PHYS 201 Mathematical Physics, Fall 2015, Homework 4
Due date: Tuesday, October 27th, 2015
1. In this exercise, we will prove the existence of a single-valued and analytic local
inverse function for any analytic function whose derivative is non-zero.
i. Prove that if w = f (z) is analytic at z0 , with w0 = f (z0 ) and f 0 (z0 ) 6= 0, then
there exists a contour C around z0 such that f (z) does not take the value w0 on
or within C except at z0 . This shows that |f (z) − w0 | has a lower bound, say m,
on C. (Hint: use the identity theorem, which states that if f is analytic, we have
f ≡ 0 in a domain D if and only if you can construct a sequence an → a within D
such that f (an ) = 0 for all n and a lies in D.)
ii. Next, consider w such that |w − w0 | < m. Show that for any such w, say w1 , there
is only one point inside C such that w1 = f (z1 ). Conclude the proof. (Hint: use
Rouché’s theorem)
2. Most of the special functions in mathematical physics can be generated by a generating function of the form
X
g(t, x) =
fn (x)tn
n
Show that we can give an integral representation of the special function fn (x) as
I
1
g(t, x)t−n−1 dt
fn (x) =
2πi
where the contour encloses the origin and no other singular points. The generating
function for Bessel functions Jn (x) is given by gB (t, x) = e(x/2)(t−1/t) . Expand gB (t, x)
as a Laurent series near the origin and find the first three terms of Jn (x) by computing
the residue.
3. Evaluate using rectangular contours (or otherwise):
R∞
2
i. −∞ e−αx eikx dx for real α, k.
R ∞ eax
ii. −∞ 1+e
x dx for 0 < a < 1. By making an appropriate substitution, show that this
integral is the beta function B(a, 1 − a) = Γ(a)Γ(1 − a).
4. Show that π cot πz has simple poles at integer values of z. Consider the integral
I
1
f (z)π cot(πz)dz
2πi CN
1
with f (z) = 1/z 2 and CN a rectangular contour with diagonally opposite vertices
−N − 1/2 andPN + 1/2. Argue why the integral goes to zero and consequently show
π2
1
that the sum ∞
n=1 n2 = 6 by computing the residue at 0.
5. Evaluate the integral
π
Z
0
cos nθ
dθ
1 − 2a cos θ + a2
R
−iz n−1 dz
iθ
for a real and > 1. (Hint: Consider the real part of 1−a(z+z
−1 )+a2 where z = e .
Many integrals involving trigonometric integrands can be solved this way.)
6. Show that, for α > 0,
Z
∞
0
t sin αt
π
dt = e−α
2
1+t
2
7. The function f (z) = (1 − z 2 )1/2 can be made single-valued using cuts running along
the real axis for |x| > 1. Using these cuts and a suitable contour, evaluate the integral
Z ∞
dx
2
x(x − 1)1/2
1
Verify your answer by substituting x = sec θ.
8. Use a semicircular contour in the upper half plane (with a bump at the origin) to
compute the integral
Z ∞
(ln x)2
dx
1 + x2
0
and deduce, as a byproduct of your calculation, that
Z ∞
ln x
dx = 0
1 + x2
0
9. Use a keyhole contour to show that
Z ∞
√
ln x
dx = − 2π 2
3/4
x (1 + x)
0
10. Evaluate the integral
1
x1/2 (1 − x)1/2
dx
2−x
0
(Hint: Apply Cauchy’s Integral Formula to the entire complex plane excluding the
branch cut between 0 and 1. Be careful about your choice of branch - verify that the
branch you’ve chosen is continuous across the negative real line and the positive real
line > 1. To compute the residue at infinity, use the formula Res(f (z)) at z = ∞ is
equal to Res(−z −2 f (1/z)) at z = 0.)
Z
2