# 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 1591719 Aug 30 '12 at 10:30
• @Norbert: thanks. It would be interesting if such integrals may possibly be solved by some real techniques. – user 1591719 Aug 30 '12 at 10:53

## 3 Answers

The answer below will restrict itself to the evaluation of integrals of the form $$\int_{a}^{b}\mathrm{d}x\,\frac{\ln{\left(cx+d\right)}}{px^{2}+qx+r}$$ for the case in which $$q^{2}-4pr<0$$ since the alternative cases are more than adequately addressed elsewhere.

Define the function $$\mathcal{I}:\mathbb{R}\times\mathbb{R}_{>0}\times\mathbb{R}_{\ge0}\times\mathbb{R}_{>0}\rightarrow\mathbb{R}$$ via the definite integral

\begin{align} \mathcal{I}{\left(a,b,c,z\right)} &:=\int_{0}^{z}\mathrm{d}x\,\frac{2b\ln{\left(x+c\right)}}{\left(x+a\right)^{2}+b^{2}}.\\ \end{align}

Also, define the Clausen function (of order 2) for real arguments by the integral representation

$$\operatorname{Cl}_{2}{\left(\theta\right)}:=-\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(\left|2\sin{\left(\frac{\varphi}{2}\right)}\right|\right)};~~~\small{\theta\in\mathbb{R}}.$$

We show below that the integral $$\mathcal{I}$$ can be evaluated in terms of Clausen functions and elementary functions.

Suppose $$\left(a,b,c,z\right)\in\mathbb{R}\times\mathbb{R}_{>0}\times\mathbb{R}_{\ge0}\times\mathbb{R}_{>0}$$, and set

$$\alpha:=\arctan{\left(\frac{a}{b}\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right),$$

$$\beta:=\arctan{\left(\frac{z+a}{b}\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right),$$

$$\gamma:=\arctan{\left(\frac{c-a}{b}\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$$

Then,

\begin{align} \mathcal{I}{\left(a,b,c,z\right)} &=\left[\arctan{\left(\frac{z+a}{b}\right)}-\arctan{\left(\frac{a}{b}\right)}\right]\ln{\left(\left(a-c\right)^{2}+b^{2}\right)}\\ &~~~~~+\operatorname{Cl}_{2}{\left(2\alpha+2\gamma\right)}-\operatorname{Cl}_{2}{\left(2\beta+2\gamma\right)}+\operatorname{Cl}_{2}{\left(\pi-2\alpha\right)}-\operatorname{Cl}_{2}{\left(\pi-2\beta\right)}.\\ \end{align}

Proof:

\begin{align} \mathcal{I}{\left(a,b,c,z\right)} &=\int_{0}^{z}\mathrm{d}x\,\frac{2b\ln{\left(x+c\right)}}{\left(x+a\right)^{2}+b^{2}}\\ &=\int_{0}^{z}\mathrm{d}x\,\frac{2b\ln{\left(\left|x+c\right|\right)}}{\left(x+a\right)^{2}+b^{2}}\\ &=\int_{\frac{a}{b}}^{\frac{z+a}{b}}\mathrm{d}y\,\frac{2b^{2}\ln{\left(\left|by-a+c\right|\right)}}{\left(by-a+a\right)^{2}+b^{2}};~~~\small{\left[x=by-a\right]}\\ &=\int_{\frac{a}{b}}^{\frac{z+a}{b}}\mathrm{d}y\,\frac{2\ln{\left(\left|by+c-a\right|\right)}}{y^{2}+1}\\ &=\int_{\tan{\left(\alpha\right)}}^{\tan{\left(\beta\right)}}\mathrm{d}y\,\frac{2\ln{\left(\left|by+b\tan{\left(\gamma\right)}\right|\right)}}{y^{2}+1}\\ &=\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\frac{2\sec^{2}{\left(\varphi\right)}\ln{\left(\left|b\tan{\left(\varphi\right)}+b\tan{\left(\gamma\right)}\right|\right)}}{\tan^{2}{\left(\varphi\right)}+1};~~~\small{\left[y=\tan{\left(\varphi\right)}\right]}\\ &=2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\left|b\tan{\left(\varphi\right)}+b\tan{\left(\gamma\right)}\right|\right)}\\ &=2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(b\right)}+2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\left|\tan{\left(\varphi\right)}+\tan{\left(\gamma\right)}\right|\right)}\\ &=2\ln{\left(b\right)}\int_{\alpha}^{\beta}\mathrm{d}\varphi+2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\left|\frac{\sin{\left(\varphi\right)}\cos{\left(\gamma\right)}+\cos{\left(\varphi\right)}\sin{\left(\gamma\right)}}{\cos{\left(\varphi\right)}\cos{\left(\gamma\right)}}\right|\right)}\\ &=2\left(\beta-\alpha\right)\ln{\left(b\right)}+2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\left|\frac{\sin{\left(\varphi+\gamma\right)}}{\cos{\left(\varphi\right)}\cos{\left(\gamma\right)}}\right|\right)}\\ &=2\left(\beta-\alpha\right)\ln{\left(b\right)}+2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\left|\frac{1}{\cos{\left(\gamma\right)}}\right|\right)}\\ &~~~~~+2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\left|\frac{\sin{\left(\varphi+\gamma\right)}}{\cos{\left(\varphi\right)}}\right|\right)}\\ &=2\left(\beta-\alpha\right)\ln{\left(b\right)}+2\left(\beta-\alpha\right)\ln{\left(\left|\frac{1}{\cos{\left(\gamma\right)}}\right|\right)}\\ &~~~~~+2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\left|\frac{2\sin{\left(\varphi+\gamma\right)}}{2\cos{\left(\varphi\right)}}\right|\right)}\\ &=\left(\beta-\alpha\right)\ln{\left(b^{2}\right)}+\left(\beta-\alpha\right)\ln{\left(\frac{1}{\cos^{2}{\left(\gamma\right)}}\right)}\\ &~~~~~+2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\frac{\left|2\sin{\left(\varphi+\gamma\right)}\right|}{\left|2\cos{\left(\varphi\right)}\right|}\right)}\\ &=\left(\beta-\alpha\right)\ln{\left(b^{2}\right)}+\left(\beta-\alpha\right)\ln{\left(\sec^{2}{\left(\gamma\right)}\right)}\\ &~~~~~+2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\left|2\sin{\left(\varphi+\gamma\right)}\right|\right)}\\ &~~~~~-2\int_{\alpha}^{\beta}\mathrm{d}\varphi\,\ln{\left(\left|2\cos{\left(\varphi\right)}\right|\right)}\\ &=\left(\beta-\alpha\right)\ln{\left(b^{2}\right)}+\left(\beta-\alpha\right)\ln{\left(1+\tan^{2}{\left(\gamma\right)}\right)}\\ &~~~~~+2\int_{\alpha+\gamma}^{\beta+\gamma}\mathrm{d}\varphi\,\ln{\left(\left|2\sin{\left(\varphi\right)}\right|\right)};~~~\small{\left[\varphi\mapsto\varphi-\gamma\right]}\\ &~~~~~+2\int_{\frac{\pi}{2}-\beta}^{\frac{\pi}{2}-\alpha}\mathrm{d}\varphi\,(-1)\ln{\left(\left|2\sin{\left(\varphi\right)}\right|\right)};~~~\small{\left[\varphi\mapsto\frac{\pi}{2}-\varphi\right]}\\ &=\left(\beta-\alpha\right)\ln{\left(b^{2}\right)}+\left(\beta-\alpha\right)\ln{\left(1+\left(\frac{c-a}{b}\right)^{2}\right)}\\ &~~~~~+\int_{2\beta+2\gamma}^{2\alpha+2\gamma}\mathrm{d}\varphi\,(-1)\ln{\left(\left|2\sin{\left(\frac{\varphi}{2}\right)}\right|\right)}\\ &~~~~~+\int_{\pi-2\beta}^{\pi-2\alpha}\mathrm{d}\varphi\,(-1)\ln{\left(\left|2\sin{\left(\frac{\varphi}{2}\right)}\right|\right)};~~~\small{\left[\varphi\mapsto\frac{\varphi}{2}\right]}\\ &=\left(\beta-\alpha\right)\ln{\left(\left(a-c\right)^{2}+b^{2}\right)}\\ &~~~~~+\operatorname{Cl}_{2}{\left(2\alpha+2\gamma\right)}-\operatorname{Cl}_{2}{\left(2\beta+2\gamma\right)}\\ &~~~~~+\operatorname{Cl}_{2}{\left(\pi-2\alpha\right)}-\operatorname{Cl}_{2}{\left(\pi-2\beta\right)}\\ &=\left[\arctan{\left(\frac{z+a}{b}\right)}-\arctan{\left(\frac{a}{b}\right)}\right]\ln{\left(\left(a-c\right)^{2}+b^{2}\right)}\\ &~~~~~+\operatorname{Cl}_{2}{\left(2\alpha+2\gamma\right)}-\operatorname{Cl}_{2}{\left(2\beta+2\gamma\right)}+\operatorname{Cl}_{2}{\left(\pi-2\alpha\right)}-\operatorname{Cl}_{2}{\left(\pi-2\beta\right)}\blacksquare.\\ \end{align}

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