# Methods to evaluate $\int _{a }^{b }\!{\frac {\ln \left( tx + u \right) }{m{x}^{2}+nx +p}}{dx}$

Today I saw a question with an answer that made me rethink of the following question, since it's not the first time I try to find an answer to it. If you look at the answer of Mhenni Benghorbal
here you'll see $2$ interesting integrals, namely: $$\int _{0 }^{\infty }\!{\frac {\ln \left( u \right) }{2+{u}^{2}- 2\,u}}{du} ; \int _{0}^{\infty }\!{\frac {\ln \left( z \right) }{2+{z}^{2}+2\,z}}dz$$ I try to find out if there is a well defined strategy to tackle such integrals. In a more general sense, we have to deal with:

$$\int _{a }^{b }\!{\frac {\ln \left( tx + u \right) }{m{x}^{2}+nx +p}}{dx}$$

Could you help here? Thanks.

• Do you know anything about residues and Cauchy formula? – Norbert Aug 30 '12 at 10:29
• @ Norbert: of course, but less practice. – user 1357113 Aug 30 '12 at 10:30
• @Norbert: thanks. It would be interesting if such integrals may possibly be solved by some real techniques. – user 1357113 Aug 30 '12 at 10:53

## 2 Answers

Related problems: (I), (II). You can use the partial fraction technique combined with the use the dilogarithm function $\operatorname{Li}_{2}(x)$, which is defined by

$$\operatorname{Li}_{2}(x) = \int_{1}^{x} \frac{\ln(t)}{1-t} \,dt \,.$$

Here is an example,

$$\int_{a}^{b} \frac{\ln(x)}{cx+d}dx =- \frac{1}{d}\left( \operatorname{Li}_{2}\left( {\frac {c+da}{c}} \right) +\ln \left(a\right) \ln \left( {\frac {c+da}{c}} \right) -\operatorname{Li}_{2} \left( { \frac {c+bd}{c}} \right) -\ln \left( b \right) \ln \left( {\frac {c+ bd}{c}} \right) \right)$$

Note that the above integral is undefined for $$\left(a < -\frac{c}{d}, -\frac{c}{d} < b \right)$$

For partial response