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Does $$\int_0^\infty \frac{x^a\ln(x)}{(x^2-bx+1)^{a+1}}{\rm d}x = 0$$ for all real numbers $a > 0$ and $b < 2$?

I came across this conjecture by showing its validity for the positive integer values of $a$ only.

To derive the result for positive integer $a$, make the substitution $u=\frac1x$ on $$ \int_0^\infty \frac{\ln(1+x^k)}{x^2-bx+1}{\rm d}x $$

and we get $$ \int_0^\infty \frac{\ln(1+x^k)}{x^2-bx+1}{\rm d}x = \int_0^\infty \frac{\ln(1+x^k)}{x^2-bx+1}{\rm d}x - k \int_0^\infty \frac{\ln(x)}{x^2-bx+1}{\rm d}x \\ \implies \int_0^\infty \frac{\ln(x)}{x^2-bx+1}{\rm d}x = 0 $$

From here, we can use Leibniz's rule of integration (differentiating with respect to $b$) $n$ times to retrieve $$\int_0^\infty \frac{x^n\ln(x)}{(x^2-bx+1)^{n+1}}{\rm d}x = 0$$

Yet I'm guessing that it is also valid for any real positive value of $a$ from numerical evidence. Complex methods are welcome but I won't really be familiar with them, so I would prefer sticking to real methods.

Jia Ming جيا ميڠ
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1 Answers1

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Of course it does. Just substitute $\displaystyle{x=\frac{1}{t}}$ to get: $$I=\int_0^\infty \frac{x^a\ln(x)}{(x^2-bx+1)^{a+1}}{\rm d}x=\int_\infty^0 \frac{1}{t^a} \frac{\ln\left(\frac{1}{t}\right)}{\left(\frac{1}{t^2}-\frac{b}{t}+1\right)^{a+1}}\frac{-dt}{t^2}$$ $$=\int_0^\infty \frac{t^{2a+2}\ln\left(\frac{1}{t}\right)}{\left(1-bt+t^2\right)^{a+1}}\frac{dt}{t^{a+2}}=\int_0^\infty \frac{t^a\ln\left(\frac{1}{t}\right)}{(t^2-bt+1)^{a+1}}dt=-I$$ $$I=-I\Rightarrow I=0$$

Zacky
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  • Omg I feel so dumb, thank you so much. Standard substitutions slip by me occasionally... – Jia Ming جيا ميڠ Dec 30 '18 at 14:51
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    I like your derivation in your question! A small typo:$$\int_0^\infty \frac{\ln(1+x^k)}{x^2-bx+1}{\rm d}x = \int_0^\infty \frac{\ln(1+x^k)}{x^2-bx+1}{\rm d}x - \color{red}k \int_0^\infty \frac{\ln(x)}{x^2-bx+1}{\rm d}x \implies \int_0^\infty \frac{\ln(x)}{x^2-bx+1}{\rm d}x = 0$$ – Zacky Dec 30 '18 at 14:53