Show an Inequality $\int_{0}^\infty\frac{\sqrt x\ln(1+x+x^2)}{1+x^2}<(1+\sqrt 3)\pi$

Question

Update

Thanks everyone in the thread answer me this question by evaluating the integral on the left explicitly, I really appreciate that!

But on the other hand, may I still request a method to prove this inequality without evaluating the integral explicitly? (For example, using some tricks for integral or write $(1+\sqrt 3)\pi$ as another integral of function $g(x)$ and $g(x)>\frac{\sqrt x\ln(1+x+x^2)}{1+x^2}$ for all $x$...?)

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A few days ago, someone has asked me to do this integral $$\int_{0}^\infty\frac{\sqrt x\ln(1+x+x^2)}{1+x^2}\mathrm dx\approx8.58193$$

I previously want to solve it by complex analysis (using contour integral) but failed... I first want to try the residues, and I divided $\pi$ for the numeral value, it is close to $2.7317\dots$ and I realize that the zeroes in $1+x+x^2$ and $1+x^2$ in the upper half in the complex plane is $\frac{1+\sqrt 3i}{2}$ and $i$, so this $2.7317\dots$ makes me think it is close to but a bit smaller than $1+\sqrt 3$. Is it a coincidence?

Also, may I ask here how this integral should be calculated besides the fact that the value is close to $(1+\sqrt 3)\pi$...?

Answer 1

Here is to explicitly evaluate the integral, via $\int_0^\infty \frac{\ln(a^2+x^2)}{b^2+x^2}dx=\frac \pi b\ln(a+b)$ \begin{align} & \int_{0}^\infty\frac{\sqrt x\ln(1+x+x^2)}{1+x^2}\overset{x\to x^2}{dx}\\ =&\ 2 \int_{0}^\infty\frac{x^2\ln(1+x^2+x^4)}{1+x^4}dx = 4\ \Re \int_{0}^\infty\frac{x^2\ln(e^{i\frac\pi3} +x^2)}{1+x^4}dx\\ = & \ 2\ \Re \int_{0}^\infty\frac{\ln(e^{i\frac\pi3} +x^2)}{e^{i\frac\pi2} +x^2}+ \frac{\ln(e^{i\frac\pi3} +x^2)}{e^{-i\frac\pi2} +x^2} \ dx\\ = & \ 2\pi \ \Re\bigg[e^{-i\frac\pi4} \ln(e^{i\frac\pi4}+ e^{i\frac\pi6} ) + e^{i\frac\pi4} \ln(e^{-i\frac\pi4}+ e^{i\frac\pi6} )\bigg]\\ = &\ \pi{\sqrt2}\bigg[\ln(\sqrt3+\sqrt2)+\frac\pi4\bigg] \end{align}

Answer 2

$$\int_0^\infty\frac{\sqrt x\ln(1+x+x^2)}{1+x^2}dx=\int_{0}^\infty\frac{\sqrt x\ln x}{1+x^2}dx+\int_{0}^\infty\frac{\ln(x+\frac1x+1)}{\sqrt x(x+\frac1x)}dx=I_1+I_2\tag{0}$$ Using the keyhole contour in the complex plane $$(1-e^{2\pi i})\int_0^\infty\frac{x^s}{1+x^2}dx=2\pi i\underset{x=e^\frac{\pi i}2;\,e^\frac{3\pi i}2}{\operatorname{Res}}\frac{x^s}{1+x^2}$$ $$\int_0^\infty\frac{x^s}{1+x^2}dx=\frac\pi2\frac1{\cos\frac{\pi s}2}$$ $$I_1=\frac{d}{ds}\,\bigg|_{s=\frac12}\frac\pi2\frac1{\cos\frac{\pi s}2}=\frac{\pi^2}{2\sqrt2}\tag{1}$$ $$I_2=\int_{0}^\infty\frac{\ln(x+\frac1x+1)}{\sqrt x(x+\frac1x)}dx\overset{t=\sqrt x}{=}2\int_0^\infty\frac{\ln(x^2+\frac1{x^2}+1)}{x^2+\frac1{x^2}}dx=2\int_0^\infty\frac{\ln((x-\frac1x)^2+3)}{(x-\frac1x)^2+2}dx$$ Using the Glasser' master theorem, or just making the substitution $t=\frac1x$ $$I_2=2\int_0^\infty\frac{\ln(x^2+3)}{x^2+2}dx=2\Re\int_{-\infty}^\infty\frac{\ln(\sqrt3-ix)}{x^2+2}dx$$ Closing the contour of integration in the upper half-plane (where the logarithm does not have branch points) $$I_2=2\Re\,2\pi i\underset{z=i\sqrt2}{\operatorname{Res}}\frac{\ln(\sqrt3-iz)}{(z-i\sqrt2)(z+i\sqrt2)}=\pi\sqrt2\ln(\sqrt3+\sqrt2)\tag{2}$$ $$I=I_1+I_2=\pi\sqrt2\left(\frac\pi4+\ln(\sqrt2+\sqrt3)\right)$$

Answer 3

Using $x=t^2$ $$I=\frac{\sqrt x\ln(1+x+x^2)}{1+x^2}\,dx=\int\frac{2 t^2 \log\left(t^4+t^2+1\right)}{t^4+1}\,dt$$ $$\frac{2 t^2}{t^4+1}=\frac{2 t^2}{(t^2+i)(t^2-i)}=\frac{1}{t^2+i}+\frac{1}{t^2-i}$$ $$\log\left(t^4+t^2+1\right)=\log(t^2-a)+\log(t^2-b)$$ where $$a=-\frac{1+i \sqrt{3}}{2}\qquad \qquad \text{and}\qquad \qquad b=-\frac{1-i \sqrt{3}}{2}$$ If $\Re(c)<0$ $$\int_0^\infty \frac {\log(t^2-c)}{t^2+i}=$$ $$\frac{\left(\frac{1}{4}+\frac{i}{4}\right) \pi \left(-4 \sqrt{-c} \tanh ^{-1}\left(\sqrt[4]{-1} \sqrt{c}\right)+\sqrt{c} (\pi -2 i \log (1-i c))\right)}{\sqrt{2} \sqrt{c}}$$ and $$\int_0^\infty \frac {\log(t^2-c)}{t^2-i}=$$ $$\frac{1}{4} \sqrt[4]{-1} \pi \left(2 \log (1+i c)+\frac{4 i \sqrt{c} \tanh ^{-1}\left((-1)^{3/4} \sqrt{c}\right)}{\sqrt{-c}}-i \pi \right)$$

Combining all the above, simplifying the complex numbers and using the logarithmic represntation of the hyperbolic tangent function $$I=\frac{\pi \left(\pi +4 \tanh ^{-1}\left(\sqrt{\frac{2}{3}}\right)\right)}{2 \sqrt{2}}$$

Answer 4

Via $\sqrt x\mapsto x $ transforms the integral into

$$ I=2 \int_0^{\infty} \frac{x^2 \ln \left(1+x^2+x^4\right)}{1+x^4} d x $$

Via $x\mapsto \frac{1}{x} $ transforms the integral into

$$ I=2 \int_0^{\infty} \frac{\ln \left(1+x^2+x^4\right)-4 \ln x}{x^4+1} d x $$ Averaging them gives $$ I= \underbrace{ \int_0^{\infty} \frac{\left(x^2+1\right) \ln \left(1+x^2+x^4\right)}{x^4+1} d x}_{J} -4 \underbrace{ \int_0^{\infty} \frac{\ln x}{x^4+1} d x}_{=-\frac{\pi^2}{8\sqrt 2} \textrm{ by Beta function }} $$ $$ \begin{aligned} J& =\int_0^{\infty} \frac{\left(x^2+1\right) \ln \left(x^2\left(x^2+\frac{1}{x^2}+1\right)\right)}{x^4+1} d x \\ & =\int_0^{\infty} \frac{\ln \left[\left(x-\frac{1}{x}\right)^2+3\right]}{\left(x-\frac{1}{x}\right)^2+2}d \left(x-\frac{1}{x}\right) +2 \underbrace{ \int_0^{\infty} \frac{\left(x^2+1\right) \ln x}{x^4+1}}_{=0 \textrm{ by } x\mapsto \frac{1}{x} }\\ & =\int_{-\infty}^{\infty} \frac{\ln \left(x^2+3\right)}{x^2+2} d x \quad (\textrm{ via }x-\frac{1}{x} \mapsto x) \\ & =2 \Re \int_{-\infty}^{\infty} \frac{\ln (x+\sqrt{3} i)}{x^2+2} d x \\ & =2 \Re\left[2 \pi i\left(\lim _{z \rightarrow \sqrt{2} i}(z-\sqrt{2} i) \frac{\ln (z+\sqrt{3} i)}{z^2+2}\right)\right] \\ & =4 \pi \Re\left[i \cdot \frac{\ln (\sqrt{2} i+\sqrt{3} i)}{2 \sqrt{2} i}\right] \\ & =\sqrt{2} \pi \ln (\sqrt{2}+\sqrt{3}) \end{aligned} $$

Hence $$ \boxed{\int_0^{\infty} \frac{\sqrt{x} \ln \left(1+x+x^2\right)}{1+x^2} d x =\sqrt{2} \pi\left[\ln (\sqrt{2}+\sqrt{3})+\frac{\pi}{4}\right]} $$

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Answer 5

Feynman’s trick

Via $\sqrt x\mapsto x$ transform the integral into

$$ \int_0^{\infty} \frac{\sqrt{x} \ln \left(1+x+x^2\right)}{1+x^2} d x = 2 \int_0^{\infty} \frac{x^2 \ln \left(1+x^2+x^4\right)}{1+x^4} d x $$

Differentiating the parametrized integral $$ I(a)= 2\int_0^{\infty} \frac{x^2 \ln \left(a(1+x^4)+x^2\right)}{1+x^4} d x , \textrm{ where }a\ge 0. $$ w.r.t. $a$ yields $$ \begin{aligned} I^{\prime}(a) & =2 \int_0^{\infty} \frac{x^2}{a\left(1+x^4\right)+x^2} d x \\ & =2 \int_0^{\infty} \frac{\frac{1}{x^2}}{a\left(1+\frac{1}{x^4}\right)+\frac{1}{x^2}} \frac{d x}{x^2} \quad \textrm{ via } x\mapsto \frac{1}{x} \\ & =2 \int \frac{1}{a\left(1+x^4\right)+x^2} d x \end{aligned} $$ Averaging the two versions brings us $$ \begin{aligned} 2 I^{\prime}(a) & =2 \int_0^{\infty} \frac{1+x^2}{a\left(1+x^4\right)+x^2} d x \\ & =\frac{2}{a} \int_0^{\infty} \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}+\frac{1}{a}} d x \\ & =\frac{2}{a} \int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+\left(2+\frac{1}{a}\right)} \\ & =\frac{2\pi}{a \sqrt{2+\frac{1}{a}}}\left[\tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2+\frac{1}{a}}}\right)\right]_0^{\infty} \\ I^{\prime}(a) & =\frac{\pi}{\sqrt{a(2a+1)}} \end{aligned} $$ Integrating back from $a=0$ to $1$ yields $$ I(1)-I(0)=\pi \int_0^1 \frac{d a}{\sqrt{a(2 a+1)}}=\sqrt 2 \pi\operatorname{arcsinh{\sqrt 2}}=\sqrt{2} \pi \ln (\sqrt{2}+\sqrt{3}) $$ Hence $$ \begin{aligned} \int_0^{\infty} \frac{\sqrt{x} \ln \left(1+x+x^2\right)}{1+x^2} d x & =\sqrt{2} \pi \ln (\sqrt{2}+\sqrt{3})+4 \int_0^{\infty} \frac{x^2 \ln x}{1+x^4}dx \\ & =\sqrt{2} \pi\left[\ln (\sqrt{2}+\sqrt{3})+\frac{\pi}{4}\right] \end{aligned} $$ Wish it helps.

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Footnote for $I(0)$: $$ \begin{aligned} \int_0^{\infty} \frac{x^a}{1+x^4} d x = & \frac{1}{2} \int_0^{\frac{\pi}{2}} \sin^{ \frac{a-1}{2} }\theta \cos ^{\frac{1-a}{2} }\theta d \theta\\=&\frac{1}{4} B\left(\frac{a+1}{4}, \frac{3-a}{4}\right)\\=& \frac{\pi}{4} \csc \left(\pi\left(\frac{a+1}{4}\right)\right) \end{aligned} $$ Differentiating $I(a)$ w.r.t. $a$ at $a=2$ yields $$ I(0)=4\left[-\left.\frac{\pi^2}{16} \csc \left(\pi\left(\frac{a+1}{4}\right)\right) \cot \left(\pi\left(\frac{a+1}{4}\right)\right)\right|_{a=2}\right] =\frac{\pi^2}{2\sqrt{2}} $$