Re: fish diagram calculation in Minkowski space

Raymond Manzoni a écrit :
(snip long derivation)

so that f(s)-log(s) will be -2 and finally
for m2/s > 1/4 :
I(m2,s)= log(m2) + 2*(sqrt(4*m2/s-1)*arctan(1/sqrt(4*m2/s-1))-1)

for m2/s < 1/4 we may replace artan(x) by log((1+i*x)/(1-i*x))/(2*i)
to get the corresponding formula :
I(m2,s)= log(m2) +
sqrt(1-4*m2/s)*log((sqrt(1-4*m2/s)+1)/(sqrt(1-4*m2/s)-1)) - 2


Sorry for the laborious derivation... :-(
The last equation may be found quickly by rewriting
I(m2)= int_0^1 log(m2 - s*x*(1-x)) dx
as
I(m2)= int_{-1/2}^{1/2} log(t^2-a^2) dt + log(s)
with a^2= 1/4-m2/s and t= x-1/2
I(m2)= int_{-1/2}^{1/2} log(t-a) + log(t+a) dt +log(s)
and use int log(x) dx= x*log(x)-x to get

I(m2)= (t-a)*log(t-a)+(t+a)*log(t+a)-2*t |_{-1/2}^{1/2} +log(s)
= t*log(t^2-a^2) + a*log((t+a)/(t-a)) |_{-1/2}^{1/2} -2 + log(s)
= log(1/4-a^2)+log(s)+ a*log((1/2+a)*(-1/2-a)/((-1/2+a)*(1/2-a)))-2
= log(m2)+ 2*a*log((1/2+a)/(1/2-a))- 2
up to a minus sign or too ;-)

Raymond

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