1796 CHAPTER 57. INFINITE PRODUCTS

while the residue at n is1

α2−n2 .

Therefore

0 = limN→∞

IN = limN→∞

2πi

[N

∑n=−N

1α2−n2 −

π cotπα

α

]which establishes the following formula of Mittag Leffler.

limN→∞

N

∑n=−N

1α2−n2 =

π cotπα

α.

Writing this in a slightly nicer form, you obtain 57.2.15.This is a very interesting formula. This will be used to factor sin(πz) . The zeros of

this function are at the integers. Therefore, considering 57.2.13 you can pick pn = 1 in theWeierstrass factorization formula. Therefore, by Corollary 57.2.2 there exists an analyticfunction g(z) such that

sin(πz) = zeg(z)∞

∏n=1

(1− z

zn

)ez/zn (57.2.16)

where the zn are the nonzero integers. Remember you can permute the factors in theseproducts. Therefore, this can be written more conveniently as

sin(πz) = zeg(z)∞

∏n=1

(1−( z

n

)2)

and it is necessary to find g(z) . Differentiating both sides of 57.2.16

π cos(πz) = eg(z)∞

∏n=1

(1−( z

n

)2)+ zg′ (z)eg(z)

∏n=1

(1−( z

n

)2)

+zeg(z)∞

∑n=1−(

2zn2

)∏k ̸=n

(1−( z

k

)2)

Now divide both sides by sin(πz) to obtain

π cot(πz) =1z+g′ (z)−

∑n=1

2z/n2

(1− z2/n2)

=1z+g′ (z)+

∑n=1

2zz2−n2 .

By 57.2.15, this yields g′ (z) = 0 for z not an integer and so g(z) = c, a constant. So far thisyields

sin(πz) = zec∞

∏n=1

(1−( z

n

)2)

1796 CHAPTER 57. INFINITE PRODUCTSwhile the residue at n is1OnThereforeN1 Tcotma0 = lim Iy = lim 2ai] ¥° =<N-se0 N-s00 n—_N a —n? Qawhich establishes the following formula of Mittag Leffler.N1 trakm y _ Reot morNo Nn a2 — rn? aWriting this in a slightly nicer form, you obtain 57.2.15.This is a very interesting formula. This will be used to factor sin(az). The zeros ofthis function are at the integers. Therefore, considering 57.2.13 you can pick p, = 1 in theWeierstrass factorization formula. Therefore, by Corollary 57.2.2 there exists an analyticfunction g(z) such thatsin (1z) = ze) I] (1 - =) elen (57.2.16)n=1 énwhere the z, are the nonzero integers. Remember you can permute the factors in theseproducts. Therefore, this can be written more conveniently assin(ne) = set T] (1-(£)")and it is necessary to find g(z). Differentiating both sides of 57.2.16mcos(mz) = e8®) I] (: — (:)’) + zg! (z) OTT] (: 7 ()’)n=+e (3) TT (1- (2)n=1 kAnNow divide both sides by sin (z) to obtainmcot(mz) = zt#@-) GanBy 57.2.15, this yields g’ (z) = 0 for z not an integer and so g(z) =c, a constant. So far thisyieldsoe 2sin (z) = ze° |] (1 — (<) )n=1 n