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In this question, the OP poses the following definite integral, which just happens to vanish: $$\int_1^\sqrt2 \frac{1}{x}\ln\bigg(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\bigg)dx=0$$ As noticed by one commenter to the question, the only zero of the integrand is at $x=\sqrt[3]{2}$, meaning that the integral of the integrand from $x=1$ to $x=\sqrt[3]{2}$ is the additive inverse of the integral of the integrand from $x=\sqrt[3]{2}$ to $x=\sqrt{2}$.

This suggests some sort of symmetry obtainable by a substitution, but I cannot find an appropriate substitution or cancellation. It seems like the answer should be much simpler than those posted to the linked question.

Any ideas?

EDIT: I believe that this more general integral also vanishes: $$\int_1^{\sqrt{t}}\frac{1}{x}\ln\bigg(\frac{t-sx^2+x^4}{tx-sx^2+x^3}\bigg)dx=0$$

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    $\begingroup$ @Rócherz Of course it isn't symmetric about $\sqrt[3]{2}$; I simply mean that perhaps an appropriate substitution might lead to a cancellation. $\endgroup$ – Frpzzd Nov 20 '18 at 0:47
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We can indeed prove the result by symmetry. Put in other words we desire to show that: $$I=\int_1^\sqrt2 \frac{1}{x}\ln\bigg(\frac{x^4-2x^2+2}{x^2-2x+2}\bigg)dx=\color{green}{\int_1^\sqrt 2\frac{1}{x}\ln x dx=\frac18\ln^2 2}$$ First let us take the LHS integral and split it in two parts: $$I=\color{red}{\int_1^\sqrt2 \frac{\ln(x^4-2x^2+2)}{x}dx}-\color{blue}{\int_1^\sqrt2 \frac{\ln(x^2-2x+2)}{x}dx}$$ For the second one $(I_2)$ we will substitute $\displaystyle{x=\frac{2}{t}\rightarrow dx=-\frac{2}{t^2}dt}$: $$I_2=\int_{\sqrt 2}^2 \frac{\ln\left(\frac{2(t^2-2t+2)}{t^2}\right)}{\frac{2}{t}}\frac{2}{t^2}dt\overset{t=x}=\int_{\sqrt 2}^2 \frac{\ln(x^2-2x+2)}{x}dx+\int_{\sqrt 2}^2\frac{\ln 2 -2\ln x}{x}dx$$ Adding with the original $I_2$ leads to: $${2I_2=\int_1^2\frac{\ln(x^2-2x+2)}{x}dx+\int_{\sqrt 2}^2\frac{\ln 2 -2\ln x}{x}dx}$$ $${\Rightarrow I_2=\frac12\int_1^2 \frac{\ln(x^2-2x+2)}{x}dx-\frac{1}{8}\ln^22}\overset{x=t^2}=\color{blue}{\int_1^\sqrt{2}\frac{\ln(t^4-2t^2+2)}{t}dt-\frac{1}{8}\ln^2 2}$$ $$I=\color{red}{\int_1^\sqrt2 \frac{\ln(x^4-2x^2+2)}{x}dx}-\color{blue}{\int_1^\sqrt{2}\frac{\ln(x^4-2x^2+2)}{x}dx+\frac{1}{8}\ln^2 2}=\color{green}{\frac18\ln^2 2}$$


Your conjecture is indeed also correct since by the exact same method we can show that: $$\int_1^{\sqrt{t}}\frac{1}{x}\ln\bigg(\frac{t-sx^2+x^4}{t-sx+x^2}\bigg)dx=\int_1^\sqrt{t} \frac{1}{x}\ln xdx=\frac{1}{8}\ln^2t$$ And this time after we split the LHS integral into two parts, we will substitute in the second integral $\displaystyle{x=\frac{t}{y}}$ ($t$ is a constant here), followed by an addition with the original $I_2$ from there and the result follows. Of course it is valid in the following form too: $$\int_1^{\sqrt{t}}\frac{1}{x}\ln\bigg(\frac{t+sx^2+x^4}{t+sx+x^2}\bigg)dx=\int_1^\sqrt{t} \frac{1}{x}\ln xdx$$

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