1812 CHAPTER 57. INFINITE PRODUCTS

Show that Ir is an entire function. The reason 0 < r < 2π is that this prevents ereiθ −1from equaling zero. The above is just a precise description of the contour integral,∫

γwz−1

ew−1 dw where γ is the contour shown below.

in which on the integrals along the real line, the argument is different in going fromr to ∞ than it is in going from ∞ to r. Now I have not defined such contour integralsover contours which have infinite length and so have chosen to simply write out ex-plicitly what is involved. You have to work with these integrals given above anywaybut the contour integral just mentioned is the motivation for them. Hint: You maywant to use convergence theorems from real analysis if it makes this more convenientbut you might not have to.

20. ↑ In the context of Problem 19 define for small δ > 0

Irδ (z)≡∫

γr,δ

wz−1

ew−1dw

where γrδ is shown below.

2δ

r

x

Show that limδ→0 Irδ (z) = Ir (z) . Hint: Use the dominated convergence theorem ifit makes this go easier. This is not a hard problem if you use these theorems but youcan probably do it without them with more work.

21. ↑ In the context of Problem 20 show that for r1 < r, Irδ (z)− Ir1δ (z) is a contourintegral, ∫

γr,r1 ,δ

wz−1

ew−1dw