# Help for zeta functional equation

1. Sep 23, 2009

### ilario980

hi,
i'm studying the functional equation of riemann zeta function for Re(s)>1;
my book(complex analysis by T. Gamelin) use contour integral in the proof, where the contour is taken on the usual 3 curves (real axis and a small circle $$C\epsilon$$ around the origin). i'm not able to figure why the integral on the circle vanish as epsilon->0; the text report:

since $$e^{z -1}$$ has a simple zero at z=0, the integrand is bounded on the circle |z|=r by C $$\epsilon^{re(s)-2}$$

wich is the estimate that the author use in this assertion?
i'm new to complex analysis and i want to say (if possible) what argument i've got to study
thanks

I.M.

2. Sep 23, 2009

### Petek

This appears to be a typo for ${e^z}$ - 1.

What's the integrand?

3. Sep 23, 2009

### ilario980

the integral that vanish is

$$\frac{1}{2\pi i}\oint{\frac{(-z)^{s-1}}{e^{z}-1}dz}$$

taken on the circle |z|=epsilon as epsilon->0

a little snapshot from Complex analysis by T. Gamelin:

Last edited by a moderator: May 4, 2017
4. Sep 23, 2009

### Petek

OK, let's concentrate on estimating the integral

$$\frac{1}{2\pi i}\oint{\frac{(-z)^{s-1}}{e^{z}-1}dz}$$

We want to show that the integrand is bounded by $C\epsilon^{Re(s) - 2}$, where C is a constant that doesn't depend on $\epsilon$.

The usual strategy for finding a bound on a fraction is to find an upper bound on the numerator and a lower bound on the denominator. I'll sketch the arguments, leaving it to you to fill in the details. Let me know if you want more details. I'll write $s = \sigma + it$. First we consider the numerator:

$$|(-z)^{s-1}| = |exp((s - 1)log (-z))| = |exp((s - 1) (log |z| + i arg (-z)))| = |exp((\sigma - 1) + it)(log |z| + i arg (-z))|$$

Now multiply out to show that this last expression equals

$$|z}|^{\sigma-1} exp(-t arg(z))$$

and conclude that this expression is

$$\leq C_1\epsilon^{\sigma-1}$$

for some constant $C_1$ independent of $\epsilon$.

Next we consider the denominator $e^{z} - 1$:

Write $f(z) = e^{z} - 1$. Since f(z) has a simple zero at z = 0, there exists a function g(z) such that

$$f(z) = zg(z)$$

such that g is analytic and non-zero at z = 0. In fact, g is non-zero on some neighborhood of z = 0, which implies that g is nonzero on the circle |z| = $\epsilon$ for $\epsilon$ sufficiently small. (This is a standard result in complex analysis and should be in your book.)

Now use standard theorems about continuous functions on compact sets to conclude that there exists a constant $C_2$ such that g(z) > $C_2$ on |z| = $\epsilon$. Therefore

$$|e^{z} - 1| > C_2|z|$$

Finally, combine these inequalities to conclude that

$$\frac{|-z|^{s-1}}{|e^{z} - 1|} \leq C\epsilon^{s-2}$$

for some constant C.

HTH

Petek

5. Sep 23, 2009

### ilario980

hi petek,
i have some questions:

why we can't use the ML estimate? (since L->0 as espilon->0 )

i've multiplied but at the end i have:

$$|z}|^{\sigma-1} exp(-t arg(-z)) exp(i (\sigma-1)arg(-z) + it log|z| )$$

what i'm missing?

$$\frac{|-z|^{s-1}}{|e^{z} - 1|} \leq C\epsilon^{s-2}$$

why this yields that the integral is less than $$C\epsilon^{s-1}$$?

thanks

I.M.

6. Sep 24, 2009

### Petek

The above expression should be enclosed in "absolute value" marks. Then note that if x is a real number,

$$|exp(ix)| = 1.$$

Use the right side of this inequality as the M in the ML estimate. As you noted, L goes to zero.

Please let me know if my replies are unclear or if you have any other questions.

Petek

7. Sep 24, 2009

### ilario980

hi Petek,
thank you very much for your help - very clear.

I have one more question:

if epsilon->0 the curve reduce to a point, and an every integral evaulated on a single point must be 0 since dz=0: this condition is not enough to proof that this curve integral is 0 (even if Re(s)<1)?

I.M.

8. Sep 25, 2009

### Petek

I'm not sure that I understand your argument. However, in general, there's a difference between epsilon approaching zero and letting epsilon equal zero.

Petek

9. Sep 25, 2009

### ilario980

the fabulous world of discontinuity

thank you very much.

Ilario M.