Let \(f\) be an entire function. If there exist \(\rho\in\mathbb R_+\) and constants \(A, B\in\mathbb R_+\) such that \[|\:f(z)| \leq A e^{B|z|^{\rho}} \quad \text { for all } \quad z \in \mathbb{C},\] then we say that \(f\) has an order of growth \(\leq \rho\).
We define the order of growth of \(f\) as \[\rho_{f}=\inf \rho,\] where the infimum is taken over all \(\rho>0\) such that \(f\) has an order of growth \(\leq \rho\).
For example, the order of growth of the function \(e^{z^{2}}\) is \(2\).
Suppose \(f\) is entire and has growth order \(\rho_{0}\). Let \(k\in\mathbb Z\) be so that \(k \leq \rho_{0}<k+1\). If \(a_{1}, a_{2}, \ldots\) denote the (non-zero) zeros of \(f\), then \[f(z)=e^{P(z)} z^{m} \prod_{n=1}^{\infty} E_{k}\left(z / a_{n}\right),\] where \(P\) is a polynomial of degree \(\leq k\), and \(m\) is the order of the zero of \(f\) at \(z=0\), and \(E_k\) are the canonical factors for \(k\in\mathbb N\).
The function \(\sin \pi z\) is entire and of order one, and its zeros are at \(z=0, \pm 1, \pm 2, \ldots\), and so, by Hadamard’s theorem we can write \[\sin \pi z=z e^{H(z)} \prod_{n=1}^{\infty}\left(1-\frac{z^{2}}{n^{2}}\right),\] where \(H(z)=a z+b\).
Taking the logarithmic derivative of this equation, we find that \[\pi \frac{\cos \pi z}{\sin \pi z}=\frac{1}{z}+H^{\prime}(z)-\sum_{n=1}^{\infty} \frac{2 z}{n^{2}-z^{2}} .\]
Passage to the limit as \(z \rightarrow 0\) gives \(a=0\), and so \(H(z)=b\). Thus, \[\frac{\sin \pi z}{z}=c \prod_{n=1}^{\infty}\left(1-\frac{z^{2}}{n^{2}}\right) .\]
Passing again to the limit as \(z \rightarrow 0\) gives \(c=\pi\), i.e. \[\sin \pi z=\pi z \prod_{n=1}^{\infty}\left(1-\frac{z^{2}}{n^{2}}\right).\]
Equivalently, we have \[\frac{\sin \pi z}{\pi z}= \prod_{n=1}^{\infty}\left(1-\frac{z^{2}}{n^{2}}\right).\]
The Euler gamma function \(\Gamma(z)\) is defined by the equation \[\frac{1}{\Gamma(z)}=z e^{\gamma z} \prod_{n=1}^{\infty}\left(1+\frac{z}{n}\right) e^{-z / n}\] where \(\gamma\) is Euler’s constant.
It follows from the definition that \(\Gamma^{-1}(z)\) is an entire function of order one. Prove it! In, fact one can show that there are \(A, B\in\mathbb R_+\) so that \[\frac{1}{|\Gamma(z)|}\le Ae^{B|z|\log|z|}.\]
Moreover, \(\Gamma(z)\) is an analytic function in the entire \(\mathbb C\) except for the points \(s=0,-1,-2, \ldots\), where it has simple poles.
For every \(z\in\mathbb C\setminus\{-n: n\in\mathbb N\}\), we have \[\Gamma(z)=\frac{1}{z} \prod_{n=1}^{\infty}\left(1+\frac{1}{n}\right)^{z}\left(1+\frac{z}{n}\right)^{-1}.\] In other words, \(\Gamma(z)\) is a meromorphic function on \(\mathbb C\) with simple poles at \(0\) and at the negative integers and with no zeros.
From the definition of an infinite product and from the definition of the function \(\Gamma(z)\), we obtain \[\begin{aligned} \frac{1}{\Gamma(z)} & =z \lim _{m \rightarrow \infty} e^{z\left(1+\frac{1}{2}+\ldots+\frac{1}{m}-\log m\right)} \cdot \lim _{m \rightarrow \infty} \prod_{n=1}^{m}\left(1+\frac{z}{n}\right) e^{-\frac{z}{n}} \\ & =z \lim _{m \rightarrow \infty} m^{-z} \prod_{n=1}^{m}\left(1+\frac{z}{n}\right)\\ &=z \lim _{m \rightarrow \infty} \prod_{n=1}^{m-1}\left(1+\frac{1}{n}\right)^{-z} \prod_{n=1}^{m}\left(1+\frac{z}{n}\right) \\ & =z \lim _{m \rightarrow \infty} \prod_{n=1}^{m}\left(1+\frac{1}{n}\right)^{-z}\left(1+\frac{z}{n}\right)\left(1+\frac{1}{m}\right)^{z} \\ & =z \prod_{n=1}^{\infty}\left(1+\frac{1}{n}\right)^{-z}\left(1+\frac{z}{n}\right). \qquad \end{aligned} \tag*{$\blacksquare$}\]
For every \(z\in\mathbb C\setminus\{-n: n\in\mathbb N\}\), we have \[\Gamma(z)=\lim _{n \rightarrow \infty} \frac{(n-1)! \cdot n^{z}}{z(z+1) \cdot \ldots \cdot (z+n-1)}.\]
Proof: From the previous theorem we have \[\begin{aligned} \Gamma(z)&=\lim_{n\to \infty}z^{-1} \prod_{m=1}^{n-1}\left(1+\frac{1}{m}\right)^{z}\left(1+\frac{z}{m}\right)^{-1}\\ &=\lim_{n\to \infty}\frac{2^z\cdot\frac{3^z}{2^z}\cdot\ldots\cdot\frac{n^z}{(n-1)^z}}{z\cdot \frac{(z+1)}{1}\cdot\ldots\cdot \frac{(z+n-1)}{n-1}} =\lim_{n\to \infty}\frac{1\cdot 2\cdot\ldots\cdot (n-1)n^z}{z\cdot (z+1)\cdot\ldots\cdot (z+n-1)}. \end{aligned}\tag*{$\blacksquare$}\]
We also have \(\Gamma(1)=\Gamma(2)=1\).
We have \(\Gamma(z+1)=z \Gamma(z)\) for all \(z\in\mathbb C\setminus\{-n: n\in\mathbb N\}\).
In particular, \(\Gamma(n+1)=n!\) for all \(n\in\mathbb N\), and \({\rm res}_{z=-m}\Gamma(z)=\frac{(-1)^m}{m!}\).
Proof: We have \[\begin{aligned} \frac{\Gamma(z+1)}{\Gamma(z)}&=\frac{z}{z+1} \lim _{m \rightarrow \infty} \prod_{n=1}^{m} \frac{\left(1+\frac{1}{n}\right)^{z+1}\left(1+\frac{z+1}{n}\right)^{-1}}{\left(1+\frac{1}{n}\right)^{z}\left(1+\frac{z}{n}\right)^{-1}} \\ & =\frac{z}{z+1} \lim _{m \rightarrow \infty} \prod_{n=1}^{m} \frac{n+1}{n} \cdot \frac{n+z}{n+z+1}\\ &=\frac{z}{z+1} \lim _{m \rightarrow \infty} \frac{(m+1)(z+1)}{m+1+z}=z . \end{aligned}\] This completes the proof. $$\tag*{$\blacksquare$}$$
\[\Gamma(2z)\Gamma\left(1/2\right)= 2^{2z-1} \Gamma\left(z\right) \Gamma\left(z+1/2\right) \quad \text{ for all }\quad z\in\mathbb C\setminus(-\mathbb N).\]
\[\frac{\sin \pi z}{\pi}=\frac{1}{\Gamma(z) \Gamma(1-z)} \quad \text{ for all }\quad z\in\mathbb C.\]
Proof: We know that \(\frac{\sin \pi z}{\pi z}= \prod_{n=1}^{\infty}\left(1-\frac{z^{2}}{n^{2}}\right)\). On the other hand, \[\frac{1}{\Gamma(z)\Gamma(-z)}=-z^2\prod_{n=1}^{\infty}\left(1-\frac{z^{2}}{n^{2}}\right)\] But we also know that \(\Gamma(1-z)=-z\Gamma(-z)\), and the result follows.
As a corollary we obtain that \(\Gamma(1/2)=\sqrt{\pi}\).
Suppose that \(\operatorname{Re}(z)>0\). Then \[\begin{aligned} \Gamma(s)=\int_{0}^{\infty} e^{-t} t^{z-1} d t. \end{aligned}\]
Proof: We know that \[\Gamma(s)=\lim _{n \rightarrow \infty} \frac{n!\cdot n^{z}}{z(z+1)(z+2) \cdots(z+n)}.\]
We have to establish two things. Firstly, we will show that \[\int_{0}^{n}\left(1-\frac{t}{n}\right)^{n} t^{z-1} d t=\frac{n!\cdot n^{z}}{z(z+1)\cdot \ldots \cdot (z+n)} \quad \text{ for all }\quad n\in\mathbb Z_+.\]
Secondly, we will show that
\[\lim _{n \rightarrow \infty} \int_{0}^{n}\left(1-\frac{t}{n}\right)^{n} t^{z-1} d t=\int_{0}^{\infty} e^{-t} t^{z-1} d t.\]
Indeed, when \(s>0\) the above integral converges and we have \[\begin{aligned} \int_{0}^{n}\left(1-\frac{t}{n}\right)^{n} t^{z-1} d t & =n^{z} \int_{0}^{1}(1-u)^{n} u^{z-1} d u =n^{z} \frac{n}{z} \int_{0}^{1}(1-u)^{n-1} u^{z} d u \\ & =n^{z} \frac{n(n-1)}{z(z+1)} \int_{0}^{1}(1-u)^{n-2} u^{z+1} d u \\ & \hspace{2cm} \vdots \\ & =n^{s} \frac{n(n-1) \cdot \ldots\cdot 1}{z(z+1) \cdot \ldots\cdot (z+n-1)} \int_{0}^{1} u^{z+n-1} d u \\ & =\frac{n!\cdot n^{z}}{z(z+1)(z+2) \cdot \ldots\cdot (z+n)} . \end{aligned}\]
Thus, it suffices to prove that \[\lim _{n \rightarrow \infty} \int_{0}^{n}\left(1-\frac{t}{n}\right)^{n} t^{z-1} d t=\int_{0}^{\infty} e^{-t} t^{z-1} d t.\]
To this end, we consider the functions \[f_{n}(t)= \begin{cases}(1-t / n)^{n} t^{z-1} & \text { if } 0 \leq t \leq n, \\ 0 & \text { if } t>n .\end{cases}\]
Each of these functions is in \(L^{1}([0, \infty))\) and satisfies the inequality \[\left|\: f_{n}(t)\right| \leq e^{-t} t^{\sigma-1}, \quad \text{ where } \quad \sigma=\operatorname{Re}(z).\]
The last inequality is easily verified by taking logarithms and noting \[n \log \left(1-\frac{t}{n}\right)=-t-\frac{t^{2}}{2 n}-\frac{t^{3}}{3 n^{2}}-\cdots<-t .\]
Furthermore, \[\lim _{n \rightarrow \infty} f_{n}(t)=t^{s-1} \lim _{n \rightarrow \infty}\left(1-\frac{t}{n}\right)^{n}=e^{-t} t^{s-1} .\]
Since the function \(e^{-t} t^{\sigma-1}\) is in \(L^{1}([0, \infty))\), the dominated convergence theorem yields \[\lim _{n \rightarrow \infty} \int_{0}^{\infty} f_{n}(t) d t =\int_{0}^{\infty} \lim _{n \rightarrow \infty} f_{n}(t) d t =\int_{0}^{\infty} e^{-t} t^{s-1} d t,\] which completes the proof of the lemma.$$\tag*{$\blacksquare$}$$
For a complex number \(x\in\mathbb C\), we observe that \[u=u(x, z):=\frac{z e^{x z}}{e^{z}-1}\] is analytic in \(|z|<2 \pi\).
Therefore, it has power series expansion around \(z=0\) given by \[u=u(x, z)=\sum_{k=0}^{\infty} \frac{B_{k}(x)}{k!} z^{k} \quad \text { in } \quad |z|<2 \pi,\] where \(B_{k}(x)\) are polynomials in the variable \(x\in\mathbb C\) with rational coefficients, known as the Bernoulli polynomials.
Further \(B_{k}=B_{k}(0)\in\mathbb Q\) are called the Bernoulli numbers given by \[\frac{z}{e^{z}-1}=\sum_{k=0}^{\infty} \frac{B_{k}}{k!} z^{k}\] derived from the above with \(x=0\).
Differentiating the power series \(k\) times with respect to \(z\), we obtain \[\left.\frac{\partial^{k} u}{\partial z^{k}}\right|_{z=0}=B_{k}(x) \quad \text { for }\quad k \geq 0.\]
We have \[\frac{z}{e^{z}-1}=\frac{z}{\sum_{k=1}^{\infty} \frac{z^{k}}{k!}}=\left(\sum_{k=0}^{\infty} \frac{z^{k}}{(k+1)!}\right)^{-1} .\]
Then we see from these expansions that \(B_{0}=1\), \(B_{1}=-\frac{1}{2}\), and \[\left(\sum_{k=0}^{\infty} \frac{z^{k}}{(k+1)!}\right)\left(\sum_{k=0}^{\infty} \frac{B_{k}}{k!} z^{k}\right)=1.\]
The left-hand side is equal to \(\sum_{m=1}^{\infty} c_{m-1} z^{m-1}\), where \[c_{m-1}=\sum_{k=0}^{m-1} \frac{B_{k}}{k!} \frac{1}{(m-k)!}.\]
Hence, we obtain the following reccurence \[\frac{B_{0}}{0!m!}+\frac{B_{1}}{1!(m-1)!}+\cdots+\frac{B_{m-1}}{(m-1)!1!} = \begin{cases}1 & \text { if } m=1, \\ 0 & \text { if } m>1. \end{cases}\]
Further, we see that \[\frac{z}{e^{z}-1}+\frac{z}{2}=1+\sum_{k=2}^{\infty} \frac{B_{k}}{k!} z^{k}.\]
Since the left-hand side is an even function of \(z\), we derive that \[B_{k}=0 \quad \text { for } \quad k=2m+1 \text { and } m\in\mathbb N.\]
Further, we compute \(B_{k}\) for \(2 \leq k \leq 14\) as follows: \(B_{2}=\frac{1}{6}\), and \[B_{4}=-\frac{1}{30}, B_{6}=\frac{1}{42}, B_{8}=-\frac{1}{30}, B_{10}=\frac{5}{66}, B_{12}=-\frac{691}{2730}, B_{14}=\frac{7}{6} .\]
The Bernoulli polynomials \(B_{k}(x)\) with \(k \geq 0\) satisfy the following:
\(B_{k}(x)\) is a monic polynomial of degree \(k\) given by \[B_{k}(x)=\sum_{m=0}^{k}\binom{k}{m} B_{m} x^{k-m}.\]
We have \[B_{k}(1)-B_{k}(0)= \begin{cases}1 & \text { if } k=1 \\ 0 & \text { if } k \neq 1 .\end{cases}\]
Also, \(B_{k}(1-x)=(-1)^{k} B_{k}(x)\).
Finally, \(B_{k}^{\prime}(x)=k B_{k-1}(x)\) for \(k>0\).
Proof of (a): We have \[\sum_{k=0}^{\infty} \frac{B_{k}(x)}{k!} z^{k}=\frac{z}{e^{z}-1} e^{x z}=\left(\sum_{k=0}^{\infty} \frac{B_{k}}{k!} z^{k}\right)\left(\sum_{k=0}^{\infty} \frac{x^{k}}{k!} z^{k}\right).\]
Now the assertion follows immediately by comparing the coefficients of \(z^{k}\) on both sides. Further, the coefficient of \(x^{k}\) in \(B_{k}(x)\) is \(\binom{k}{0} B_{0}=1\). Hence, \(B_{k}(x)\) is a monic polynomial of degree \(k\).$$\tag*{$\blacksquare$}$$
Proof of (b): Note that \[u(1, z)-u(0, z)=\frac{z}{e^{z}-1} e^{z}-\frac{z}{e^{z}-1}=z.\]
By differentiating both sides \(k\) times, we see that \(B_{k}(1)-B_{k}(0)=1\) if \(k=1\) and \(0\) otherwise.$$\tag*{$\blacksquare$}$$
Proof of (c): Note that \[u(1-x, z)=\frac{z}{e^{z}-1} e^{(1-x) z}=\frac{-z}{e^{-z}-1} e^{-x z}=u(x,-z).\]
By differentiating both sides \(k\) times, we derive that \(B_{k}(1-x)= (-1)^{k} B_{k}(x)\).$$\tag*{$\blacksquare$}$$
Proof of (d): We have \(\frac{\partial}{\partial x} u(x, z)=z u(x, z)\), which we rewrite as \[\sum_{k=1}^{\infty} \frac{B_{k}^{\prime}(x)}{k!} z^{k}=\sum_{k=1}^{\infty} \frac{B_{k-1}(x)}{(k-1)!} z^{k}.\]
By comparing the coefficients of \(z^{k}\) on both sides, we obtain \[\frac{B_{k}^{\prime}(x)}{k!}=\frac{B_{k-1}(x)}{(k-1)!} \quad \text { for }\quad k>0,\] which implies the assertion.$$\tag*{$\blacksquare$}$$
Let \(b>a\) and \(q \geq 1\) be integers. Let \(f\in C^q([a, b])\), then \[\sum_{n=a+1}^{b} f(n)=\int_{a}^{b} f(x) d x+\sum_{r=1}^{q}(-1)^{r} \frac{B_{r}}{r!}\left(f^{(r-1)}(b)-f^{(r-1)}(a)\right)+R_{q},\] where \[R_{q}=\frac{(-1)^{q+1}}{q!} \int_{a}^{b} B_{q}(x-\lfloor x\rfloor) f^{(q)}(x) d x.\]
Proof: Let \(F\) be \(q\) times continuously differentiable in \([0,1]\).
By the previous lemma (a) and (d) with \(k=1\), we have \[B_{1}(x)=B_{0} x+B_{1}=x-\frac{1}{2},\quad \text{ and } \quad B_{1}^{\prime}(x)=1.\]
Then \[\int_{0}^{1} F(x) d x=\int_{0}^{1} F(x) B_{1}^{\prime}(x) d x.\]
Integrating the right-hand side by parts, we derive from the previous lemma (d) with \(k=2\) that \[\begin{aligned} \int_{0}^{1} F(x) B_{1}^{\prime}(x) d x & =\frac{F(1)+F(0)}{2}-\int_{0}^{1} F^{\prime}(x) B_{1}(x) d x \\ & =\frac{F(1)+F(0)}{2}-\frac{1}{2} \int_{0}^{1} F^{\prime}(x) B_{2}^{\prime}(x) d x. \end{aligned}\]
By the previous lemma (b) with \(k>1\) and (d) with \(k=3\), we obtain \[\begin{aligned} \int_{0}^{1} F^{\prime}(x) B_{2}^{\prime}(x) d x & =B_{2}\left(F^{\prime}(1)-F^{\prime}(0)\right)-\int_{0}^{1} F^{\prime \prime}(x) B_{2}(x) d x \\ & =B_{2}\left(F^{\prime}(1)-F^{\prime}(0)\right)-\frac{1}{3} \int_{0}^{1} F^{\prime \prime}(x) B_{3}^{\prime}(x) d x. \end{aligned}\]
Now we proceed inductively, as above, for obtaining \[\begin{gathered} \int_{0}^{1} F(x) d x=\frac{F(1)+F(0)}{2}+\sum_{r=2}^{q}(-1)^{r-1} \frac{B_{r}}{r!}\left(F^{(r-1)}(1)-F^{(r-1)}(0)\right) \\ +\frac{(-1)^{q}}{q!} \int_{0}^{1} B_{q}(x) F^{(q)}(x) d x. \end{gathered}\]
Since \(B_{1}=-\frac{1}{2}\), we obtain \[\frac{F(1)+F(0)}{2}=F(1)-\frac{F(1)-F(0)}{2}=F(1)+B_{1}(F(1)-F(0)) .\]
Consequently, we have \[\begin{gathered} F(1)=\int_{0}^{1} F(x) d x+\sum_{r=1}^{q}(-1)^{r} \frac{B_{r}}{r!}\left(F^{(r-1)}(1)-F^{(r-1)}(0)\right) \\ +\frac{(-1)^{q+1}}{q!} \int_{0}^{1} B_{q}(x) F^{(q)}(x) d x. \end{gathered}\]
For positive integer \(n\) with \(a \leq n \leq b\), let \(F(x)=f(n-1+x)\).
Then \(F(x)\) is \(q\) times continuously differentiable in \([0,1]\) since \(f\) is \(q\) times continuously differentiable in \([a, b]\).
Then we derive from the above formula that \[\begin{aligned} f(n)=\int_{n-1}^{n} f(x) d x+ & \sum_{r=1}^{q}(-1)^{r} \frac{B_{r}}{r!}\left(f^{(r-1)}(n)-f^{(r-1)}(n-1)\right) \\ & +\frac{(-1)^{q+1}}{q!} \int_{0}^{1} B_{q}(x) f^{(q)}(n-1+x) d x. \end{aligned}\]
Letting \(n\) run from \(a+1\) to \(b\), we obtain \[\sum_{n=a+1}^{b} f(n)=\int_{a}^{b} f(x) d x+\sum_{r=1}^{q}(-1)^{r} \frac{B_{r}}{r!}\left(f^{(r-1)}(b)-f^{(r-1)}(a)\right)+R_{q},\] where \[R_{q}=\frac{(-1)^{q+1}}{q!} \sum_{n=a+1}^{b} \int_{0}^{1} B_{q}(x) f^{(q)}(n-1+x) d x.\]
For \(n=a+r\) with \(1 \leq r \leq b-a\), we have \[\int_{0}^{1} B_{q}(x) f^{(q)}(n-1+x) d x=\int_{0}^{1} B_{q}(x) f^{(q)}(a+r-1+x) d x\]
Putting \(a+r-1+x=y\), the above integral is equal to \[\int_{a+r-1}^{a+r} B_{q}(y-\lfloor y\rfloor) f^{(q)}(y) d y .\]
Hence \[R_{q}=\frac{(-1)^{q+1}}{q!} \int_{a}^{b} B_{q}(x-\lfloor x\rfloor) f^{(q)}(x) d x.\]
This completes the Euler–Maclaurin–Jacobi summation formula.$$\tag*{$\blacksquare$}$$
Let \(m\in\mathbb N\). For all \(z\in\mathbb C\setminus\{-n\in\mathbb Z: n\ge0\}\), we have \[\log \Gamma(z)=\left(z-\frac{1}{2}\right) \log z-z+\frac{1}{2} \log 2 \pi+K_{m}(z),\tag{*}\] where logarithm has principal value and \[K_{m}(z)=\sum_{j=1}^{m} \frac{B_{2 j}}{(2 j-1) 2 j} \frac{1}{z^{2 j-1}}-\frac{1}{2 m} \int_{0}^{\infty} \frac{B_{2 m}(x-\lfloor x\rfloor)}{(x+z)^{2 m}} d x.\]
Proof: We check that both the sides in are holomorphic functions of \(z\) in the region \(\mathbb C\setminus(-\infty, 0]\). Therefore, by the identity theorem, it suffices to prove (*) for all real numbers \(z \geq z_{0}\) where \(z_{0}>0\) is sufficiently large.
By the properties of the \(\Gamma\) function, we have \[\begin{aligned} \Gamma(z) & =(z-1) \Gamma(z-1)=\prod_{n=1}^{\infty}\left(\left(1+\frac{z-1}{n}\right)^{-1}\left(1+\frac{1}{n}\right)^{z-1}\right) \\ & =\lim _{N \rightarrow \infty}\left((N+1)^{z-1} \prod_{n=1}^{N}\left(1+\frac{z-1}{n}\right)^{-1}\right). \end{aligned}\]
Since \(\Gamma(z)\) has no zero and it has pole at zero and at negative integer and none of the term in the above product vanishes, we derive that \[\log \Gamma(z)=\lim _{N \rightarrow \infty}\left((z-1) \log N-\sum_{n=1}^{N} \log \left(\frac{n+z-1}{n}\right)\right).\]
Now we apply the Euler–Maclaurin–Jacobi formula to the sum above with \(a=1\),\(b=N\), \(f(x)=\log (x+z-1)-\log x\) and \(q=2 m\).
For \(x \in[a, b]\) and \(1 \leq r \leq 2 m\), we observe that \[f^{(r)}(x)=(-1)^{r-1}(r-1)!\left(\frac{1}{(x+z-1)^{r}}-\frac{1}{x^{r}}\right).\]
Hence we conclude \[\begin{aligned} &\sum_{n=1}^{N} \log \left(\frac{n+z-1}{n}\right)= \log z+\sum_{n=2}^{N} \log \left(\frac{n+z-1}{n}\right) = \log z\\ &+\int_{1}^{N}(\log (x+z-1)-\log x) d x+\frac{1}{2}(\log (N+z-1)-\log N -\log z)\\ &+\sum_{j=1}^{m} \frac{B_{2 j}}{(2 j-1) 2 j}\left(\frac{1}{(N+z-1)^{2 j-1}}-\frac{1}{N^{2 j-1}}-\frac{1}{z^{2 j-1}}+1\right) \\ & +\frac{1}{2 m} \int_{1}^{N} B_{2 m}(x-\lfloor x\rfloor)\left(\frac{1}{(x+z-1)^{2 m}}-\frac{1}{x^{2 m}}\right) d x, \end{aligned}\tag{**}\] since \(B_{1}=-\frac{1}{2}\) and \(B_{3}=B_{5}=\cdots=B_{2 m-1}=0\).
Integrating by parts the first integral in (**). we obtain \[\int_{1}^{N} \log (x+z-1) d x=(N+z-1) \log (N+z-1)-z \log z-N+1,\] and \[\int_{1}^{N} \log x d x=N \log N-N+1.\]
Further \[\lim_{N\to \infty}\frac{1}{2}(\log (N+z-1)-\log N -\log z)=-\frac{1}{2} \log z.\]
Also note that \[\int_{1}^{\infty} \frac{B_{2 m}(x-\lfloor x\rfloor)}{(x+z-1)^{2 m}} d x=\int_{0}^{\infty} \frac{B_{2 m}(x-\lfloor x\rfloor)}{(x+z)^{2 m}} d x\]
Hence, we obtain \[\begin{aligned} \lim_{N\to\infty}&\Bigg(\sum_{j=1}^{m} \frac{B_{2 j}}{(2 j-1) 2 j}\left(\frac{1}{(N+z-1)^{2 j-1}}-\frac{1}{N^{2 j-1}}-\frac{1}{z^{2 j-1}}+1\right) \\ & +\frac{1}{2 m} \int_{1}^{N} B_{2 m}(x-\lfloor x\rfloor)\left(\frac{1}{(x+z-1)^{2 m}}-\frac{1}{x^{2 m}}\right) d x\Bigg)\\ &=-K_m(z)-L_m', \end{aligned}\] where \[K_m(z)=\sum_{j=1}^{m} \frac{B_{2 j}}{(2 j-1) 2 j}\frac{1}{z^{2 j-1}}-\frac{1}{2 m}\int_{0}^{\infty} \frac{B_{2 m}(x-\lfloor x\rfloor)}{(x+z)^{2 m}} d x,\] and \[L_m'=\frac{1}{2 m} \int_{1}^{\infty} \frac{B_{2 m}(x-\lfloor x\rfloor)}{x^{2 m}} d x-\sum_{j=1}^{m} \frac{B_{2 j}}{(2 j-1) 2 j}.\]
Therefore, we conclude that \[\log \Gamma(z)=A+\lim _{n \rightarrow \infty} B(N),\] where \[A=\left(z-\frac{1}{2}\right) \log z+K_{m}(z)+L_{m}^{\prime},\] and \[\begin{aligned} B(N) & =-(N+z-1) \log (N+z-1)+(N+z-1) \log N \\ & =-(N+z-1) \log \left(1+\frac{z-1}{N}\right), \end{aligned}\] satisfying \[\lim _{N \rightarrow \infty} B(N)=-z+1 .\]
Hence, we have proved that \[\log \Gamma(z)=\left(z-\frac{1}{2}\right) \log z-z+K_{m}(z)+L_{m},\] where \(L_{m}=L_{m}^{\prime}+1\). Since \(\lim _{z \rightarrow \infty} K_{m}(z)=0\), we have \[\lim _{z \rightarrow \infty}\left(\log \Gamma(z)-\left(z-\frac{1}{2}\right) \log z+z\right)=L_{m}.\]
The proof will be competed if we prove that \(L_{m}=\frac{1}{2} \log 2 \pi.\)
Taking \(z=N+1\) and using \(\Gamma(N+1)=N!\), we have \[\begin{aligned} L_{m}&=\lim _{N \rightarrow \infty}\left(\log N!-\left(N+\frac{1}{2}\right) \log (N+1)+N+1\right)\\ &=\lim _{N \rightarrow \infty}\left(\log N!-\left(N+\frac{1}{2}\right) \log N+N\right), \end{aligned}\] since \(\log (N+1)=\log N+O(1 / N)\).
Therefore \[\lim _{N \rightarrow \infty} \frac{N!}{N^{N} N^{1 / 2} e^{-N}}=e^{L m}.\]
Setting \(z=\frac{1}{2}\) in both sides of \(\frac{\sin \pi z}{\pi z}= \prod_{n=1}^{\infty}\left(1-\frac{z^{2}}{n^{2}}\right)\), we obtain \[\prod_{n=1}^{\infty}\left(1-\frac{1}{4 n^{2}}\right)^{-1}=\frac{\pi}{2}.\]
Thus \[\frac{\pi}{2}=\lim _{N \rightarrow \infty} \prod_{n=1}^{N} \frac{4 n^{2}}{(2 n-1)(2 n+1)}=\lim _{N \rightarrow \infty} \frac{4^{2 N}(N!)^{4}}{((2 N)!)^{2}(2 N+1)}\] by rewriting \[\prod_{n=1}^{N}(2 n-1)(2 n+1)=\frac{((2 N)!)^2(2 N+1)}{(2 \cdot 4 \cdots 2 N)^{2}}=\frac{((2 N)!)^2(2 N+1)}{4^{N}(N!)^{2}}.\]
Now, we derive that \[\frac{\pi}{2}=\lim _{N \rightarrow \infty} \frac{4^{2 N} N^{4 N} e^{-4 N} N^{2} e^{4 L_{m}}}{(2 N)^{4 N} e^{-4 N} 2 N(2 N+1) e^{2 L_{m}}}=\frac{1}{4} e^{2 L_{m}},\] which implies that \(L_{m}=\frac{1}{2} \log (2 \pi)\) as desired.$$\tag*{$\blacksquare$}$$
Let \(0<\delta<\pi\), then for any \(z\in\mathbb C\) so that \(|\arg z|<\pi-\delta\), we have \[\log \Gamma(z)=\left(z-\frac{1}{2}\right) \log z-z+\frac{1}{2} \log 2 \pi+O(|z|^{-1}),\tag{**}\] uniformly as \(|z|\to\infty\), where logarithm has principal value, and the implicit constant depend at most on \(\delta\).
Proof: If we apply the previous theorem with \(m=1\), we obtain \[\log \Gamma(z)=\left(z-\frac{1}{2}\right) \log z-z+\frac{1}{2} \log 2 \pi+K_{1}(z),\] where \[\begin{aligned} K_{1}(z)=\frac{1}{12z}-\frac{1}{2} \int_{0}^{\infty} \frac{B_{2}(x-\lfloor x\rfloor)}{(x+z)^{2}} d x=-\frac{1}{2} \int_{0}^{\infty} \frac{B_{2}(x-\lfloor x\rfloor)-B_2}{(x+z)^{2}} d x, \end{aligned}\] and \(B_2(x)-B_2=x^2-x\).
Note that \[\bigg|\int_{0}^{\infty} \frac{B_{2}(x-\lfloor x\rfloor)-B_2}{(x+z)^{2}} d x\bigg|\le \frac{1}{4}\int_{0}^{\infty} \frac{d x}{|x+z|^{2}} ,\] since \(B_2(x-\lfloor x\rfloor)-B_2\le 1/4\).
If \(\theta=\arg z\), then \(z=re^{i\theta}\) and we can write \[\begin{aligned} \int_{0}^{\infty} \frac{d x}{|x+z|^{2}} &=\int_{0}^{\infty} \frac{d x}{x^2+r^2+2xr\cos\theta}\\ &\le \int_{0}^{\infty} \frac{d x}{x^2+r^2-2xr\cos\delta}, \end{aligned}\] since \(|\theta|=|\arg z|<\pi-\delta\).
Since \(2xr\le x^2+r^2\), then \[\begin{aligned} \int_{0}^{\infty} \frac{d x}{x^2+r^2-2xr\cos\delta}\le \frac{1}{1-\cos\delta} \int_{0}^{\infty} \frac{d x}{x^2+r^2}\le \frac{\pi}{2(1-\cos\delta)r}. \end{aligned}\]
In fact we proved that \[|K_1(z)|\le \frac{\pi}{16(1-\cos\delta)r}.\]
This completes the proof.$$\tag*{$\blacksquare$}$$
Let \(0<\delta<\pi\), then for any \(z\in\mathbb C\) so that \(|\arg z|<\pi-\delta\), we have \[\frac{\Gamma'(z)}{\Gamma(z)}=\log z-\frac{1}{2z}+O(|z|^{-2}),\tag{***}\] uniformly as \(|z|\to\infty\), where logarithm has principal value, and the implicit constant depend at most on \(\delta\).
Proof: It suffices to differentiate the formula \[\log \Gamma(z)=\left(z-\frac{1}{2}\right) \log z-z+\frac{1}{2} \log 2 \pi+K_{1}(z),\] where \[\begin{aligned} K_{1}(z)=-\frac{1}{2} \int_{0}^{\infty} \frac{B_{2}(x-\lfloor x\rfloor)-B_2}{(x+z)^{2}} d x, \quad \text{ and } \quad B_2(x)-B_2=x^2-x. \end{aligned}\]
Then we have \[\begin{aligned} \frac{\Gamma'(z)}{\Gamma(z)}=\log z-\frac{1}{2z} +K_{1}'(z), \end{aligned}\] where \[\begin{aligned} K_{1}'(z)=\int_{0}^{\infty} \frac{B_{2}(x-\lfloor x\rfloor)-B_2}{(x+z)^{3}} d x, \end{aligned}\]
Observe that \[\begin{aligned} |K_{1}'(z)|\le \frac{1}{4}\int_{0}^{\infty} \frac{dx}{|x+z|^{3}}. \end{aligned}\]
If \(\theta=\arg z\), then \(z=re^{i\theta}\) and \(|\theta|=|\arg z|<\pi-\delta\), and we can write \[\begin{aligned} \int_{0}^{\infty} \frac{d x}{|x+z|^{3}} &=\int_{0}^{\infty} \frac{d x}{(x^2+r^2+2xr\cos\theta)^{3/2}}\\ &\le \int_{0}^{\infty} \frac{d x}{(x^2+r^2-2xr\cos\delta)^{3/2}}. \end{aligned}\]
Since \(2xr\le x^2+r^2\), then \[\begin{aligned} \int_{0}^{\infty} \frac{d x}{(x^2+r^2-2xr\cos\delta)^{3/2}}&\le \frac{1}{(1-\cos\delta)^{3/2}} \int_{0}^{\infty} \frac{d x}{(x^2+r^2)^{3/2}}\\ &\le \frac{\pi}{2(1-\cos\delta)^{3/2}r^2}. \end{aligned}\]
In fact we proved that \[|K_1'(z)|\le \frac{\pi}{8(1-\cos\delta)^{3/2}r^2}.\]
This completes the proof.$$\tag*{$\blacksquare$}$$
Let \(z=\sigma+it\) with \(z \neq 0\) such that either \(\sigma>0, t=0\) or \(t \neq 0\). Then \[\left|K_{1}(z)\right| \leq \begin{cases}\frac{1}{8 \sigma} & \text { if } \sigma>0, t=0, \\ \frac{1}{8|t|}\arctan \frac{|t|}{\sigma}&\text { if } t\neq 0, \end{cases}\] where \[0 \leq \arctan \frac{|t|}{\sigma}=|\arg z|<\pi .\]
Proof: By the previous theorem with \(m=1\), we have \[K_{1}(z)=-\frac{1}{2} \int_{0}^{\infty} \frac{B_{2}(x-\lfloor x\rfloor)}{(x+z)^{2}} d x+\frac{B_{2}}{2 z}=-\frac{1}{2} \int_{0}^{\infty} \frac{B_{2}(x-\lfloor x\rfloor)-B_{2}}{(x+z)^{2}} d x.\]
By the properties of Bernoulli’s polynomials \(B_{2}(x)-B_{2}=x^{2}-x.\)
Therefore \[\left|B_{2}(x-\lfloor x\rfloor)-B_{2}\right| \leq \frac{1}{4}.\] and hence \[\left|K_{1}(z)\right| \leq \frac{1}{8} \int_{0}^{\infty} \frac{d x}{(\sigma+x)^{2}+t^{2}}.\]
We may assume that \(t>0\). Let \(\sigma>0\). By putting \(\sigma+x=t \tan \theta\), the integral is equal to \[\frac{1}{t} \int_{\frac{\pi}{2}-\arctan \frac{t}{\sigma}}^{\frac{\pi}{2}} d \theta=\frac{1}{t} \arctan \frac{t}{\sigma},\] and the assertion follows. Here, we have used the identity \[\arctan (u)=\frac{\pi}{2}-\arctan \left(\frac{1}{u}\right)\] for \(0<u \leq 1\). The proof for the case \(\sigma \leq 0\) is similar.$$\tag*{$\blacksquare$}$$
Let \(a, b\in\mathbb R\) be fixed and \(a \leq b\).
Then for every \(z=\sigma+ it\) with \(\sigma\in[a, b]\) and \(|t| \geq 1\), we have \[\Gamma(z)=\sqrt{2 \pi} e^{-\frac{\pi}{2}|t|}|t|^{\sigma-\frac{1}{2}} e^{i|t|(\log |t|-1)} e^{\frac{\pi i}{2}\left(\sigma-\frac{1}{2}\right)}\left(1+O\left(\frac{1}{|t|}\right)\right).\]
Moreover, \(|\Gamma(z)|=\sqrt{2 \pi} e^{-\frac{\pi}{2}|t|}|t|^{\sigma-\frac{1}{2}}\left(1+O\left(\frac{1}{|t|}\right)\right)\).
This implies \(|\Gamma(z)|=O\left(e^{-\frac{\pi}{2}|t|}|t|^{\sigma-\frac{1}{2}}\right)\) and
\(\frac{1}{|\Gamma(z)|}=O\left(e^{\frac{\pi}{2}|t|}|t|^{\frac{1}{2}-\sigma}\right)\).
Proof: Since \(\Gamma(\overline{z})=\overline{\Gamma(z)}\), we may suppose that \(t \geq 1\). Further we may assume that \(t\) exceeds sufficiently large number depending only on \(a\) and \(b\), otherwise Corollary follows. It suffices to prove (i), which implies immediately (ii), (iii) and (iv).
By the previous theorem and the previous lemma, we have \[\begin{aligned} \log \Gamma(z)=&\frac{1}{2} \log (2 \pi)+\left(\sigma+i t-\frac{1}{2}\right) \log (\sigma+i t)-(\sigma+i t)\\ &+ \frac{\theta}{8 t} \arctan \frac{t}{\sigma}, \qquad \text{where} \quad |\theta| \leq 1. \end{aligned}\]
The second term on the right-hand side above is equal to \[\left(\sigma-\frac{1}{2}+i t\right)\left(\log \left(\sqrt{\sigma^{2}+t^{2}}\right)+i \arctan \frac{t}{\sigma}\right).\]
Then \[\log \left(\sqrt{\sigma^{2}+t^{2}}\right)=\log t+\frac{1}{2} \log \left(1+\frac{\sigma^{2}}{t^{2}}\right)=\log t+O\left(\frac{1}{t^{2}}\right).\]
Also \[\arctan \frac{t}{\sigma}=\frac{\pi}{2}-\arctan \frac{\sigma}{t}=\frac{\pi}{2}-\frac{\sigma}{t}+O\left(\frac{1}{t^{3}}\right) .\]
Further, we have \[\frac{\theta}{8 t} \arctan \frac{t}{\sigma}= \frac{\theta}{8 t}\left(\frac{\pi}{2}-\arctan \frac{\sigma}{t}\right)=O\left(\frac{1}{t}\right)\]
Therefore, we conclude \[\begin{aligned} \log \Gamma(z)=&\frac{1}{2} \log (2 \pi)+\frac{\pi i}{2}\left(\sigma-\frac{1}{2}\right)-\frac{\pi}{2} t+\left(\sigma-\frac{1}{2}\right) \log t\\ &+i t(\log t-1)+O\left(\frac{1}{t}\right). \end{aligned}\]
Hence \(\Gamma(z)=\sqrt{2 \pi} e^{\frac{\pi i}{2}\left(\sigma-\frac{1}{2}\right)} e^{-\frac{\pi}{2} t} t^{\sigma-\frac{1}{2}} e^{i t(\log t-1)}\left(1+O\left(\frac{1}{t}\right)\right)\).$$\tag*{$\blacksquare$}$$
For \(n \geq 0\), we have
\((z+n)^{-2}=(z+n)^{-1}-(z+n+1)^{-1}+(z+n)^{-2}(z+n+1)^{-1}\).
Further, we have \[\begin{aligned} &(z+n)^{-2}(z+n+1)^{-1}\\ &=\frac{1}{2}(z+n)^{-2}-\frac{1}{2}(z+n+1)^{-2}+\frac{1}{2}(z+n)^{-2}(z+ n+1)^{-2}. \end{aligned}\]
We also have \[\begin{aligned} &(z+n)^{-2}(z+n+1)^{-2}\\ &=\frac{1}{3}(z+n)^{-3}-\frac{1}{3}(z+n+1)^{-3}-\frac{1}{3}(z+n)^{-3}(z+ n+1)^{-3}. \end{aligned}\]
Proof of (i): We have \[\begin{aligned} (z+n)^{-2}&-(z+n)^{-2}(z+n+1)^{-1} =(z+n)^{-2}\left(1-(z+n+1)^{-1}\right) \\ & =(z+n)^{-1}(z+n+1)^{-1}=(z+n)^{-1}-(z+n+1)^{-1}.\qquad \end{aligned} \tag*{$\blacksquare$}\]
Proof of (ii): We have \[\begin{aligned} (z+n)^{-2}&(z+n+1)^{-1} -\frac{1}{2}(z+n)^{-2}(z+n+1)^{-2}\\ &=\frac{1}{2}(z+n)^{-2}(z+n+1)^{-2}(2 z+2 n+1) \\ & =\frac{1}{2}(z+n)^{-2}(z+n+1)^{-2}\left((z+n+1)^{2}-(z+n)^{2}\right) \\ & =\frac{1}{2}(z+n)^{-2}-\frac{1}{2}(z+n+1)^{-2} .\qquad \end{aligned}\tag*{$\blacksquare$}\]
Proof of (iii): By multiplying both sides of \[\begin{aligned} &(z+n)^{-2}(z+n+1)^{-2}\\ &=\frac{1}{3}(z+n)^{-3}-\frac{1}{3}(z+n+1)^{-3}-\frac{1}{3}(z+n)^{-3}(z+ n+1)^{-3}, \end{aligned}\] by \(3(z+n)^{3}(z+n+1)^{3}\), we prove that \[3(z+n+1)(z+n)=(z+n+1)^{3}-(z+n)^{3}-1.\] The right-hand side is equal to \[\begin{aligned} (z+n+1)^{2}&+(z+n)^{2}+(z+n)(z+n+1)-1\\ &= (z+n+2)(z+n)+(z+n)^{2} +(z+n+1)(z+n) \\ = & 3(z+n+1)(z+n). \qquad \end{aligned}\tag*{$\blacksquare$}\]
For \(z>0\), we have \[\sqrt{\frac{2 \pi}{z}}\left(\frac{z}{e}\right)^{z} \leq \Gamma(z) \leq \sqrt{\frac{2 \pi}{z}}\left(\frac{z}{e}\right)^{z} e^{\frac{1}{12 z}}.\]
Proof: For \(z>0\), we observe that the last inequality is equivalent to \[0 \leq \log \Gamma(z)-\left(z-\frac{1}{2}\right) \log z+z-\frac{1}{2} \log (2 \pi) \leq \frac{1}{12} z^{-1}.\] by taking logarithms.
Note that \[\frac{d^{2}}{d z^{2}} \log \Gamma(z)=\frac{d}{d z}\bigg(\frac{\Gamma'(z)}{\Gamma(z)}\bigg)=\sum_{n=0}^{\infty}(z+n)^{-2}>0.\]
By the previous Lemma (i), the last sum in is equal to \[z^{-1}+\sum_{n=0}^{\infty}(z+n)^{-2}(z+n+1)^{-1}.\]
Now we apply the last Lemma (ii) to each term of the above sum, and we obtain \[\sum_{n=0}^{\infty}(z+n)^{-2}(z+n+1)^{-1}=\frac{1}{2} z^{-2}+\frac{1}{2} \sum_{n=0}^{\infty}(z+n)^{-2}(z+n+1)^{-1} .\]
Thus \[\frac{d^{2}}{d z^{2}} \log \Gamma(z)=z^{-1}+\frac{1}{2} z^{-2}+\frac{1}{2} \sum_{n=0}^{\infty}(z+n)^{-2}(z+n+1)^{-2},\] and since \(z>0\), we conclude that \[\frac{d^{2}}{d z^{2}} \log \Gamma(z) \geq z^{-1}+\frac{1}{2} z^{-2}.\]
Further, by the previous Lemma (iii), we have \[\begin{aligned} \frac{d^{2}}{d z^{2}} \log \Gamma(z) & =z^{-1}+\frac{1}{2} z^{-2}+\frac{1}{6} z^{-3}-\frac{1}{6} \sum_{n=0}^{\infty}(z+n)^{-3}(z+n+1)^{-3} \\ & \leq z^{-1}+\frac{1}{2} z^{-2}+\frac{1}{6} z^{-3}. \end{aligned}\]
Combining the above two inequalities, we obtain \[0 \leq \frac{d^{2}}{d z^{2}} \log \Gamma(z)-z^{-1}-\frac{1}{2} z^{-2} \leq \frac{1}{6} z^{-3} .\]
Let \[F(z)=\frac{d}{d z} \log \Gamma(z)-\log z+\frac{1}{2} z^{-1},\] so that \[0 \leq F^{\prime}(z) \leq \frac{1}{6} z^{-3} .\]
This implies that \(F\) is non-decreasing. Further, by integrating from \(z_{0}\) to \(z\) with \(z>z_{0}>1\), we have \[0 \leq \int_{z_{0}}^{z} F^{\prime}(\zeta) d \zeta \leq \frac{1}{6} \int_{z_{0}}^{z} \zeta^{-3} d \zeta.\]
Thus \[0 \leq F(z)-F\left(z_{0}\right) \leq \frac{1}{12}\left(z_{0}^{-2}-z^{-2}\right) \leq \frac{1}{12} z_{0}^{-2} .\]
Therefore, \(F(z)\) for \(z>1\) is bounded above and hence \[\lim _{z \rightarrow \infty} F(z)=c\] exists.
Further, by letting \(z\) tend to infinity and taking \(z_{0}=z\), we have \[-\frac{1}{12} z^{-2} \leq F(z)-c \leq 0.\]
Now we define \[g(z)=\log \Gamma(z)-\left(z-\frac{1}{2}\right) \log z+z-c z \tag{6.13.15}\] so that we see that \[g^{\prime}(z)=F(z)-c .\]
Further we derive that \[-\frac{1}{12} z^{-2} \leq g^{\prime}(z) \leq 0,\] which implies \[\lim _{z \rightarrow \infty} g(z)=c_{1}\] exists and \(0 \leq g(z)-c_{1} \leq \frac{1}{12} z^{-1}\).
Next we consider \[g(z+1)-g(z)=-\left(z+\frac{1}{2}\right) \log \left(\frac{z+1}{z}\right)+1-c.\]
By letting \(z\) tend to infinity on both sides, we derive that \(c=0\).
Now, we note that \[\begin{aligned} g(2 z)&-g(z)-g\left(z+\frac{1}{2}\right)\\ &= \log \frac{\Gamma(2 z)}{\Gamma(z) \Gamma\left(z+\frac{1}{2}\right)}-\left(2 z-\frac{1}{2}\right) \log 2-\left(2 z-\frac{1}{2}\right) \log z \\ & +\left(z-\frac{1}{2}\right) \log z+z \log \left(z+\frac{1}{2}\right)-\frac{1}{2} \\ = & \log \frac{2^{\frac{1}{2}-2 z} \Gamma(2 z)}{\Gamma(z) \Gamma\left(z+\frac{1}{2}\right)}+z \log \left(\frac{z+\frac{1}{2}}{z}\right)-\frac{1}{2}. \end{aligned}\]
By letting \(z\) tend to infinity on both sides, and using duplication formula \[\Gamma(2z)\Gamma\left(1/2\right)= 2^{2z-1} \Gamma\left(z\right) \Gamma\left(z+1/2\right),\] we conclude that \[-c_{1}=\lim _{z \rightarrow \infty}\left(g(2 z)-g(z)-g\left(z+\frac{1}{2}\right)\right)=-\frac{1}{2} \log (2 \pi),\] and hence \[c_{1}=\frac{1}{2} \log (2 \pi).\]
This completes the proof of the theorem. $$\tag*{$\blacksquare$}$$