Special Functions

Transforms

remark:

ˆf(ξ):=Reiξxf(x)dxf(x)=12πReiξxˆf(ξ)dξ.

figures/2021-12-20_07-55-38.png

The Gamma Function

definition (Gamma function):

Γ(z)=0tz1etdt.

remark:

Some interesting properties of Γ: Γ(z+1)=zΓ(z) and has simple poles at z=0,1,2, with residues Resz=mΓ(z)=(1)m/m!. There is also a factorization Γ(z)=1zeγzn=1(1+zn)ezn where γ:=limNn=1N1nlog(N)

Γ(z)Γ(1z)=πsin(πz), which yields a product factorization for sin(πz).

L(tz1,s=1)=Γ(z) and L(tn,s=1)=Γ(n+1).

The residues:

figures/2021-12-19_19-59-45.png

proposition (Γ is holomorphic on the right half-plane):

figures/2021-12-19_19-58-16.png

proposition (Functional equation for Γ):

figures/2021-12-19_19-58-44.png

proposition (Meromorphic continuation of Γ):

figures/2021-12-19_19-59-05.png

The Beta Function

definition (Beta function):

B(z,w)=10tz1(1t)w1 dt.

exercise (?):

Show that B(z,w)=Γ(z)Γ(w)Γ(z+w).

Hint: find L(tz1) and L(tz1tw1).

Riemann Zeta

definition (The Riemann Zeta Function):

figures/2021-12-19_20-00-12.png

There is a product expansion ζ(s)=pSpecZ11ps, where SpecZ are the primes of Z.

proposition (Meromorphic continuation of ζ):

figures/2021-12-19_20-00-54.png

There is a functional equation:

figures/2021-12-19_20-03-08.png

proposition (Zeros of ζ):

figures/2021-12-19_20-02-43.png

Weierstrass

definition (The Weierstrass function):

figures/2021-12-19_22-33-34.png

remark:

figures/2021-12-19_22-34-18.png

Elliptic Functions

#todo

Infinite Series and Products

fact (Infinite products):

figures/2021-12-14_17-36-04.png

theorem (Hadamard factorization):

Write Ep(z)={1zn=0(1z)exp(z+z22++znn) otherwise , and define the order of an entire function f to be the infimum over p where there exists some R such that |f(z)|e|z|p for |z|>R. Suppose f is entire of order p, write {zk}kn for its set of nonzero zeros repeated with multiplicity, and suppose z=0 is a zero of f of order m. Then there is a decomposition f(z)=zmeg(z)k1Ep(zzk), where deg(g)p.

exercise (Hadamard expansion of sin):

Find a Hadamard expansion of sin(πz).

solution:

sin(πz) has order 1, and its zero set is zk=k for kZ. So one can write sin(πz)=zeaz+bkZ{0}(1zk)ezk=zeaz+bk1(1z2k2). Determine eb=π by considering sin(πz)/z as z0, and use that sin(πz) is odd and the product factor is even to conclude eaz is even and thus equal to 1. This yields sin(πz)=πzk11z2k2.

theorem (Weierstrass factorization):

figures/2021-12-14_17-36-26.png

remark:

An interesting way to sum infinite series:

n=f(n)=(sum of  residues  of πcotπzf(z))n=(1)nf(n)=(sum of  residues  of πcscπzf(z))n=f(2n+12)=(sum of  residues of πtanπzf(z))n=(1)nf(2n+12)=(sum of residues of πsecπzf(z))..

exercise (Sum formulas: 1/(n-a)^2):

Show that kZ1(zk)2=(πcsc(πz))2.

#complex/exercise/work

exercise (Sum formulas: 1/n2):

Show n11n2=π26 by integrating πcot(πz)z2.

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exercise (Sum formulas: 1/n2+a2):

Show that kZ1k2+a2=πcoth(πa)afor a>0.

#complex/exercise/work

exercise (?):

Show that k11k2+a2=12πcoth(πa)a12a2a>0.

#complex/exercise/work

exercise (?):

Show that kZ1(k12)2=π2.

#complex/exercise/work

exercise (?):

Show that kZ(1)k(k+a)2=π2cos(πa)csc2(πa)for aRZ.

#complex/exercise/work

exercise (Computing ζ(2) by integration):

By computing 12πicot(πz)z2dz, say using a large rectangle, show that ζ(2)=k01k2=π26.

#complex/exercise/work

#todo #complex/exercise/work