Help for zeta functional equation

[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 [tex]C\epsilon[/tex] around the origin). I'm not able to figure why the integral on the circle vanish as epsilon->0; the text report:

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

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.

 

[Petek]

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 [tex]C\epsilon[/tex] around the origin). I'm not able to figure why the integral on the circle vanish as epsilon->0; the text report:

since [tex]e^{z -1}[/tex]

This appears to be a typo for [itex]{e^z}[/itex] - 1.

has a simple zero at z=0, the integrand

What's the integrand?

is bounded on the circle |z|=r by C [tex]\epsilon^{re(s)-2}[/tex]

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.

 

[ilario980]

the integral that vanish is [tex]\frac{1}{2\pi i}\oint{\frac{(-z)^{s-1}}{e^{z}-1}dz}[/tex]taken on the circle |z|=epsilon as epsilon->0a little snapshot from Complex analysis by T. Gamelin:

http://ilario.mazzei.googlepages.com/Immagine.GIF

 

[Petek]

OK, let's concentrate on estimating the integral

[tex]
\frac{1}{2\pi i}\oint{\frac{(-z)^{s-1}}{e^{z}-1}dz}
[/tex]

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

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 [itex]s = \sigma + it[/itex]. First we consider the numerator:

[tex]|(-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))|[/tex]

Now multiply out to show that this last expression equals

[tex]|z}|^{\sigma-1} exp(-t arg(z))[/tex]

and conclude that this expression is

[tex]\leq C_1\epsilon^{\sigma-1}[/tex]

for some constant [itex]C_1[/itex] independent of [itex]\epsilon[/itex].

Next we consider the denominator [itex]e^{z} - 1[/itex]:

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

[tex]f(z) = zg(z)[/tex]

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| = [itex]\epsilon[/itex] for [itex]\epsilon[/itex] 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 [itex]C_2[/itex] such that g(z) > [itex]C_2[/itex] on |z| = [itex]\epsilon[/itex]. Therefore

[tex]|e^{z} - 1| > C_2|z|[/tex]

Finally, combine these inequalities to conclude that

[tex]\frac{|-z|^{s-1}}{|e^{z} - 1|} \leq C\epsilon^{s-2}[/tex]

for some constant C.

HTH

Petek

 

[ilario980]

hi petek,
thanks for your reply.
i have some questions:

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

Now multiply out to show that this last expression equals

[tex]|z}|^{\sigma-1} exp(-t arg(z))[/tex]

i've multiplied but at the end i have:

[tex]|z}|^{\sigma-1} exp(-t arg(-z)) exp(i (\sigma-1)arg(-z) + it log|z| )[/tex]


what I'm missing?



about the inequalities:


[tex]\frac{|-z|^{s-1}}{|e^{z} - 1|} \leq C\epsilon^{s-2}[/tex]


why this yields that the integral is less than [tex]C\epsilon^{s-1}[/tex]?



thanks


I.M.

 

[Petek]

hi petek,
thanks for your reply.
i have some questions:

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

We will. See my reply to your last question.

i've multiplied but at the end i have:

[tex]|z}|^{\sigma-1} exp(-t arg(-z)) exp(i (\sigma-1)arg(-z) + it log|z| )[/tex]


what I'm missing?

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

[tex]|exp(ix)| = 1.[/tex]

about the inequalities:

[tex]\frac{|-z|^{s-1}}{|e^{z} - 1|} \leq C\epsilon^{s-2}[/tex]


why this yields that the integral is less than [tex]C\epsilon^{s-1}[/tex]?

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

thanks


I.M.

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

Petek

 

[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.

 

[Petek]

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.

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

 

[ilario980]

the fabulous world of discontinuity

thank you very much.


Ilario M.