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Evaluate $$\int_0^1 \frac{\ln{\left(ax^2+2bx+a\right)}}{x^2+1} \; \mathrm{d}x, \;a>b>0$$

I tried- $$=\frac{\pi \ln{a}}{4}+\int_0^1 \frac{\ln{\left(x^2+2tx+1\right)}}{x^2+1} \; \mathrm{d}x$$ Where $t=\frac{b}{a}$ $$=\frac{\pi \ln{a}}{4}+\int_0^1 \frac{\ln{\left({\left(x+t\right)}^2+c\right)}}{x^2+1} \; \mathrm{d}x$$ Where $c=1-t^2$. I dont think this helps much. I also tried analyzing things like $x \to \frac{1}{x}$ in the integral in the second line $$I=\int_0^1 \frac{\ln{\left(x^2+2tx+1\right)}}{x^2+1} \; \mathrm{d}x=\int_1^{\infty} \frac{\ln{\left(x^2+2tx+1\right)}}{x^2+1} \; \mathrm{d}x - 2C$$ $$2I=\int_0^{\infty} \frac{\ln{\left(x^2+2tx+1\right)}}{x^2+1} \; \mathrm{d}x - 2C$$ C is catalans constant. Does this help? Any ideas. (I sense complex analysis is the best here

vitamin d
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1 Answers1

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Concerning the antiderivative, you can write $$\log(a x^2+2bx+a)=\log(a)+\log(x-s)+\log(x-t)$$ where $s$ and $t$ are the complex roots of $ax^2+2bx+a=0$.

On the other side $$\frac 1{x^2+1}=\frac 1{(x+i)(x-i)}=\frac i 2\left(\frac 1{x+i}-\frac 1{x-i}\right)$$ So, we have $$\frac{\log(a x^2+2bx+a)}{x^2+1}=\frac{\log(a)}{x^2+1}+\frac i 2\left(\frac 1{x+i}-\frac 1{x-i}\right)\log(x-s)+\frac i 2\left(\frac 1{x+i}-\frac 1{x-i}\right)\log(x-t)$$ which means four integrals looking like $$\int \frac{\log(x-\alpha)}{x-\beta}\,dx$$ which are not difficult (one integration by part).

Later, make a lot of simplifications to get rid of most of the complex numbers.

Claude Leibovici
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  • I see that the log terms simplify [here][https://www.wolframalpha.com/input/?i=integral+of+ln%28x-s%29%2F%28x%2Bi%29-ln%28x-s%29%2F%28x-i%29%2Bln%28x-t%29%2F%28x%2Bi%29-ln%28x-t%29%2F%28x-i%29++dx+]. But Wolfram didn't simplify the 8 Li terms at x=1 minus at x=0. Is there a way to simplify those? –  Aug 04 '20 at 13:46