57.6. EXERCISES 1813

where the oriented contour is shown below.

γr,r1,δ

In this contour integral, wz−1 denotes e(z−1) log(w) where log(w) = ln |w|+ iarg(w)for arg(w) ∈ (0,2π) . Explain why this integral equals zero. From Problem 20 itfollows that Ir = Ir1 . Therefore, you can define an entire function, I (z)≡ Ir (z) for allr positive but sufficiently small. Hint: Remember the Cauchy integral formula foranalytic functions defined on simply connected regions. You could argue there is asimply connected region containing γr,r1,δ

.

22. ↑ In case Rez > 1, you can get an interesting formula for I (z) by taking the limit asr→ 0. Recall that

Ir (z) ≡∫ 2π

0

e(z−1)(lnr+iθ)

ereiθ −1ireiθ dθ +

∫∞

r

e(z−1)(ln t+2πi)

et −1dt (57.6.25)

+∫ r

∞

e(z−1) ln t

et −1dt

and now it is desired to take a limit in the case where Rez > 1. Show the first integralabove converges to 0 as r→ 0. Next argue the sum of the two last integrals convergesto (

e(z−1)2πi−1)∫ ∞

0

e(z−1) ln(t)

et −1dt.

Thus

I (z) =(ez2πi−1

)∫ ∞

0

e(z−1) ln(t)

et −1dt (57.6.26)

when Rez > 1.

23. ↑ So what does all this have to do with the zeta function and the gamma function?The zeta function is defined for Rez > 1 by

∞

∑n=1

1nz ≡ ζ (z) .

By Problem 15, whenever Rez > 0,

Γ(z) =∫

∞

0e−ttz−1dt.