Matematikte , Wallis çarpımı ,
π
{\displaystyle \pi }
sonsuz çarpım olarak veren bir ifadedir. 1656'da John Wallis tarafından yayınlanmıstır ve şu şekilde ifade edilmektedir:[ 1]
π
2
=
∏
n
=
1
∞
4
n
2
4
n
2
−
1
=
∏
n
=
1
∞
(
2
n
2
n
−
1
⋅
2
n
2
n
+
1
)
=
(
2
1
⋅
2
3
)
⋅
(
4
3
⋅
4
5
)
⋅
(
6
5
⋅
6
7
)
⋅
(
8
7
⋅
8
9
)
⋅
⋯
{\displaystyle {\begin{aligned}{\frac {\pi }{2}}&=\prod _{n=1}^{\infty }{\frac {4n^{2}}{4n^{2}-1}}=\prod _{n=1}^{\infty }\left({\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}\right)\\[6pt]&={\Big (}{\frac {2}{1}}\cdot {\frac {2}{3}}{\Big )}\cdot {\Big (}{\frac {4}{3}}\cdot {\frac {4}{5}}{\Big )}\cdot {\Big (}{\frac {6}{5}}\cdot {\frac {6}{7}}{\Big )}\cdot {\Big (}{\frac {8}{7}}\cdot {\frac {8}{9}}{\Big )}\cdot \;\cdots \\\end{aligned}}}
Wallis bu sonsuz çarpımı enterpolasyon kullanarak türetmiştir; ancak, yöntemi titiz olarak kabul edilmemektedir. Daha modern bir çıkarım,
∫
0
π
sin
n
x
d
x
{\displaystyle \int _{0}^{\pi }\sin ^{n}x\,dx}
Wallis integrallerinin bir hali olan
I
(
n
)
=
∫
0
π
sin
n
x
d
x
{\displaystyle I(n)=\int _{0}^{\pi }\sin ^{n}x\,dx}
tanımlayalım. Kısmi integral yöntemi kullanarak
u
=
sin
n
−
1
x
⇒
d
u
=
(
n
−
1
)
sin
n
−
2
x
cos
x
d
x
d
v
=
sin
x
d
x
⇒
v
=
−
cos
x
{\displaystyle {\begin{aligned}u&=\sin ^{n-1}x\\\Rightarrow du&=(n-1)\sin ^{n-2}x\cos x\,dx\\dv&=\sin x\,dx\\\Rightarrow v&=-\cos x\end{aligned}}}
⇒
I
(
n
)
=
∫
0
π
sin
n
x
d
x
=
−
sin
n
−
1
x
cos
x
|
0
π
−
∫
0
π
(
−
cos
x
)
(
n
−
1
)
sin
n
−
2
x
cos
x
d
x
=
0
+
(
n
−
1
)
∫
0
π
cos
2
x
sin
n
−
2
x
d
x
,
n
>
1
=
(
n
−
1
)
∫
0
π
(
1
−
sin
2
x
)
sin
n
−
2
x
d
x
=
(
n
−
1
)
∫
0
π
sin
n
−
2
x
d
x
−
(
n
−
1
)
∫
0
π
sin
n
x
d
x
=
(
n
−
1
)
I
(
n
−
2
)
−
(
n
−
1
)
I
(
n
)
=
n
−
1
n
I
(
n
−
2
)
⇒
I
(
n
)
I
(
n
−
2
)
=
n
−
1
n
{\displaystyle {\begin{aligned}\Rightarrow I(n)&=\int _{0}^{\pi }\sin ^{n}x\,dx\\[6pt]{}&=-\sin ^{n-1}x\cos x{\Biggl |}_{0}^{\pi }-\int _{0}^{\pi }(-\cos x)(n-1)\sin ^{n-2}x\cos x\,dx\\[6pt]{}&=0+(n-1)\int _{0}^{\pi }\cos ^{2}x\sin ^{n-2}x\,dx,\qquad n>1\\[6pt]{}&=(n-1)\int _{0}^{\pi }(1-\sin ^{2}x)\sin ^{n-2}x\,dx\\[6pt]{}&=(n-1)\int _{0}^{\pi }\sin ^{n-2}x\,dx-(n-1)\int _{0}^{\pi }\sin ^{n}x\,dx\\[6pt]{}&=(n-1)I(n-2)-(n-1)I(n)\\[6pt]{}&={\frac {n-1}{n}}I(n-2)\\[6pt]\Rightarrow {\frac {I(n)}{I(n-2)}}&={\frac {n-1}{n}}\\[6pt]\end{aligned}}}
Şimdi, kolaylık olması açısından iki değişken ikâmesi yaparak şunu elde edelim:
I
(
2
n
)
=
2
n
−
1
2
n
I
(
2
n
−
2
)
{\displaystyle I(2n)={\frac {2n-1}{2n}}I(2n-2)}
I
(
2
n
+
1
)
=
2
n
2
n
+
1
I
(
2
n
−
1
)
{\displaystyle I(2n+1)={\frac {2n}{2n+1}}I(2n-1)}
I
(
0
)
{\displaystyle I(0)}
I
(
1
)
{\displaystyle I(1)}
I
(
0
)
=
∫
0
π
d
x
=
x
|
0
π
=
π
I
(
1
)
=
∫
0
π
sin
x
d
x
=
−
cos
x
|
0
π
=
(
−
cos
π
)
−
(
−
cos
0
)
=
−
(
−
1
)
−
(
−
1
)
=
2
{\displaystyle {\begin{aligned}I(0)&=\int _{0}^{\pi }dx=x{\Biggl |}_{0}^{\pi }=\pi \\[6pt]I(1)&=\int _{0}^{\pi }\sin x\,dx=-\cos x{\Biggl |}_{0}^{\pi }=(-\cos \pi )-(-\cos 0)=-(-1)-(-1)=2\\[6pt]\end{aligned}}}
Çift değerler için hesaplamak için
I
(
2
n
)
{\displaystyle I(2n)}
I
(
0
)
{\displaystyle I(0)}
I
(
2
n
)
=
∫
0
π
sin
2
n
x
d
x
=
2
n
−
1
2
n
I
(
2
n
−
2
)
=
2
n
−
1
2
n
⋅
2
n
−
3
2
n
−
2
I
(
2
n
−
4
)
{\displaystyle I(2n)=\int _{0}^{\pi }\sin ^{2n}x\,dx={\frac {2n-1}{2n}}I(2n-2)={\frac {2n-1}{2n}}\cdot {\frac {2n-3}{2n-2}}I(2n-4)}
=
2
n
−
1
2
n
⋅
2
n
−
3
2
n
−
2
⋅
2
n
−
5
2
n
−
4
⋅
⋯
⋅
5
6
⋅
3
4
⋅
1
2
I
(
0
)
=
π
∏
k
=
1
n
2
k
−
1
2
k
{\displaystyle ={\frac {2n-1}{2n}}\cdot {\frac {2n-3}{2n-2}}\cdot {\frac {2n-5}{2n-4}}\cdot \cdots \cdot {\frac {5}{6}}\cdot {\frac {3}{4}}\cdot {\frac {1}{2}}I(0)=\pi \prod _{k=1}^{n}{\frac {2k-1}{2k}}}
Tek değerler için de benzer bir yöntem takip edilebilir ve
I
(
1
)
{\displaystyle I(1)}
I
(
2
n
+
1
)
=
∫
0
π
sin
2
n
+
1
x
d
x
=
2
n
2
n
+
1
I
(
2
n
−
1
)
=
2
n
2
n
+
1
⋅
2
n
−
2
2
n
−
1
I
(
2
n
−
3
)
{\displaystyle I(2n+1)=\int _{0}^{\pi }\sin ^{2n+1}x\,dx={\frac {2n}{2n+1}}I(2n-1)={\frac {2n}{2n+1}}\cdot {\frac {2n-2}{2n-1}}I(2n-3)}
=
2
n
2
n
+
1
⋅
2
n
−
2
2
n
−
1
⋅
2
n
−
4
2
n
−
3
⋅
⋯
⋅
6
7
⋅
4
5
⋅
2
3
I
(
1
)
=
2
∏
k
=
1
n
2
k
2
k
+
1
{\displaystyle ={\frac {2n}{2n+1}}\cdot {\frac {2n-2}{2n-1}}\cdot {\frac {2n-4}{2n-3}}\cdot \cdots \cdot {\frac {6}{7}}\cdot {\frac {4}{5}}\cdot {\frac {2}{3}}I(1)=2\prod _{k=1}^{n}{\frac {2k}{2k+1}}}
Ayrıca,
sin
x
≤
x
{\displaystyle \sin {x}\leq x}
sin
2
n
+
1
x
≤
sin
2
n
x
≤
sin
2
n
−
1
x
,
0
≤
x
≤
π
{\displaystyle \sin ^{2n+1}x\leq \sin ^{2n}x\leq \sin ^{2n-1}x,0\leq x\leq \pi }
⇒
I
(
2
n
+
1
)
≤
I
(
2
n
)
≤
I
(
2
n
−
1
)
{\displaystyle \Rightarrow I(2n+1)\leq I(2n)\leq I(2n-1)}
olduğunu gözlemliyoruz. Her iki tarafı
I
(
2
n
+
1
)
{\displaystyle I(2n+1)}
I
(
2
n
)
=
2
n
−
1
2
n
I
(
2
n
−
2
)
{\displaystyle I(2n)={\frac {2n-1}{2n}}I(2n-2)}
I
(
2
n
+
1
)
=
2
n
2
n
+
1
I
(
2
n
−
1
)
{\displaystyle I(2n+1)={\frac {2n}{2n+1}}I(2n-1)}
⇒
1
≤
I
(
2
n
)
I
(
2
n
+
1
)
≤
I
(
2
n
−
1
)
I
(
2
n
+
1
)
=
2
n
+
1
2
n
{\displaystyle \Rightarrow 1\leq {\frac {I(2n)}{I(2n+1)}}\leq {\frac {I(2n-1)}{I(2n+1)}}={\frac {2n+1}{2n}}}
elde ediyoruz. Sıkıştırma teoremi ile
⇒
lim
n
→
∞
I
(
2
n
)
I
(
2
n
+
1
)
=
1
{\displaystyle \Rightarrow \lim _{n\rightarrow \infty }{\frac {I(2n)}{I(2n+1)}}=1}
elde edilir. Yani,
lim
n
→
∞
I
(
2
n
)
I
(
2
n
+
1
)
=
π
2
lim
n
→
∞
∏
k
=
1
n
(
2
k
−
1
2
k
⋅
2
k
+
1
2
k
)
=
1
{\displaystyle \lim _{n\rightarrow \infty }{\frac {I(2n)}{I(2n+1)}}={\frac {\pi }{2}}\lim _{n\rightarrow \infty }\prod _{k=1}^{n}\left({\frac {2k-1}{2k}}\cdot {\frac {2k+1}{2k}}\right)=1}
⇒
π
2
=
∏
k
=
1
∞
(
2
k
2
k
−
1
⋅
2
k
2
k
+
1
)
=
2
1
⋅
2
3
⋅
4
3
⋅
4
5
⋅
6
5
⋅
6
7
⋅
⋯
{\displaystyle \Rightarrow {\frac {\pi }{2}}=\prod _{k=1}^{\infty }\left({\frac {2k}{2k-1}}\cdot {\frac {2k}{2k+1}}\right)={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot \cdots }
elde edilir.
değiştir
Yukarıdaki ispat genellikle modern kalkülüs ders kitaplarında yer alsa da, geriye dönüp bakıldığında Wallis çarpımı, sinüs fonksiyonu için daha sonra elde edilmişl olan Euler sonsuz çarpımının kolay bir sonucu olarak ortaya çıkar.
sin
x
x
=
∏
n
=
1
∞
(
1
−
x
2
n
2
π
2
)
{\displaystyle {\frac {\sin x}{x}}=\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{n^{2}\pi ^{2}}}\right)}
x
=
π
2
{\displaystyle x={\frac {\pi }{2}}}
⇒
2
π
=
∏
n
=
1
∞
(
1
−
1
4
n
2
)
⇒
π
2
=
∏
n
=
1
∞
(
4
n
2
4
n
2
−
1
)
=
∏
n
=
1
∞
(
2
n
2
n
−
1
⋅
2
n
2
n
+
1
)
=
2
1
⋅
2
3
⋅
4
3
⋅
4
5
⋅
6
5
⋅
6
7
⋯
{\displaystyle {\begin{aligned}\Rightarrow {\frac {2}{\pi }}&=\prod _{n=1}^{\infty }\left(1-{\frac {1}{4n^{2}}}\right)\\[6pt]\Rightarrow {\frac {\pi }{2}}&=\prod _{n=1}^{\infty }\left({\frac {4n^{2}}{4n^{2}-1}}\right)\\[6pt]&=\prod _{n=1}^{\infty }\left({\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}\right)={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdots \end{aligned}}}
[ 1]
değiştir
Riemann zeta fonksiyonu ve Dirichlet eta fonksiyonu şu şekilde tanımlanabilir:[ 1]
ζ
(
s
)
=
∑
n
=
1
∞
1
n
s
,
ℜ
(
s
)
>
1
η
(
s
)
=
(
1
−
2
1
−
s
)
ζ
(
s
)
=
∑
n
=
1
∞
(
−
1
)
n
−
1
n
s
,
ℜ
(
s
)
>
0
{\displaystyle {\begin{aligned}\zeta (s)&=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}},\Re (s)>1\\[6pt]\eta (s)&=(1-2^{1-s})\zeta (s)\\[6pt]&=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n^{s}}},\Re (s)>0\end{aligned}}}
Son seriye Euler dönüşümü uygulandığında aşağıdaki elde edilir:
η
(
s
)
=
1
2
+
1
2
∑
n
=
1
∞
(
−
1
)
n
−
1
[
1
n
s
−
1
(
n
+
1
)
s
]
,
ℜ
(
s
)
>
−
1
⇒
η
′
(
s
)
=
(
1
−
2
1
−
s
)
ζ
′
(
s
)
+
2
1
−
s
(
ln
2
)
ζ
(
s
)
=
−
1
2
∑
n
=
1
∞
(
−
1
)
n
−
1
[
ln
n
n
s
−
ln
(
n
+
1
)
(
n
+
1
)
s
]
,
ℜ
(
s
)
>
−
1
{\displaystyle {\begin{aligned}\eta (s)&={\frac {1}{2}}+{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[{\frac {1}{n^{s}}}-{\frac {1}{(n+1)^{s}}}\right],\Re (s)>-1\\[6pt]\Rightarrow \eta '(s)&=(1-2^{1-s})\zeta '(s)+2^{1-s}(\ln 2)\zeta (s)\\[6pt]&=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[{\frac {\ln n}{n^{s}}}-{\frac {\ln(n+1)}{(n+1)^{s}}}\right],\Re (s)>-1\end{aligned}}}
⇒
η
′
(
0
)
=
−
ζ
′
(
0
)
−
ln
2
=
−
1
2
∑
n
=
1
∞
(
−
1
)
n
−
1
[
ln
n
−
ln
(
n
+
1
)
]
=
−
1
2
∑
n
=
1
∞
(
−
1
)
n
−
1
ln
n
n
+
1
=
−
1
2
(
ln
1
2
−
ln
2
3
+
ln
3
4
−
ln
4
5
+
ln
5
6
−
⋯
)
=
1
2
(
ln
2
1
+
ln
2
3
+
ln
4
3
+
ln
4
5
+
ln
6
5
+
⋯
)
=
1
2
ln
(
2
1
⋅
2
3
⋅
4
3
⋅
4
5
⋅
⋯
)
=
1
2
ln
π
2
⇒
ζ
′
(
0
)
=
−
1
2
ln
(
2
π
)
{\displaystyle {\begin{aligned}\Rightarrow \eta '(0)&=-\zeta '(0)-\ln 2=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[\ln n-\ln(n+1)\right]\\[6pt]&=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\ln {\frac {n}{n+1}}\\[6pt]&=-{\frac {1}{2}}\left(\ln {\frac {1}{2}}-\ln {\frac {2}{3}}+\ln {\frac {3}{4}}-\ln {\frac {4}{5}}+\ln {\frac {5}{6}}-\cdots \right)\\[6pt]&={\frac {1}{2}}\left(\ln {\frac {2}{1}}+\ln {\frac {2}{3}}+\ln {\frac {4}{3}}+\ln {\frac {4}{5}}+\ln {\frac {6}{5}}+\cdots \right)\\[6pt]&={\frac {1}{2}}\ln \left({\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot \cdots \right)={\frac {1}{2}}\ln {\frac {\pi }{2}}\\\Rightarrow \zeta '(0)&=-{\frac {1}{2}}\ln \left(2\pi \right)\end{aligned}}}
![{\displaystyle {\begin{aligned}{\frac {\pi }{2}}&=\prod _{n=1}^{\infty }{\frac {4n^{2}}{4n^{2}-1}}=\prod _{n=1}^{\infty }\left({\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}\right)\\[6pt]&={\Big (}{\frac {2}{1}}\cdot {\frac {2}{3}}{\Big )}\cdot {\Big (}{\frac {4}{3}}\cdot {\frac {4}{5}}{\Big )}\cdot {\Big (}{\frac {6}{5}}\cdot {\frac {6}{7}}{\Big )}\cdot {\Big (}{\frac {8}{7}}\cdot {\frac {8}{9}}{\Big )}\cdot \;\cdots \\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/df59bf8aa67b6dff8be6cffb4f59777cea828454)
![{\displaystyle {\begin{aligned}\Rightarrow I(n)&=\int _{0}^{\pi }\sin ^{n}x\,dx\\[6pt]{}&=-\sin ^{n-1}x\cos x{\Biggl |}_{0}^{\pi }-\int _{0}^{\pi }(-\cos x)(n-1)\sin ^{n-2}x\cos x\,dx\\[6pt]{}&=0+(n-1)\int _{0}^{\pi }\cos ^{2}x\sin ^{n-2}x\,dx,\qquad n>1\\[6pt]{}&=(n-1)\int _{0}^{\pi }(1-\sin ^{2}x)\sin ^{n-2}x\,dx\\[6pt]{}&=(n-1)\int _{0}^{\pi }\sin ^{n-2}x\,dx-(n-1)\int _{0}^{\pi }\sin ^{n}x\,dx\\[6pt]{}&=(n-1)I(n-2)-(n-1)I(n)\\[6pt]{}&={\frac {n-1}{n}}I(n-2)\\[6pt]\Rightarrow {\frac {I(n)}{I(n-2)}}&={\frac {n-1}{n}}\\[6pt]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb327a4dcf3b4321e18cd38e4ab967de34205e3d)
![{\displaystyle {\begin{aligned}I(0)&=\int _{0}^{\pi }dx=x{\Biggl |}_{0}^{\pi }=\pi \\[6pt]I(1)&=\int _{0}^{\pi }\sin x\,dx=-\cos x{\Biggl |}_{0}^{\pi }=(-\cos \pi )-(-\cos 0)=-(-1)-(-1)=2\\[6pt]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/61d7bb90cabaa8b31f60711a286f7d3905dc4470)
![{\displaystyle {\begin{aligned}\Rightarrow {\frac {2}{\pi }}&=\prod _{n=1}^{\infty }\left(1-{\frac {1}{4n^{2}}}\right)\\[6pt]\Rightarrow {\frac {\pi }{2}}&=\prod _{n=1}^{\infty }\left({\frac {4n^{2}}{4n^{2}-1}}\right)\\[6pt]&=\prod _{n=1}^{\infty }\left({\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}\right)={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdots \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/74ce7488d92c2708916455e66c7770dbfed21150)
![{\displaystyle n!={\sqrt {2\pi n}}{\left({\frac {n}{e}}\right)}^{n}\left[1+O\left({\frac {1}{n}}\right)\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96fbe5666b3943b49f7279545f2d83f745c8bed2)
![{\displaystyle {\begin{aligned}p_{k}&={1 \over {2k+1}}\prod _{n=1}^{k}{\frac {(2n)^{4}}{[(2n)(2n-1)]^{2}}}\\[6pt]&={1 \over {2k+1}}\cdot {{2^{4k}\,(k!)^{4}} \over {[(2k)!]^{2}}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4425b5472edd553ad732621d722aa0a7ddf13ab)
![{\displaystyle {\begin{aligned}\zeta (s)&=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}},\Re (s)>1\\[6pt]\eta (s)&=(1-2^{1-s})\zeta (s)\\[6pt]&=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n^{s}}},\Re (s)>0\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fdf32ebbc781cbf33667c496c10e1231e0a10c3e)
![{\displaystyle {\begin{aligned}\eta (s)&={\frac {1}{2}}+{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[{\frac {1}{n^{s}}}-{\frac {1}{(n+1)^{s}}}\right],\Re (s)>-1\\[6pt]\Rightarrow \eta '(s)&=(1-2^{1-s})\zeta '(s)+2^{1-s}(\ln 2)\zeta (s)\\[6pt]&=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[{\frac {\ln n}{n^{s}}}-{\frac {\ln(n+1)}{(n+1)^{s}}}\right],\Re (s)>-1\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e62998eb05e0fccc87195869efc243c134bad554)
![{\displaystyle {\begin{aligned}\Rightarrow \eta '(0)&=-\zeta '(0)-\ln 2=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\left[\ln n-\ln(n+1)\right]\\[6pt]&=-{\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n-1}\ln {\frac {n}{n+1}}\\[6pt]&=-{\frac {1}{2}}\left(\ln {\frac {1}{2}}-\ln {\frac {2}{3}}+\ln {\frac {3}{4}}-\ln {\frac {4}{5}}+\ln {\frac {5}{6}}-\cdots \right)\\[6pt]&={\frac {1}{2}}\left(\ln {\frac {2}{1}}+\ln {\frac {2}{3}}+\ln {\frac {4}{3}}+\ln {\frac {4}{5}}+\ln {\frac {6}{5}}+\cdots \right)\\[6pt]&={\frac {1}{2}}\ln \left({\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot \cdots \right)={\frac {1}{2}}\ln {\frac {\pi }{2}}\\\Rightarrow \zeta '(0)&=-{\frac {1}{2}}\ln \left(2\pi \right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3803d63abff3794e62b95627aefe28f36754d24f)