Primes, Complex-Contours, and Reimann

[saltydog]

I'm working on understanding the following relation which was referenced in the Number Theory Forum some time ago:

$$x-\text{ln}(2\pi)-\sum_{\rho} \frac{x^{\rho}}{\rho}-\frac{1}{2}\text{ln}(1-\frac{1}{x^2})= -\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\zeta^{'}(z)}{\zeta(z)}\frac{x^z}{z}dz$$

where:

$$\zeta(z)$$

is the Reimann Zeta function.

(c is taken to be right of the critical strip although I'm barely qualified to even say that let alone evaluate it but I digress)

It's encountered in one of the proofs of the Prime Number Theorem. It's an interesting contour-integral to solve and involves lots of different concepts in math.

I'm having some problems applying Residue Integration to it, not to mention figuring out the limits for the residues but believe I can work them out with some time. Just thought I'd post the problem in case others are interested.

Oh yea: Here's the reference

:http://www.maths.ex.ac.uk/~mwatkins/zeta/pntproof.htm

 

[Hurkyl]

Is the goal to prove the left equals the right? Or are we granted this equation and want to evaluate the right to prove something interesting?

 

[saltydog]

Is the goal to prove the left equals the right? Or are we granted this equation and want to evaluate the right to prove something interesting?

Hello Hurkly. My goal is to prove the contour integral equals the LHS. However, and again I'm barely qualified to say this: the LHS equals the RHS only if the Reimann Hypothesis is correct.

 

[shmoe]

That holds irregardles of the truth of the Riemann Hypothesis. RH gives you better control over the sum over the zeros on the left hand side, the real part of the zeros determines the magnitude of the x^rho term. Keeping in mind that the zeros come in fours, symmetric about the critical line and the real axis, the best you can hope for is that the real part of the rho's is 1/2. Combine this with asymptotics for the number of zeros in the critical strip and you can bound this term, and get a good error for psi(x), which is the RHS by Perron's.

You want c to the right of 1 so you can apply Perron's to get psi(x).

The full derivation of this explicit formula takes some work. Roughly you look at the integral over the rectangle with corners c+iT, -R+iT, -R-iT, c-iT, where R and T are positive and evaluate using the residue theorem- you have poles at s=1, giving the x term, s=0, giving the log(2Pi) term, the zeros of zeta in the critical strip, the infinite sum term, and the trivial zeros of zeta, which sum to the remaining log term. You then estimate the integrals over the horizontal segments and over the -R part. You have to choose R and T carefully to do this. Letting R and T go to infinity will give the result.

You can find proofs of this in many books, such as Edward's, or a truncated version in Ivic, or in Ingham's lovely Distribution of Primes book. Since you're interested in filling in the details yourself, I'd suggest you pick up Ram Murty's Problem's in Analytic Number Theory. It has exercises to guide you through this (and it does have full solutions if you get really stuck).

 

[Hurkyl]

Oh!

All the zeroes on the critical line have even multiplicity? That's nifty to know! (Are they known to be multiplicity 2? Or just that they're even multiplicity?)

 

[saltydog]

You then estimate the integrals over the horizontal segments and over the -R part. You have to choose R and T carefully to do this. Letting R and T go to infinity will give the result.

Well that answers, at least conceptually, the first question I had about applying Cauchy's Integral Theorem to the problem: the remaining parts of the contour.

Thanks for your comments Shmoe. I do indeed intend to work-through (in detail) the solution to this contour integral. May take a while.

 

[shmoe]

It follows from the reflection principle that the zeros come in complex conjugates, and it follows from the functional equation that if s is a zero in the critical strip, then so is 1-s. The "4 zeros" is purely a symmetry thing, if the zero is on the critical line then there are really only 2 and I don't mean to imply anything about the multiplicity of the zero in this case (though of course if there are actually 4, they would all have the same multiplicty).

All zeros found to date are simple.

 

[shmoe]

Well that answers, at least conceptually, the first question I had about applying Cauchy's Integral Theorem to the problem: the remaining parts of the contour.

Thanks for your comments Shmoe. I do indeed intend to work-through (in detail) the solution to this contour integral. May take a while.

No problem. Estimating the integral along the rest of the contour is not an easy task and takes a fair bit of work, so I do strongly suggest you take a look at Murty's book if you can (other books will work of course, I'm a little partial to Murty though). Of course, I'd be happy to provide some distilled pointers as needed if you prefer to go at it more on your own!

 

[saltydog]

No problem. Estimating the integral along the rest of the contour is not an easy task and takes a fair bit of work, so I do strongly suggest you take a look at Murty's book if you can (other books will work of course, I'm a little partial to Murty though). Of course, I'd be happy to provide some distilled pointers as needed if you prefer to go at it more on your own!

I looked it up. It's a graduate-level book. I'll pursue it though and yes, I do like going "more on my own" to some extent as that the best approach to learning however I would be happy if on some occasion you (anyone else too of course) could look at my work (just briefly) here to make sure everything is poke-a-poke. I intend to approach it from the perspective of residue integration:

$$\oint_C f(z)dz=2\pi i\sum_{j=1}^N \text{Res}_{z=z_i} f(z)$$

I can think of two ways to do this: a rectangular contour or a "half-circle" contour but first I'll attempt a rectangular contour as shown in the attached plot.

$$\oint_C f(z)dz=\int_{s_1}f(z)dz+\int_{s_2}f(z)dz+\int_{s_3} f(z)dz+\int_{s_4}f(z)dz$$

This then, by virtue of the Residue Theorem, is also equal to the sum of the residues of all the poles contained in C (multiplied by $2\pi i$).

I can then say:

$$\int_{s_1}f(z)dz+\int_{s_2}f(z)dz+\int_{s_3} f(z)dz+\int_{s_4}f(z)dz= 2\pi i\sum_{j=1}^N \text{Res}_{z\to z_j}f(z)$$

Letting:

$$f(z)=f(x,z)=\frac{\zeta^{'}(z)}{\zeta(z)}\frac{x^z}{z}$$

and establishing the right side of the rectangle on the vertical line $c\pm i$ and then taking the limit of the integral as the size of the contour approaches infinity, I can rewrite the integral as:

$$2\pi i\sum_{j=1}^N \text{Res}_{z\to z_j}f(z)=\int_{c-i\infty}^{c+i\infty}f(z)dz+ \int_{S_2}f(z)dz+\int_{S_3} f(z)dz+\int_{S_4}f(z)dz$$

where big S represents the sides of an infinite rectangle.

Or:

$$\int_{c-i\infty}^{c+i\infty}f(z)dz=2\pi i\sum_{j=1}^N \text{Res}_{z\to z_j}f(z)-\left(\int_{S_2}f(z)dz+\int_{S_3} f(z)dz+\int_{S_4}f(z)dz\right)$$

Now the integral of interest has a $1/(2\pi i)$ factor so I multiply throughout by this factor and obtain:

$$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}f(z)dz=\sum_{j=1}^N \text{Res}_{z\to z_j}f(z)-\frac{1}{2\pi i}\left(\int_{S_2}f(z)dz+\int_{S_3} f(z)dz+\int_{S_4}f(z)dz\right)$$

. . . alright then . . . I think so far so good . . .

 

[shmoe]

I looked it up. It's a graduate-level book.

This problem is at the level of that book.

When you take the limit of your contour, your result doesn't make sense, what is this "infinite rectangle"? Can you describe what points make up S_2, S_3, and S_4? I'm sure you've seen contour integration used to solve a real integral over the x-axis. A typical method is to consider a contour from [-R,R] and the upper half of a circle of radius R centered at the origin. You then show the integral over the semicircle is O(1/R), or some other suitable bound that tends to zero as R->infinity. You are never integrating over an "infinite semi circle", rather bounding finite contours and showing their contribution tends to zero as you consider larger contours.

Your sum over the zeros has some problems too, how are you ordering them and why do you end up with only N of them? Better to separate the trivial and non-trivial zeros as they end up contributing to very different terms in the end.

 

[saltydog]

I feel I'm making progress with this: My initial concern was the evaluation of the outlying contours first:

$$I_O=\int_{S_2}f(z)dz+\int_{S_3}f(z)dz+\int_{S_4}f(z)dz$$

See the attached diagram. Note that I'm pinning the right side of the rectangle to the line Re(z)=1 and extending it uniformly so that each side is 2a. Consider first the finite rectangle with side equal to 2a and the corresponsing contour along each. The integrals would be:

$$I_{s2}=\int_{1+ai}^{(1-2a)+ai} f(z)dz$$

$$I_{s3}=\int_{(1-2a)+ai}^{(1-2a)-ai} f(z)dz$$

$$I_{s4}=\int_{(1-2a)-ai}^{1-ai} f(z)dz$$

Now I wish to convert the limits to polar coordinates:

$$z=Re^{\theta i}$$

Referring to the first diagram yields:

$$I_{s2}=\int_{R_1e^{\theta_1 i}}^{R_2e^{\theta_2 i}} f(z)dz$$

$$I_{s3}=\int_{R_2e^{\theta_2 i}}^{R_3e^{\theta_3 i}} f(z)dz$$

$$I_{s4}=\int_{R_3e^{\theta_3 i}}^{R_4e^{\theta_4 i}} f(z)dz$$

Now, because of symmetry of the contour:

$$R_1=R_4,\quad \theta_1=-\theta_4$$

$$R_2=R_3,\quad \theta_2=-\theta_3$$

The integrals then become:

$$I_{s2}=\int_{R_1e^{\theta_1 i}}^{R_2e^{\theta_2 i}} f(z)dz$$

$$I_{s3}=\int_{R_2e^{\theta_2 i}}^{R_2e^{-\theta_2 i}} f(z)dz$$

$$I_{s4}=\int_{R_2e^{-\theta_2 i}}^{R_1e^{-\theta_1 i}} f(z)dz$$

Again because of symmetry, I can write these integrals as:

$$I_{O}=\int_{R_1e^{\theta_1 i}}^{R_1 e^{-\theta_1 i}} f(z)dz$$

Now I wish to take the limit of this integral as $R_1$ goes to infinity:

$$\mathop\lim\limits_{R_1\to\infty}\int_{R_1e^{\theta_1 i}}^{R_1 e^{-\theta_1 i}} f(z)dz$$

Note the second plot which is a very, very large picture of the square as the sides approach infinity. As I extend [itex]R_1\to\infty,\;\theta_1\to\pi/2[/tex]

This then becomes:

$$I_O}=\mathop\lim\limits_{R_1\to\infty}\int_{R_1e^{\pi/2 i}}^{R_1 e^{-\pi/2 i}} f(z)dz$$

Thus I am left with evaluating the following limit:

$$\mathop\lim\limits_{k\to\infty}\int_{ik}^{-ik} \frac{\zeta^{'}(z)}{\zeta(z)}\frac{x^z}{z}dz$$

and can restate the problem as:

$$\frac{1}{2\pi i}\int_{1-i\infty}^{1+i\infty} \frac{\zeta^{'}(z)}{\zeta(z)}\frac{x^z}{z}dz=\sum_{j}^N \text{Res}_{z\to z_j} f(z)-\frac{1}{2\pi i}\mathop\lim\limits_{k\to\infty}\int_{ik}^{-ik} \frac{\zeta^{'}(z)}{\zeta(z)}\frac{x^z}{z}dz$$

I think everything is Ok so far.

I've already done some numerical estimations for the limit and it appears to approach zero. I've not yet looked at the proofs as I wish to (try to) do as much as I can without looking at the answer.

 

[saltydog]

This problem is at the level of that book.

When you take the limit of your contour, your result doesn't make sense, what is this "infinite rectangle"? Can you describe what points make up S_2, S_3, and S_4? I'm sure you've seen contour integration used to solve a real integral over the x-axis. A typical method is to consider a contour from [-R,R] and the upper half of a circle of radius R centered at the origin. You then show the integral over the semicircle is O(1/R), or some other suitable bound that tends to zero as R->infinity. You are never integrating over an "infinite semi circle", rather bounding finite contours and showing their contribution tends to zero as you consider larger contours.

Your sum over the zeros has some problems too, how are you ordering them and why do you end up with only N of them? Better to separate the trivial and non-trivial zeros as they end up contributing to very different terms in the end.

Hey Shome. Didn't see this post before I posted the last one. Would have addressed it first. Need to regroup.

Edit: Wait a minute. I think I addressed your first paragraph in the analysis above. Did I?

 

[shmoe]

$$I_{O}=\int_{R_1e^{\theta_1 i}}^{R_1 e^{-\theta_1 i}} f(z)dz$$

is not true, unless you mean something non-standard by this notation, the contour you have on the right above is just $$s_1$$, though in the opposite direction. Taking the limit as R->infinity does not let you shift this contour to the imaginary axis either. This would be a bad thing to do in any case, you want to avoid poles on your contours, which is one of the reasons you can't take the right part of the rectangle to lie on the line Re(z)=1 like you've done (this will become extremely important when it comes time to estimate f(z) on it)..

Trying to express the endpoints of a rectangle in polar coordinates is not a productive thing to do, and seems to have created more confusion.

 

[saltydog]

$$I_{O}=\int_{R_1e^{\theta_1 i}}^{R_1 e^{-\theta_1 i}} f(z)dz$$

is not true, unless you mean something non-standard by this notation, the contour you have on the right above is just $$s_1$$, though in the opposite direction. Taking the limit as R->infinity does not let you shift this contour to the imaginary axis either. This would be a bad thing to do in any case, you want to avoid poles on your contours, which is one of the reasons you can't take the right part of the rectangle to lie on the line Re(z)=1 like you've done (this will become extremely important when it comes time to estimate f(z) on it)..

Trying to express the endpoints of a rectangle in polar coordinates is not a productive thing to do, and seems to have created more confusion.

Alright . . . but I first checked this procedure with:

$$\oint \frac{1}{z+1}dz$$

over the same contour and the analytical results (the limiting cases) were in excellent agreement with numerical approximations. Can you tell me if my analysis above for this integral is acceptable at least?

Edit: I don't know Shmoe. I'm not too good at this I recognize that and don't wish to act like I do. But someone in here once said, "you don't do math by just staring at it and waiting for the answer to pop into you head, you try things". Now who said that?

I'm off to get some references . . .

 

[saltydog]

For the record I acknowledge my analysis above is ill-performed and reflects my less than proficient understanding of the subject matter. Personally a better approach for me is to become better at contour integration before tackling a tough problem like the above. I hope no one was led astray above. As a matter of closure, I intend to review one or more of the proofs cited in the references suggested above and hope to prepare a short summary of the technique. May be a while though.

 

[shmoe]

Alright . . . but I first checked this procedure with:

$$\oint \frac{1}{z+1}dz$$

over the same contour and the analytical results (the limiting cases) were in excellent agreement with numerical approximations. Can you tell me if my analysis above for this integral is acceptable at least?

For 1/(z+1) the integral along the straight segment from -iT to iT will be very close to the integral from 1-iT to 1+iT (though I should stress they are never actually equal, except the limit as T->infinity). The integral around the rectangle with these corners will be zero as there are no poles inside, and the integral over the horizontal bits will be bounded by 2/T or so. Getting a correct result in this case (for the limit) doesn't justify how you've manipulated the contours though. try the same integrals with 1/(z-(1/2+i)) for example.

Let me add some detail to my objections. When you went from having integrals over s_2, s_3, and s_4 to $$I_{O}=\int_{R_1e^{\theta_1 i}}^{R_1 e^{-\theta_1 i}} f(z)dz$$, it looks to me like you're just taking a "shortcut" and rather than the full path over s_2, s_3, s_4 you're going straight from the start of s_2 to the end of s_4. The is only going to be true when there are no residues inside the rectangle (or they happen to cancel), which is not the case here.

My next complaint was how you dealt with this $$I_O$$ integral. You're right that as R_1->infinity we have theta_1->Pi/2, but theta is dependant on R and you can't allow it to go to pi/2 on it's own.the contour for $$I_O$$ starts on the line Re(z)=1 and by allowing theta to go to Pi/2 independantly, you've shifted this contour to one on the imaginary axis, and you aren't guaranteed equality or to even be close. In this case you're passing over all the zeros in the critical strip, so they are definitely not going to be equal.

Edit: I don't know Shmoe. I'm not too good at this I recognize that and don't wish to act like I do. But someone in here once said, "you don't do math by just staring at it and waiting for the answer to pop into you head, you try things". Now who said that?

Probably someone extremely wise. I do hope to encourage what you're doing, you will learn much in the process. My criticisms are meant to be constructive, so please take them that way. This is a very complicated problem, and you need quite a bit of background material to solve it, like density estimates for the zeros in the critical strip and the hadamard factorization to get suitable estimates for your integrand. I can't recommend a book like Murty's strongly enough for your purposes. It's like you're in a very dark room and you can see a light shining on your target. Few people will be able to navigate the room without some more light, the exercises in murty shine light in some key locations to help you along the way without fully illuminating the room, so you necessarily learn some navigation techniques on your own. (Riemann was capable of seeing in the dark and first turned the light on over the target, but we can't all hope to have his vision.)


Your last reply appeared as I typed the above. That looks like a good plan. I'd argue there are few applications of contour integration this beautiful, so if ever there was a reason to learn it well, this is it!

 

[saltydog]

In my effort to understand the derivation of von Mangoldt's function:

$$\psi(x)=x-\sum \frac{x^{\rho}}{\rho}-ln(2\pi)-1/2 ln(1-1/x^2)$$

via contour integration, I reached the conclusion that I would have to become proficient in estimating upper bounds for contour integrals. This is conveniently undertaken from the perspective of the following theorem:

If for a function of a complex variable f(z), there exists a constant M such that:

$$|f(z)|\leq M$$

everywhere along the path C of length L, then:

$$\left|\int_C f(z)dz\right|\leq ML$$

So if I have an integrand:

$$\frac{f(z)}{g(z)}$$

and I can show that along the path of integration:

$$|f(z)|\leq h(p)$$

$$|g(z)|\geq w(p)$$

for some parameter p, then:

$$\left|\frac{f(z)}{g(z)}\right|\leq \frac{h(p)}{w(p)}$$

and thus:

$$\left|\int_C\frac{f(z)}{g(z)}dz\right|\leq L\frac{h(p)}{w(p)}$$

I'd like to put this into practice with a simple example:

$$\int_{c-i\infty}^{c+i\infty}\frac{k^z}{z(z+1)}dz,\;k>1,\;c>1$$

by first considering a finite contour integral along the path shown in the first plot. Note that I choose R (the radius of the circle) so that all poles of the integrand are contained within the contour. In the plot shown, R=3 and c=2. By the Residue Theorem:

$$\int_{C}\frac{k^z}{z(z+1)}dz=\sum\Text{Res}f(z)=\int_{\Gamma_1} f(z)dz+\int_{\Gamma_2}f(z)dz$$

where big C is the entire closed-contour, $\Gamma_1$ is the straight-line contour and $\Gamma_2$ is the contour around the arc of the circle. Now, I eventually wish to take the limit of the last expression as t goes to infinity and I wish to know how will this affect the integral around $\Gamma_2$. I can do this by estimating an upper bound for this integral and taking the limit of that expression as t goes to infinity.

First consider the numerator of the integrand and how it varies along $\Gamma_2$:

$$\begin{align*} e^{zln(k)}&=e^{(x+iy)ln(k)}\\ &=e^{xln(k)+iln(k)y}\\ &=e^{xln(k)}\left[Cos(yln(k))+iSin(yln(k))\right]\\ &=k^x\left[Cos(yln(k))+iSin(yln(k))\right]\\ \end{align}$$

Note that along $\Gamma_2$ the real component of z, (x), will always be less than or equal to c so that I can write:

$$|k^z|\leq k^c$$

Now for the denominator:

$$z(z+1)$$

I wish to find some upper bound in terms of the parameter t. Note that R in the diagram below is larger in terms of absolute value than the largest pole of the integrand (which is -1). That means along $\Gamma_2$:

$$|z|\geq R$$

and:

$$|z+1|\geq R$$

so that:

$$|z(z+1)|\geq R^2$$

Thus I can say:

$$\left|\int_{\Gamma_2}\frac{k^z}{z(z+2)}dz\right|\leq\int_{\Gamma_2}\left|\frac{k^z}{z(z+1)} \right|dz\leq 2\pi R\frac{k^c}{R^2}=2\pi\frac{k^c}{R}$$

Now R>t from the diagram below so that:

$$\left|\int_{\Gamma_2}\frac{k^z}{z(z+2)}dz\right|<2\pi\frac{k^c}{t}$$

As t goes to infinity, the contribution from this integral approaches zero as shown by the inequality above. Therefore:

$$\int_{c-i\infty}^{c+i\infty}\frac{k^z}{z(z+1)}dz=2\pi i \sum\text{Res} f(z)=\pi i$$

Anyway, I need to get good at this if I'm to have any chance of understanding the formula above. Makin' progress though and learning a lot about some parts of Analytic Number Theory and the Zeta function . . . interesting.

 

[shmoe]

Looks good.

Your bound $$|z+1|\geq R$$ isn't quite right, the best you can get on your contour is $$|z+1|\geq R-1$$.

Note that you can take any c>0 here, you aren't restricted to c>1.

Something seems to have gone astray in the end when you summed the residues? Were you looking at the special case x=2 perhaps?

Keep it up!

 

[saltydog]

Looks good.

Your bound $$|z+1|\geq R$$ isn't quite right, the best you can get on your contour is $$|z+1|\geq R-1$$.

Note that you can take any c>0 here, you aren't restricted to c>1.

Something seems to have gone astray in the end when you summed the residues? Were you looking at the special case x=2 perhaps?

Keep it up!

Thanks for clearing that up for me . I see now the error I made by looking at the inequality in terms of:

$$\sqrt{(x+1)^2+y^2}$$

and noting that x ranges from -3 to 0 in the contour I studied and therefore as you stated:

$$|z+1|\geq R-1$$

I see also that c need only be greater than zero and yes I was studying the case of c=2, k=2.

I'm currently working with several books, some of which you suggested. It's tough but I seem to learn something new each day. For example, several months ago I worked on inverting a Laplace Transform over an expanding keyhole contour and at the time I was unable to estimate an upper bound for two of the contours. After working on the problem above I went back and was able to do so. Funny how things work out that way.

 

[shmoe]

You saw through my slip that turned your k to an x. I've written "x/n" where you have "k" so many times in similar integrals, my hand is just desperate for an "x" to appear somewhere.

Of course your residue calculation is perfectly correct in this case

 

[saltydog]

Continuing to move towards the integral of interest, I thought it useful to solve the following
first:

$$\int_{a-i\infty}^{a+i\infty}\frac{x^z}{z}dz,\quad x>1,\;a=2$$

by considering the (finite) rectangular contour in the attached plot and evaluating the limit of
the integral as I extend the contour towards infinity. Note the vertices extend from:

$$a\pm ih\quad\text{and}\quad -K\pm ih$$

Thus by the theory of Residues, integration around this contour in a counter-clockwise direction

yields:

$$\oint_C\frac{x^z}{z}dz=\int_{a-ih}^{a+ih}+ \int_{a+ih}^{-K+ih}+ \int_{-K+ih}^{-K-ih}+ \int_{-K-ih}^{a-ih}=2\pi i\left(\text{Res}\frac{x^z}{z}\right)_{z\to 0}= 2\pi i$$

Now, this is important: being a novice at this that I am, I would initially just proceed to
consider the integration in a counter-clockwise fashion. However, in the case of vertical and
horizontal contours, it makes the problem more intutitive (for me anyway) to switch the order of integrations so that the integration is done from "bottom up" and "left to right". Thus:

$$\int_{a-ih}^{a+ih}-\int_{-K+ih}^{a+ih} -\int_{-K-ih}^{-K+ih} +\int_{-K-ih}^{a-ih}=2\pi i$$

or:

$$\int_{a-ih}^{a+ih}=\int_{-K+ih}^{a+ih} +\int_{-K-ih}^{-K+ih} -\int_{-K-ih}^{a-ih}+2\pi i$$

Since:

$$\int_{a-i\infty}^{a+\infty}\frac{x^z}{z}dz= \mathop\lim\limits_{h\to\infty}\int_{a-ih}^{a+ih}\frac{x^z}{z}dz$$

Thus the objective now is to estimate an upper bound for the RHS above and then consider it's limit as both K and h go to infinity in an iterative fashion: first letting K go to infinity
then h.

First consider the left side of the contour:

$$\int_{-K-ih}^{-K+ih} \frac{x^z}{z}dz$$

Similar to the post above, it can be shown that along this contour:

$$|x^z|\leq x^{-K}\quad\text{and}|z|\geq K$$

thus:

$$\left|\int_{-K-ih}^{-K+ih}\frac{x^z}{z}dz\right|\leq 2h\frac{x^{-K}}{K}$$

Now, the top horizontal contour,

$$\int_{-K+ih}^{a+ih} \frac{x^z}{z}dz$$

can be bounded by a real integral as follows:

Along this path, the imaginary part of z does not change; it remains at ih and since:

$$x^z=e^{(\rho+ih)\text{ln}(x)}$$

then:

$$|x^z|\leq x^{\rho}\quad -K\leq\rho\leq h$$

Note also the top diagonal line in the figure. This shows that $|z|\geq h$.

Thus I can write:

$$\left|\int_{-K+ih}^{a+ih}\frac{x^z}{z}dz\right|\leq \int_{-K}^{a}\frac{x^{\rho}}{h}d\rho$$

Evaluating the integral on the RHS I obtain:

$$\left|\int_{-K+ih}^{a+ih}\frac{x^z}{z}dz\right|\leq \left(\frac{x^a-x^{-K}}{hln(x)}\right)$$

Interesting isn't it

Putting this all together I obtain:

$$\left|\int_{a-ih}^{a+ih}\frac{x^z}{z}dz \right|\leq\left\{\frac{2(x^a-x^{-K})}{hln(x)}+ 2h\frac{x^{-K}}{K}+2\pi i\right\}$$

(I need to double check this expression)

Now here comes that part that I'm not very confident of: I first wish to consider the limit of
the RHS above as K goes to infinity (extending the contour first, infinitely far to the left), then take the limit as h goes to infinity (extending the contour from bottom to top).

Taking the first limit and noting that x>1, I obtain:

$$\left|\int_{a-ih}^{a+ih}\frac{x^z}{z}dz \right|\leq\left\{\frac{2x^{a}}{hln(x)}+2\pi i\right\}$$

Now, I extend the contour infinitely far to the top and bottom and consider the limit:

$$\mathop\lim\limits_{h\to\infty}\left|\int_{a-ih}^{a+ih}\frac{x^z}{z}dz\right|= \left|\int_{a-i\infty}^{a+i\infty}\frac{x^z}{z}dz\right|\leq \mathop\lim\limits_{h\to\infty}\left\{ \frac{2x^a}{hln(x)}+2\pi i\right\}=2\pi i$$

Thus:

$$\int_{a-i\infty}^{a+i\infty} \frac{x^z}{z}dz=2\pi i$$

 

[shmoe]

$$\left|\int_{a-ih}^{a+ih}\frac{x^z}{z}dz \right|\leq\left\{\frac{2(x^a-x^{-K})}{hln(x)}+ 2h\frac{x^{-K}}{K}+2\pi i\right\}$$

(I need to double check this expression)

I haven't checked you bounds too carefully (they look good though), this should be something like:
$$\left|\int_{a-ih}^{a+ih}\frac{x^z}{z}dz -2\pi i\right|\leq\left\{\frac{2(x^a-x^{-K})}{hln(x)}+ 2h\frac{x^{-K}}{K}\right\}$$

You took the limits just fine. It's sometimes usefull to have 'truncated' versions, let K go to infinity then leaving h as a parameter to be choosen depending on your needs. In the case of the explicit forumula you'd have a finite sum over the zeros this way, only picking up residues of the zeros whose imaginary part is between -h and h.

This problem can also be handled with the same contour as the last one, the semicircle instead of a rectangle, but more care is required.

 

[saltydog]

Thought I'd post another related problem I'm working on in case others are interested:

$$\int_{1-i\infty}^{1+i\infty}\text{Cot}(z)\frac{x^z}{z}dz,\quad x=2$$

Note it's similar in form to the integral containing the Zeta function:

$$\int_{a-i\infty}^{a+i\infty}\frac{f^{'}(z)}{f(z)}\frac{x^z}{z}dz$$

and can be solved analytically in a similar way. Interesting I think.

Anyway, try figuring it out for fun if you like. The answer is approx 4.1 i.

Regarding von Mangoldt's function: it's tough but getting better each day I work on it. I really need to study the properties of the Zeta function and related Number Theory properties to approach it appropriately. Takes time. Interesting though.