Transforms
ˆf(ξ):=∫Re−iξxf(x)dxf(x)=12π∫Reiξxˆf(ξ)dξ.
The Gamma Function
Γ(z)=∫∞0tz−1e−tdt.
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)=1zeγz∏∞n=1(1+zn)e−zn where γ:=limN→∞∑n=1N1n−log(N)
Γ(z)Γ(1−z)=πsin(πz), which yields a product factorization for sin(πz).
L(tz−1,s=1)=Γ(z) and L(tn,s=1)=Γ(n+1).
The residues:
The Beta Function
B(z,w)=∫10tz−1(1−t)w−1 dt.
Show that B(z,w)=Γ(z)Γ(w)Γ(z+w).
Hint: find L(tz−1) and L(tz−1∗tw−1).
Riemann Zeta
There is a product expansion ζ(s)=∏p∈SpecZ11−p−s, where SpecZ are the primes of Z.
There is a functional equation:
Weierstrass ℘
Elliptic Functions
Infinite Series and Products
Write Ep(z)={1−zn=0(1−z)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}k≤n 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)∏k≥1Ep(zzk), where deg(g)≤p.
Find a Hadamard expansion of sin(πz).
solution:
sin(πz) has order 1, and its zero set is zk=k for k∈Z. So one can write sin(πz)=zeaz+b∏k∈Z∖{0}(1−zk)ezk=zeaz+b∏k≥1(1−z2k2). Determine eb=π by considering sin(πz)/z as z→0, 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)=πz∏k≥11−z2k2.
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))..
By computing 12πi∮cot(πz)z2dz, say using a large rectangle, show that ζ(2)=∑k≥01k2=π26.