- #1
parton
- 83
- 1
I read the following expression in a book:
[tex] \int_{-\infty}^{\infty} \dfrac{1}{t(1-t)} \log \left| \dfrac{t^{2} q^{2}}{(p-tq)^{2}} \right| ~ dt = - \pi^{2} [/tex]
p and q are both timelike four-vectors, so p², q² > 0
This integral was solved by using the identity
[tex] \lim_{s \to \infty} \int_{-s}^{s} \dfrac{\log \vert t-a \vert}{t-b} ~ dt = \dfrac{\pi^{2}}{2} sign(b - a) [/tex]
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.
Could anyone help me please?
[tex] \int_{-\infty}^{\infty} \dfrac{1}{t(1-t)} \log \left| \dfrac{t^{2} q^{2}}{(p-tq)^{2}} \right| ~ dt = - \pi^{2} [/tex]
p and q are both timelike four-vectors, so p², q² > 0
This integral was solved by using the identity
[tex] \lim_{s \to \infty} \int_{-s}^{s} \dfrac{\log \vert t-a \vert}{t-b} ~ dt = \dfrac{\pi^{2}}{2} sign(b - a) [/tex]
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.
Could anyone help me please?