On 10 23 , 11 46 , Raymond Manzoni wrote:
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

Many thanks for your clear and elegant derivation! You are so cool~!
I'm really appreciated.

However, I still have some problems to consult you (I'm bad at complex
analysis :-( )
Actually, first time I met the tough integral, i.e. I(m2), I tried to
rewrite the integrand as log[(x-a)(x-b)], then to integrate it by
part. But I don't know how to deal with each situation, I mean, when s
4m^2 or s 4m^2, the singular points x = a and x = b are at
different positions, I don't know if the positions of the
singularities affect the solution.(actually, the solution in the case
of s 4m^2 should have imaginary part, but for the case of s 4m^2,
the solution is real)
I can find that, when s 4m^2, the singularities a and b are near
the real axis between the interval [0,1]. Otherwise I can do nothing.
(:-P) From your previous article, the case of s 4m^2, there is an
imaginary part if we note log(-1) = i*pi. (Here is part of your answer
log((sqrt(1-4*m2/s)+1)/(sqrt(1-4*m2/s)-1)) -- the denominator is
negative)

The sign of the imaginary part if quite important for physics since
it affects the scattering process. But I can't work it
out...Nevertheless, your instruction really helped me a lot. Again,
you are so cool!

Sincerely