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 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.

 

[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 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}

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

has a simple zero at z=0, the integrand

What's 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.

 

[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->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

<br /> \frac{1}{2\pi i}\oint{\frac{(-z)^{s-1}}{e^{z}-1}dz}<br />

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))|<br /> <br /> = |exp((s - 1) (log |z| + i arg (-z)))|<br /> <br /> = |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| &gt; 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

 

[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

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

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?



about the inequalities:


\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.

 

[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:

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


what I'm missing?

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

|exp(ix)| = 1.

about the inequalities:

\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}?

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.