a\right)!}\right]_{k=0}^{k=n+1}\\ & =\frac{d}{dx}\sum_{k=0}^{n}\frac{x^{k+a}}{\left(k+a\right)!}+\left[\frac{x^{k+a-1}}{\left(k+a-1\right)!}\right]_{k=0}^{k=n+1}\\ & =e^{x}\left(\left(1+\frac{d}{dx}\right)e^{-x}\sum_{k=0}^{\infty}\frac{x^{k+a}}{\left(k+a\right)!}+e^{-x}\left[\frac{x^{k+a-1}}{\left(k+a-1\right)!}\right]_{k=0}^{k=n+1}\right)\\ & =-e^{x}\int_{0}^{x}\left[\frac{x^{k+a-1}}{\left(k+a-1\right)!}\right]_{k=0}^{k=n+1}e^{-x}dx+\left[\sum_{k=0}^{n}\frac{x^{k+a}}{\left(k+a\right)!}\right]_{x=0}\\ & =-e^{x}\left[\frac{1}{\Gamma\left(k+a\right)}\int_{0}^{x}x^{k+a-1}e^{-x}dx\right]_{k=0}^{k=n+1}\\ & =-e^{x}\left[\frac{\gamma\left(k+a,x\right)}{\Gamma\left(k+a\right)}\right]_{k=0}^{k=n+1}\\ & =-e^{x}\left[\frac{\Gamma\left(k+a\right)-\Gamma\left(k+a,x\right)}{\Gamma\left(k+a\right)}\right]_{k=0}^{k=n+1}\\ & =e^{x}\left[\frac{\Gamma\left(k+a,x\right)}{\Gamma\left(k+a\right)}\right]_{k=0}^{k=n+1}\\ & =e^{x}\left(\frac{\Gamma\left(n+a+1,x\right)}{\Gamma\left(n+a+1\right)}-\frac{\Gamma\left(a,x\right)}{\Gamma\left(a\right)}\right) \end{align*}

(3)

\begin{align*} \sum_{k=0}^{\infty}\frac{x^{k+a}}{\left(k+a\right)!} & =\lim_{n\rightarrow\infty}\sum_{k=0}^{n}\frac{x^{k+a}}{\left(k+a\right)!}\\ & =\lim_{n\rightarrow\infty}e^{x}\left(\frac{\Gamma\left(n+a+1,x\right)}{\Gamma\left(n+a+1\right)}-\frac{\Gamma\left(a,x\right)}{\Gamma\left(a\right)}\right)\\ & =\lim_{n\rightarrow\infty}e^{x}\left(1-\frac{\gamma\left(n+a+1,x\right)}{\Gamma\left(n+a+1\right)}-\frac{\Gamma\left(a,x\right)}{\Gamma\left(a\right)}\right)\\ & =e^{x}\left(1-\lim_{n\rightarrow\infty}\frac{1}{\Gamma\left(n+a+1\right)}\int_{0}^{x}t^{n+a}e^{-t}dt-\frac{\Gamma\left(a,x\right)}{\Gamma\left(a\right)}\right)\\ & =e^{x}\left(1-\int_{0}^{x}\lim_{n\rightarrow\infty}\frac{t^{n+a}}{\left(n+a\right)!}e^{-t}dt-\frac{\Gamma\left(a,x\right)}{\Gamma\left(a\right)}\right)\\ & =e^{x}\left(1-\frac{\Gamma\left(a,x\right)}{\Gamma\left(a\right)}\right) \end{align*}

(4)

\begin{align*} a!x^{a} & =xa!x^{a-1}\\ & =x\frac{d}{dx}\left(a-1\right)!x^{a}\\ & =x\frac{d}{dx}x\left(a-1\right)!x^{a-1}\\ & =x\frac{d}{dx}x\left(\LHS-\left[k!x^{k}\right]_{k=a-1}^{k=a}\right)\\ & =x\LHS+x^{2}\frac{d}{dx}\LHS-\left[\left(k+1\right)!x^{k+1}\right]_{k=a-1}^{k=a}\\ & =\frac{x^{2}}{1-x}\frac{d}{dx}\LHS-\frac{1}{1-x}\left[\left(k+1\right)!x^{k+1}\right]_{k=a-1}^{k=a}\\ & =e^{\int_{\infty}^{x}\frac{1-x}{x^{2}}dx}\left(1+\frac{x^{2}}{1-x}\frac{d}{dx}\right)e^{-\int_{\infty}^{x}\frac{1-x^{2}}{x}dx}\LHS-\frac{1}{1-x}\left[\left(k+1\right)!x^{k+1}\right]_{k=a-1}^{k=a}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\left(1+\frac{x^{2}}{1-x}\frac{d}{dx}\right)xe^{\frac{1}{x}}\LHS-\frac{1}{1-x}\left[\left(k+1\right)!x^{k+1}\right]_{k=a-1}^{k=a}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\int_{y}^{x}xe^{\frac{1}{x}}\frac{1-x}{x^{2}}\frac{1}{1-x}\left[\left(k+1\right)!x^{k+1}\right]_{k=a-1}^{k=a}dx+\LHS\left(x\rightarrow y\right)\cnd{y=\begin{cases} -0 & a\geq0\\ \infty & a<0 \end{cases}}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\left[\left(k+1\right)!\int_{y}^{x}e^{\frac{1}{x}}x^{k}dx\right]_{k=a-1}^{k=a}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\left[\left(-1\right)^{k}\left(k+1\right)!\int_{y}^{x}e^{-\frac{-1}{x}}\left(\frac{-1}{x}\right)^{-k}\left(-\frac{1}{x}\right)^{-2}\left(\frac{1}{x^{2}}\right)dx\right]_{k=a-1}^{k=a}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\left[\left(-1\right)^{k}\left(k+1\right)!\left(\int_{\infty}^{\frac{-1}{x}}+\int_{\frac{-1}{y}}^{\infty}\right)e^{-\frac{-1}{x}}\left(\frac{-1}{x}\right)^{-k-2}d\left(\frac{-1}{x}\right)\right]_{k=a-1}^{k=a}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\left[\left(-1\right)^{k}\left(k+1\right)!\left(\Gamma\left(-k-1,-\frac{1}{x}\right)+\Gamma\left(-k-1\right)\theta\left(\frac{1}{y}+1.0\right)\right)\right]_{k=a-1}^{k=a}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\left[\left(-1\right)^{k}\left(k+1\right)!\Gamma\left(-k-1,-\frac{1}{x}\right)+\left(-1\right)^{k}\frac{\pi}{\sin\left(\pi k\right)}\theta\left(\frac{1}{y}+1.0\right)\right]_{k=a-1}^{k=a}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\left[\left(-1\right)^{k}\left(k+1\right)!\Gamma\left(-\left(k+1\right),-\frac{1}{x}\right)\right]_{k=a-1}^{k=a}\\ & =-\frac{1}{x}e^{-\frac{1}{x}}\left[\left(-1\right)^{a+1}\left(a+1\right)!\Gamma\left(-\left(a+1\right),-\frac{1}{x}\right)-\left(-1\right)^{a}a!\Gamma\left(-a,-\frac{1}{x}\right)\right] \end{align*}

(5)

\(\Re\left(a\right)>0\)のときの証明
\begin{align*} \sum_{k=0}^{n}\left(k+a\right)!x^{k+a} & =x\sum_{k=0}^{n}\left(k+a\right)!x^{k+a-1}\\ & =x\frac{d}{dx}\sum_{k=0}^{n}\left(k+a-1\right)!x^{k+a}\\ & =x\frac{d}{dx}x\sum_{k=0}^{n}\left(k+a-1\right)!x^{k+a-1}\\ & =x\frac{d}{dx}x\sum_{k=-1}^{n-1}\left(k+a\right)!x^{k+a}\\ & =x\frac{d}{dx}x\left(\sum_{k=0}^{n}\left(k+a\right)!x^{k+a}-\left[\left(k+a\right)!x^{k+a}\right]_{k=-1}^{k=n}\right)\\ & =x\sum_{k=0}^{n}\left(k+a\right)!x^{k+a}+x^{2}\frac{d}{dx}\sum_{k=0}^{n}\left(k+a\right)!x^{k+a}-\left[\left(k+a+1\right)!x^{k+a+1}\right]_{k=-1}^{k=n}\\ & =\frac{x^{2}}{1-x}\frac{d}{dx}\sum_{k=0}^{n}\left(k+a\right)!x^{k+a}-\frac{1}{1-x}\left[\left(k+a+1\right)!x^{k+a+1}\right]_{k=-1}^{k=n}\\ & =e^{\int_{0}^{x}\frac{1-x}{x^{2}}dx}\left(\frac{x^{2}}{1-x}\frac{d}{dx}+1\right)e^{-\int_{0}^{x}\frac{1-x}{x^{2}}dx}\sum_{k=0}^{n}\left(k+a\right)!x^{k+a}-\frac{1}{1-x}\left[\left(k+a+1\right)!x^{k+a+1}\right]_{k=-1}^{k=n}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\left(\frac{x^{2}}{1-x}\frac{d}{dx}+1\right)xe^{\frac{1}{x}}\sum_{k=0}^{n}\left(k+a\right)!x^{k+a}-\frac{1}{1-x}\left[\left(k+a+1\right)!x^{k+a+1}\right]_{k=-1}^{k=n}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\int_{0}^{x}\frac{1-x}{x^{2}}\frac{1}{1-x}xe^{\frac{1}{x}}\left[\left(k+a+1\right)!x^{k+a+1}\right]_{k=-1}^{k=n}dx+\left[\sum_{k=0}^{n}\left(k+a\right)!x^{k+a}\right]_{x=0}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\left[\left(k+a+1\right)!\int_{0}^{x}x^{k+a}e^{\frac{1}{x}}dx\right]_{k=-1}^{k=n}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\left[\left(k+a+1\right)!\int_{0}^{x}\left(\frac{1}{x}\right)^{-\left(k+a+2\right)}e^{\frac{1}{x}}\frac{-1}{x^{2}}dx\right]_{k=-1}^{k=n}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\left[\left(k+a+1\right)!\left(-1\right)^{k+a+2}\int_{0}^{x}\left(-\frac{1}{x}\right)^{-\left(k+a+2\right)}e^{-\frac{-1}{x}}d\frac{-1}{x}\right]_{k=-1}^{k=n}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\left[\left(-1\right)^{k+a}\left(k+a+1\right)!\gamma\left(-\left(k+a+1\right),-\frac{1}{x}\right)\right]_{k=-1}^{k=n}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\left[\left(-1\right)^{k+a}\left(k+a+1\right)!\left\{ \Gamma\left(-\left(k+a+1\right)\right)-\Gamma\left(-\left(k+a+1\right),-\frac{1}{x}\right)\right\} \right]_{k=-1}^{k=n}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\left[\left(-1\right)^{k+a}\left\{ \Gamma\left(1+k+a+1\right)!\Gamma\left(-\left(k+a+1\right)\right)-\left(k+a+1\right)!\Gamma\left(-\left(k+a+1\right),-\frac{1}{x}\right)\right\} \right]_{k=-1}^{k=n}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\left[\left(-1\right)^{k+a}\left\{ \frac{\pi}{\sin\left(-\left(k+a+1\right)\pi\right)}-\left(k+a+1\right)!\Gamma\left(-\left(k+a+1\right),-\frac{1}{x}\right)\right\} \right]_{k=-1}^{k=n}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\left[\left(-1\right)^{k+a}\left\{ \frac{-\left(-1\right)^{k+1}\pi}{\sin\left(a\pi\right)}-\left(k+a+1\right)!\Gamma\left(-\left(k+a+1\right),-\frac{1}{x}\right)\right\} \right]_{k=-1}^{k=n}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\left[\frac{\left(-1\right)^{a}\pi}{\sin\left(a\pi\right)}-\left(-1\right)^{k+a}\left(k+a+1\right)!\Gamma\left(-\left(k+a+1\right),-\frac{1}{x}\right)\right]_{k=-1}^{k=n}\\ & =\frac{1}{x}e^{-\frac{1}{x}}\left\{ \left(-1\right)^{n+a+1}\left(n+a+1\right)!\Gamma\left(-\left(n+a+1\right),-\frac{1}{x}\right)-\left(-1\right)^{a}a!\Gamma\left(-a,-\frac{1}{x}\right)\right\} \end{align*}

(6)

ダランベールの収束判定法より、
\begin{align*} \lim_{k\rightarrow\infty}\frac{\left(k+1+a\right)!x^{k+1+a}}{\left(k+a\right)!x^{k+a}} & =\lim_{k\rightarrow\infty}\left(k+1+a\right)x \end{align*} となるので\(x\ne0\)のとき発散する。

(6)-2

\(x\not=0\)のとき、
\begin{align*} \lim_{n\rightarrow\infty}\sum_{k=0}^{n}\left(k+a\right)!x^{k+a} & =\lim_{n\rightarrow\infty}\frac{1}{x}e^{-\frac{1}{x}}\left(-1\right)^{a+1}\left\{ a!\Gamma\left(-a,-\frac{1}{x}\right)+\left(-1\right)^{n}\left(n+a+1\right)!\Gamma\left(-\left(n+a+1\right),-\frac{1}{x}\right)\right\} \\ & =\frac{1}{x}e^{-\frac{1}{x}}\left(-1\right)^{a+1}\left\{ a!\Gamma\left(-a,-\frac{1}{x}\right)+\lim_{n\rightarrow\infty}\left(-1\right)^{n}\left(n+a+1\right)!\int_{-\frac{1}{x}}^{\infty}t^{-\left(n+a+1\right)-1}e^{-t}dt\right\} \\ & =\frac{1}{x}e^{-\frac{1}{x}}\left(-1\right)^{a+1}\left\{ a!\Gamma\left(-a,-\frac{1}{x}\right)+\left(-1\right)^{a+1}\lim_{n\rightarrow\infty}\int_{-\frac{1}{x}}^{\infty}\frac{\left(n+a+1\right)!}{\left(-t\right)^{n+a+1}}\frac{e^{-t}}{\left(-t\right)}d\left(-t\right)\right\} \end{align*} となるので発散。