Solving Strange Log Identity - Richard

In summary: I have 2 different PDF's of the same paper).I am not familiar with Hurwitz Zeta function, but I'm sure it is something that is only used in certain areas of physics and/or mathematics. So yes, in this context (high-energy physics) it is probably just a convention or notation used to simplify the expression. In summary, the conversation revolved around understanding the first line of equation 44 in a paper about high-energy physics, specifically the use of integrals and logarithms. The conversation also touched on the justification for omitting the m=0 term and the use of different regularization techniques. The experts discussed the physical basis for the notation and suggested further exploration in the Physics section.
  • #1
robousy
334
1
Hey folks,

I'm reading the paper: http://arxiv.org/abs/hep-ph/0301168

and I'm trying to make sense of the first line of eqtn 44 where he states that we can write:

[tex]\frac{1}{2}\sum \int\frac{d^{2n}k}{(2\pi)^{2n}}log(k^2+\frac{m^2}{L^2})[/tex]

as

[tex]-\frac{1}{2}\sum\int\frac{d^{2n}k}{(2\pi)^{2n}}\int_0^\infty ds\frac{1}{s}e^{-(k^2+\frac{m^2}{L^2})s}[/tex]

It seems to me that the last integral in 's' is somehow a way of expressing the log term, but I can't really see how. I tried the integral and it diverges. Any ideas here folks?
Thanks!

Richard
 
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  • #2
We can reduce this problem to showing if

[tex]\log t = - \int^{\infty}_0 \frac{e^{-ts}}{s} ds[/tex].

Show us how you did the integral.

EDIT: Ahh don't worry, One can verify that integral extremely quickly. Differentiate both sides with respect to t, and using the differentiation under the integral sign rule on the RHS. The original question is a valid equality.
 
Last edited:
  • #3
I the correct formula is

[tex]\log t = \int^{\infty}_0\left(e^{-s}-e^{-ts}\right) \frac{ds}{s}[/tex]

Check:

[tex] \int^{\infty}_0\left(e^{-s}-e^{-ts}\right) \frac{ds}{s} = \int_{0}^{\infty}\int_{1}^{t}e^{-ts}\, dt\, ds = \int_{1}^{t}\left[\frac{e^{-ts}}{t}\right]_{s=0}^{\infty} dt = \int_{1}^{t}\frac{dt}{t} = \log t[/tex]
 
  • #4
benorin said:
I the correct formula is

[tex]\log t = \int^{\infty}_0\left(e^{-s}-e^{-ts}\right) \frac{ds}{s}[/tex]

Check:

[tex] \int^{\infty}_0\left(e^{-s}-e^{-ts}\right) \frac{ds}{s} = \int_{0}^{\infty}\int_{1}^{t}e^{-ts}\, dt\, ds = \int_{1}^{t}\left[\frac{e^{-ts}}{t}\right]_{s=0}^{\infty} dt = \int_{1}^{t}\frac{dt}{t} = \log t[/tex]

Ahh sorry, my bad, you are correct. I shouldn't have assumed my constant was equal to zero =]

Here's how to get it using the Differentiation under the Integral sign method;

Let [tex] I(t) = - \int^{\infty}_0 \frac{e^{-ts}}{s} ds[/tex]

[tex] \frac{d}{dt} I(t) = \frac{d}{dt} \left( - \int^{\infty}_0 \frac{e^{-ts}}{s} ds \right )[/tex]

[tex] = \int^{\infty}_0 \frac{\partial}{\partial t} \left( \frac{-e^{-ts}}{s} \right) ds[/tex]

[tex] = \int^{\infty}_0 e^{-ts} ds = \frac{-e^{-ts}}{t}|^{\infty}_0 = \frac{1}{t}[/tex]

[tex] I(t) = \log_e (t) + C[/tex]

Letting t=1, [tex] C = I(1) = - \int^{\infty}_0 \frac{e^{-s}}{s} ds[/tex]

Hence, [tex]I(t) = \log_e (t) - \int^{\infty}_0 \frac{e^{-s}}{s} ds[/tex] which is the same result.
 
  • #5
Ok guys, first of all thanks for helping me work through this, Gib Z, that's a great proof of Benorins formula. Makes perfect sense. I've just worked it through twice to make sure it goes in, and I think its there to stay. :)

Something that was throwing me was an integration formula in Schaums that gave the result as infinite for the given power of 's' in the denominator.

One thing that I'm still unclear on is the 'extra term'.

e.g, my problem filtered down to showing that:

[tex]\log t = - \int^{\infty}_0 \frac{e^{-ts}}{s} ds[/tex]

as used in http://arxiv.org/abs/hep-ph/0301168 eq 44.

But the proof demonstrates that

[tex]\log t = \int^{\infty}_0\left(e^{-s}-e^{-ts}\right) \frac{ds}{s}[/tex]

Do you guys think that because the result is an energy then you can just subtract off the constant?

Thanks again for the help!

Rich
 
  • #6
I am uncertain as to what the little dash ( ' ) just after the sigma refers to. If it means derivative, that may be where the constant goes.
 
  • #7
Hey, the dash means that the m=0 term is omitted.

Rich
 
  • #8
Why do they write it like that when it's so much more nicely written

[tex]
\sum_{m=1}^{\infty} \int\frac{d^{2n}k}{(2\pi)^{2n}} \log_e \left(k^2+\frac{m^2}{ L^2}\right)
[/tex]

:(

As the where that constant goes, it definitely isn't mathematically justified so there must be some physical basis for it. I suggest you create another thread in the Physics section, link to this thread and ask them over there.
 
  • #9
Hey GibZ.
Its just part of the regularization. He did in fact originally write it as

[tex] \sum_{m=1}^{\infty} \int\frac{d^{2n}k}{(2\pi)^{2n}} \log_e \left(k^2+\frac{m^2}{ L^2}\right) [/tex]

and perform the dimensional regularization, but it led to derivatives of the Hurwitz Zeta function, so he solved it again using a different regularization process that expressed the results in terms of modified bessel functions of the 2nd kind, without any derivs of the zeta function.
 

1. What is a "strange log identity"?

A strange log identity refers to a mathematical equation or expression that involves logarithmic functions and has a unique or unexpected solution. These identities can be challenging to solve and often require advanced mathematical techniques.

2. Who is Richard and why is this identity named after him?

Richard is most likely a mathematician or scientist who discovered or popularized the specific log identity in question. Often in mathematics, equations or concepts are named after the person who first discovered or extensively studied them.

3. Why is it important to solve strange log identities?

Solving strange log identities can help expand our understanding of logarithmic functions and their properties. It also allows us to approach more complex mathematical problems and find solutions that may not be easily apparent.

4. What are some strategies for solving strange log identities?

One strategy is to try to simplify the expression by using log rules and properties. Another approach is to substitute known values or variables to see if a pattern emerges. Additionally, using graphing or computational tools can also aid in solving these identities.

5. Can strange log identities have more than one solution?

Yes, it is possible for a strange log identity to have multiple solutions. In some cases, there may be infinitely many solutions. It is important to carefully check the validity of each solution to ensure it satisfies the original equation.

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