# Expressing $\int _0^1da\int _0^{1-a}\ln ((a-1+b)^2-4 y a b )+\int_0^1 da\int _0^{1-a}\frac{(a-1+b)^2}{(a-1+b)^2-4 y a b}db$ in terms of dilogarithms

I came across these integrals and I'm trying to rewrite them in terms of Dilogarithms: $\mathrm{Li}_2(z):=-\int_0^z \frac{\mathrm{d}s}{s}\log(1-s)$. Can anyone suggest how to contunue? If there is a good change of variables to be done for instance.

$$\int _0^1\mathrm{d}a\int _0^{1-a}\ln \left((a-1+b)^2-4 y a b\right)\mathrm{d}b\\[7mm]+4 (1-y)\int_0^1 \mathrm{d}a\int _0^{1-a}\frac{(a-1+b)^2}{(a-1+b)^2-4 y a b}\mathrm{d}b$$

I see a pattern in $(a+b-1)^2$ but can't think of a good change of variables to get the second integral in a simple form.

The first integration (that with respect to $b$ say) can be done straightforward but the second integration cannot all be done in analytic form so I'm trying to write it in the form of dilogarithms. Any suggestions how to continue?

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Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. – arjafi May 31 '13 at 7:08
OK the problem was that the characters ran out. But I see your point. – Mauchy May 31 '13 at 7:38