# Homework Help: Logarithm, identity

1. Oct 27, 2009

### parton

I read the following expression in a book:

$$\int_{-\infty}^{\infty} \dfrac{1}{t(1-t)} \log \left| \dfrac{t^{2} q^{2}}{(p-tq)^{2}} \right| ~ dt = - \pi^{2}$$

p and q are both timelike four-vectors, so p², q² > 0

This integral was solved by using the identity

$$\lim_{s \to \infty} \int_{-s}^{s} \dfrac{\log \vert t-a \vert}{t-b} ~ dt = \dfrac{\pi^{2}}{2} sign(b - a)$$

But I don't know how I can apply this identity to the integral above. I don't find any substitution or anything else to 'convert' the integral
in such a shape where I can use these formula.

2. Oct 27, 2009

### tiny-tim

Hi parton!

Try splitting the 1/t(1-t) in partical fractions, and the log into the difference of two logs, and then apply the identity several times.

3. Oct 28, 2009

### parton

I tried to split the integral as you suggested:

$$\int_{-\infty}^{\infty} \dfrac{1}{t(1-t)} \log \left| \dfrac{t^{2} q^{2}}{(p-tq)^{2}} \right| ~ dt = \int_{-\infty}^{\infty} \left[ \dfrac{1}{t} - \dfrac{1}{t-1} \right] \left[ \log \vert t^{2} q^{2} \vert - \log \vert (p-tq)^{2} \vert \right] ~ dt$$

For the first part, I've got:

$$\int_{-\infty}^{\infty} \left[ \dfrac{1}{t} - \dfrac{1}{t-1} \right] \log \vert t^{2} q^{2} \vert \right] ~ dt= - \pi^{2}$$

Here, I decomposed the logarithm again, i.e.

$$\log \vert t^{2} q^{2} \vert^{2} = 2 \log \vert t \vert + \log \vert q^{2} \vert$$

I just used the identity for computing
$$- 2 \int_{-\infty}^{\infty} \dfrac{1}{t-1} \log \vert t \vert = - \pi^{2}$$

The other integrals coming from this part will vanish, for example:

$$2 \int_{-\infty}^{\infty} \dfrac{1}{t} \log \vert t \vert ~ dt = \left[ \log \left( \vert t \vert \right) ^{2} \right]_{-\infty}^{\infty} = 0$$

I argued that: $$\left[ \log \left( \vert t \vert \right) ^{2} \right]_{-s}^{s} = 0$$, so it has also to be valid for s--> infinity. But I'm not sure wheter this argument is correct.

Nevertheless, I don't know how to compute the second part of the originial integral:

$$- \int_{-\infty}^{\infty} \left[ \dfrac{1}{t} - \dfrac{1}{t-1} \right] \log \vert (p-tq)^{2} \vert ~ dt$$

There is this nasty square appearing in the logarithm, and now I don't know how to continue

To you know some way to solve this???

4. Oct 28, 2009

### Hurkyl

Staff Emeritus
Don't you know something about log xe?

5. Oct 28, 2009

### parton

ok, maybe I could simply write:

$$-\int_{-\infty}^{\infty} \left[ \dfrac{1}{t} - \dfrac{1}{t-1} \right] \log \vert (p-tq)^{2} \vert ~ dt = -2 \int_{-\infty}^{\infty} \left[ \dfrac{1}{t} - \dfrac{1}{t-1} \right] \log \vert p -t q \vert ~ dt$$

But how do I solve that? p and q are four-vectors, so I can't simply apply a substitution !?

6. Oct 28, 2009

### tiny-tim

Hi parton!

The difficulty is that p and q are 4-vectors.

But (p - tq)2 is a quadratic in t (with coefficients which are combinations of the coordinates of p and q), so it must be of the form (t - a)(t - b), where (p - aq)2 = (p - bq)2 = 0.

Does that help?

7. Oct 28, 2009

### parton

Yes, it does help, thanks a lot

So, I wrote:

$$(p-tq)^{2} = (t-t_{1}) (t-t_{2})$$

where $$t_{1,2} = \dfrac{1}{q^{2}} \left[ pq \pm \sqrt{ (pq)^{2} - p^{2} q^{2} } \right]$$

And I've got:

$$- \int_{-\infty}^{\infty} \dfrac{1}{t} \left[ \log \vert t - t_{1} \vert + \log \vert t - t_{2} \vert \right] = - \dfrac{\pi^{2}}{2} \left[ sign(-t_{1}) + sign(-t_{2}) \right] = 0$$

This is of course equal to zero, because of the opposite signs of $$t_{1}$$ and $$t_{2}$$.

But the next one isn't so easy:

$$+ \int_{-\infty}^{\infty} \dfrac{1}{t-1} \left[ \log \vert t - t_{1} \vert + \log \vert t - t_{2} \vert \right] = \dfrac{\pi^{2}}{2} \left[ sign(1-t_{1}) + sign(1-t_{2}) \right]$$

Ok, I know that if $$t_{2} < 0 \Rightarrow 1-t_{2} >0 \Rightarrow sign(1-t_{2}) = 1$$

But what is $$sign(1-t_{1})$$ ?

8. Oct 28, 2009

### parton

The equation above is not equal to zero.

If I choose a frame where $$p = (p_{0}, \vec{0})$$ than I have:

$$t_{1,2} = \dfrac{p_{0}}{q^{2}} \left[ q^{0} \pm \vert \vec{q} \vert \right] < 0$$ because of $$q^{2} = q_{0}^{2} - \vec{q} \, ^{2} > 0$$ and $$q_{0} < 0$$.

So I obtain:

$$- \int_{-\infty}^{\infty} \dfrac{1}{t} \left[ \log \vert t - t_{1} \vert + \log \vert t - t_{2} \vert \right] ~ dt = - \dfrac{\pi^{2}}{2} \left[ sign(-t_{1}) + sign(-t_{2}) \right] = - \pi^{2}$$

and

$$+ \int_{-\infty}^{\infty} \dfrac{1}{t-1} \left[ \log \vert t - t_{1} \vert + \log \vert t - t_{2} \vert \right] ~ dt = \dfrac{\pi^{2}}{2} \left[ sign(1-t_{1}) + sign(1-t_{2}) \right] = \pi^{2}$$

and finally:

$$-\int_{-\infty}^{\infty} \left[ \dfrac{1}{t} - \dfrac{1}{t-1} \right] \log \vert (p-tq)^{2} \vert ~ dt = 0$$

Is anything wrong in my argumentation or is it correct?

Last edited: Oct 28, 2009
9. Oct 29, 2009

### parton

By the way,

does anyone know where I can find such identities like

$$\lim_{s \to \infty} \int_{-s}^{s} \dfrac{\log \vert t-a \vert}{t-b} ~ dt = \dfrac{\pi^{2}}{2} sign(b - a)$$

Is there any book where such formulas are "collected"? I looked in different tables of integrals, but couldn't find them.