Example 3.3.1 Gamma function

is of $C^{\mathrm{\infty }}$ with respect to $\alpha >0$.

Let $f(x,\alpha )={\text{e}}^{-x}{x}^{\alpha -1}$, then

If $\frac{{\partial }^{k}f}{\partial {\alpha }^k}$ is uniformly convergent with respect to $\alpha$ in a certain neighborhood $U({\alpha }_0)$, then

We use a strong Weierstrass function (with no respect to $\alpha$) to control the integrand. Consider $$\begin{array}{rl}\text{I. }& {\int }_0^1{\text{e}}^{-x}{x}^{\alpha -1}(\ln x{)}^k\text{d}x\\ \text{II. }& {\int }_1^{+\mathrm{\infty }}{\text{e}}^{-x}{x}^{\alpha -1}(\ln x{)}^k\text{d}x\end{array}$$

• Take any ${\delta }_0\in (0,1)$, it could be proved by L’Hôpital that

  

  So $x^{{\delta }_0}(\ln x{)}^k$ is bounded on $(0,1]$, meaning $\mathrm{\exists }{M}_1{\delta }_0>0$ such that $|x^{{\delta }_0}(\ln x{)}^k|\le M_1$. Hence, for $\mathrm{\forall }\alpha \in [2{\delta }_0,+\mathrm{\infty })$,

  

  According to Weierstrass Criterion, I is uniformly convergent for $\alpha \in [2{\delta }_0,+\mathrm{\infty })$.

• Take any ${\delta }_0\in (0,1)$, for $\mathrm{\forall }\alpha \in (0,1+\frac{1}{{\delta }_0}]$,

  

  According to Weierstrass Criterion, II is uniformly convergent for $\alpha \in (0,1+\frac{1}{{\delta }_0}]$.

Therefore, $\frac{{\partial }^{k}f}{\partial {\alpha }^k}$ is uniformly convergent with respect to $\alpha \in [2{\delta }_0,1+\frac{1}{{\delta }_0}]$, meaning $\mathrm{\Gamma }(\alpha )$ is $k^{\text{th}}$-order derivable in $(2{\delta }_0,1+\frac{1}{{\delta }_0})$.

Example 3.3.2 Beta function $B(\alpha ,\beta )$.

is of $C^{\mathrm{\infty }}$ with respect to $(\alpha ,\beta )$ and

When $m,n\in {\mathbb{N}}^{\ast }$,

Let

We need to prove that ${\int }_0^1{f}_{k,l}(x)\text{d}x$ is uniformly convergent with respect to $(\alpha ,\beta )$ in any bounded closed set of $(0,+\mathrm{\infty })\times (0,+\mathrm{\infty })$.

Take any ${\delta }_0\in (0,1)$, $\mathrm{\forall }(\alpha ,\beta )\in [2{\delta }_0,\frac{2}{{\delta }_0}]\times [2{\delta }_0,\frac{2}{{\delta }_0}]$, $\mathrm{\exists }M>0$ such that

Therefore,

• $x\in (0,\frac{1}{2}]$,

  

  and ${\int }_0^{\frac{1}{2}}\frac{\text{d}x}{x^{1-{\delta }_0}}$ is convergent.

• $x\in [\frac{1}{2},1)$,

  

  and ${\int }_{\frac{1}{2}}^1\frac{\text{d}x}{(1-x{)}^{1-{\delta }_0}}$ is convergent.

According to Weierstrass Criterion, ${\int }_0^1{f}_{k,l}(x)\text{d}x$ is uniformly convergent with respect to $(\alpha ,\beta )\in [2{\delta }_0,\frac{2}{{\delta }_0}]\times [2{\delta }_0,\frac{2}{{\delta }_0}]$, meaning $B(\alpha ,\beta )$ is $k^{\text{th}}$-order derivable with respect to $\alpha$ and $l^{\text{th}}$-order derivable with respect to $\beta$ in $(2{\delta }_0,\frac{2}{{\delta }_0})\times (2{\delta }_0,\frac{2}{{\delta }_0})$.

Prove the relation between Gamma function and Beta function. $$\begin{array}{rl}B(\alpha ,\beta )& ={\int }_0^1{x}^{\alpha -1}(1-x{)}^{\beta -1}\text{d}x\underset{y\in (0,+\mathrm{\infty })}{\overset{x=1/(y+1)}{\to }}{\int }_0^{+\mathrm{\infty }}\frac{y^{\beta -1}}{(1+y{)}^{\alpha +\beta }}\text{d}y\\ \mathrm{\Gamma }(\alpha )& ={\int }_0^{+\mathrm{\infty }}{t}^{\alpha -1}{\text{e}}^{-t}\text{d}t\underset{y\text{ is const}}{\overset{t=xy}{\to }}{y}^{\alpha }{\int }_0^{+\mathrm{\infty }}{x}^{\alpha -1}{\text{e}}^{-xy}\text{d}x\\ \mathrm{\Gamma }(\alpha +\beta )& =(1+y{)}^{\alpha +\beta }{\int }_0^{+\mathrm{\infty }}{x}^{\alpha +\beta -1}{\text{e}}^{-x(1+y)}\text{d}x\\ \mathrm{\Gamma }(\alpha +\beta )B(\alpha ,\beta )& ={\int }_0^{+\mathrm{\infty }}\frac{\mathrm{\Gamma }(\alpha +\beta )y^{\beta -1}}{(1+y{)}^{\alpha +\beta }}\text{d}y={\int }_0^{+\mathrm{\infty }}{\int }_0^{+\mathrm{\infty }}{x}^{\alpha +\beta -1}{y}^{\beta -1}{\text{e}}^{-x(1+y)}\text{d}x\text{d}y\\ & ={\int }_0^{+\mathrm{\infty }}{x}^{\alpha +\beta -1}{\text{e}}^{-x}\left[{\int }_0^{+\mathrm{\infty }}{y}^{\beta -1}{\text{e}}^{-xy}\text{d}y\right]\text{d}x\stackrel{t=xy}{\to }\\ & ={\int }_0^{+\mathrm{\infty }}{x}^{\alpha -1}{\text{e}}^{-x}\left[{\int }_0^{+\mathrm{\infty }}{t}^{\beta -1}{\text{e}}^{-t}\text{d}t\right]\text{d}x\\ & ={\int }_0^{+\mathrm{\infty }}{x}^{\alpha -1}{\text{e}}^{-x}\text{d}x{\int }_0^{+\mathrm{\infty }}{t}^{\beta -1}{\text{e}}^{-t}\text{d}t=\mathrm{\Gamma }(\alpha )\mathrm{\Gamma }(\beta )\end{array}$$

Example 3.3.3 Dirichlet function $D(\alpha )$.

Let

Take any $\delta >0$, then for any $y\in [\delta ,+\mathrm{\infty })$, we have

convergent. According to Weierstrass Criterion, $g$ is uniformly convergent with respect to $y\in [\delta ,+\mathrm{\infty })$.

is uniformly consistent with respect to $y\in [\delta ,+\mathrm{\infty })$, meaning $g$ is derivable in $(0,+\mathrm{\infty })$. $$\begin{array}{rl}{g}^{\prime }(y)& =-{\int }_0^{+\mathrm{\infty }}{\text{e}}^{-yx}\sin x\text{d}x={\int }_0^{+\mathrm{\infty }}{\text{e}}^{-yx}\text{d}\cos x\\ & ={{\text{e}}^{-yx}\cos x|}_0^{+\mathrm{\infty }}-{\int }_0^{+\mathrm{\infty }}\cos x{\text{d}}_x{\text{e}}^{-yx}=-1+y{\int }_0^{+\mathrm{\infty }}{\text{e}}^{-yx}\cos x\text{d}x\\ & =-1+{\int }_0^{+\mathrm{\infty }}{\text{e}}^{-yx}\text{d}\sin x=-1+(y{\text{e}}^{-yx}\sin x{)}_0^{+\mathrm{\infty }}+y^2{\int }_0^{+\mathrm{\infty }}{\text{e}}^{-yx}\sin x\text{d}x\\ g^{\prime }(y)& =-{\int }_0^{+\mathrm{\infty }}{\text{e}}^{-yx}\sin x\text{d}x=-\frac{1}{y^2+1}\Rightarrow g(y)=-\arctan y+C\\ \underset{y\to +\mathrm{\infty }}{lim}g(y)& =\underset{y\to +\mathrm{\infty }}{lim}{\int }_0^{+\mathrm{\infty }}{\text{e}}^{-yx}\frac{\sin x}{x}\text{d}x={\int }_0^{+\mathrm{\infty }}\left(\underset{y\to +\mathrm{\infty }}{lim}{\text{e}}^{-yx}\frac{\sin x}{x}\right)\text{d}x=0\\ \underset{y\to +\mathrm{\infty }}{lim}g(y)& =\underset{y\to +\mathrm{\infty }}{lim}(-\arctan y+C)=C-\frac{\pi }{2}=0\Rightarrow C=\frac{\pi }{2}\\ g(0)& ={\int }_0^{+\mathrm{\infty }}\frac{\sin x}{x}\text{d}x=\frac{\pi }{2}\end{array}$$