# Strange log identity

1. Apr 21, 2008

### robousy

Hey folks,

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

$$\frac{1}{2}\sum \int\frac{d^{2n}k}{(2\pi)^{2n}}log(k^2+\frac{m^2}{L^2})$$

as

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

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

2. Apr 21, 2008

### Gib Z

We can reduce this problem to showing if

$$\log t = - \int^{\infty}_0 \frac{e^{-ts}}{s} ds$$.

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: Apr 21, 2008
3. Apr 21, 2008

### benorin

I the correct formula is

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

Check:

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

4. Apr 21, 2008

### Gib Z

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 $$I(t) = - \int^{\infty}_0 \frac{e^{-ts}}{s} ds$$

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

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

$$= \int^{\infty}_0 e^{-ts} ds = \frac{-e^{-ts}}{t}|^{\infty}_0 = \frac{1}{t}$$

$$I(t) = \log_e (t) + C$$

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

Hence, $$I(t) = \log_e (t) - \int^{\infty}_0 \frac{e^{-s}}{s} ds$$ which is the same result.

5. Apr 22, 2008

### robousy

Ok guys, first of all thanks for helping me work through this, Gib Z, thats 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:

$$\log t = - \int^{\infty}_0 \frac{e^{-ts}}{s} ds$$

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

But the proof demonstrates that

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

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. Apr 22, 2008

### Gib Z

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. Apr 23, 2008

### robousy

Hey, the dash means that the m=0 term is omitted.

Rich

8. Apr 23, 2008

### Gib Z

Why do they write it like that when it's so much more nicely written

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

:(

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. Apr 23, 2008

### robousy

Hey GibZ.
Its just part of the regularization. He did in fact originally write it as

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

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.