What is value of this integral? $\int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^{2})}dx$

Question

What is value of this integral $$I=\int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^2)}dx$$

My work : \begin{align*}I&=\int_{0}^{\infty}\frac{\log(1+4x^2)}{9+x^2}dx+\int_{0}^{\infty}\frac{\log(4+x^2)(10+10x^2)}{(9+x^2)(1+9x^2)}dx\\ &=\int_{0}^{\infty}\frac{\log(1+4x^2)}{9+x^2}dx+\int_{0}^{\infty}\frac{\log(4+x^2)}{1+9x^2}dx+\int_{0}^{\infty}\frac{\log(4+x^2)}{9+x^2}dx\\ &=j_{1}+j_{2}+j_{3}\\ \end{align*} $$j_{1}=\int_{0}^{\infty}\frac{\log(1+4x^2)}{9+x^2}dx=\sum_{n=0}^{\infty}\frac{(-1)^n(2^{2(n+1)})}{n+1}\int_{0}^{\infty}\frac{x^{2(n+1)}}{9+x^2}dx$$
$$j_{2}=\log(4)\int_{0}^{\infty}\frac{1}{1+(3x)^2}dx+\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)(2^{2(n+1)})}\int_{0}^{\infty}\frac{x^{2(n+1)}}{1+9x^2}dx=\frac{\log(4)\pi}{6}+\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)2^{2(n+1)}}\int_{0}^{\infty}\frac{x^{2(n+1)}}{1+9x^2}dx$$
$$j_{3}=\int_{0}^{\infty}\frac{\log(4+x^2)}{9+x^2}dx=\frac{\log(4)\pi}{54}+\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)2^{2(n+1)}}\int_{0}^{\infty}\frac{x^{2(n+1)}}{9+x^2}dx$$
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Answer 1

Consider using differentiation under the integral sign. Parameterize the integral as the following: $$I(a)=\int_0^{\infty} \frac{\ln{\left(1+ax^2\right)}}{9+x^2} + \frac{\ln{\left(a+x^2\right)}}{1+9x^2}+ \frac{\ln{\left(a+x^2\right)}}{9+x^2} \; \mathrm{d}x$$ The integral in question is $I(4)$. First, differentiate $I(a)$ with respect to $a$: \begin{align*} I'(a)&=\int_0^{\infty} \frac{x^2}{(9+x^2)(1+ax^2)}+\frac{1}{(1+9x^2)(a+x^2)}+\frac{1}{(9+x^2)(a+x^2)} \; \mathrm{d}x \\ &=\int_0^{\infty} -\frac{10 (a - 1)}{(a - 9) (9 a - 1) (a + x^2)} + \frac{2 (9 a - 41)}{(a - 9) (9 a - 1) (x^2 + 9)} + \frac{9}{(9 a - 1) (9 x^2 + 1)} + \frac{1}{(1 - 9 a) (a x^2 + 1)} \; \mathrm{d}x \\ &=\frac{\pi}{6} \left(\frac{\frac{19}{\sqrt{a}}+9}{3a+10\sqrt{a}+3}\right)\\ \end{align*} Now, $I'(a)$ with respect to $a$ from $0$ to $4$: \begin{align*} I(4)&=\frac{\pi}{6} \int_0^4 \frac{\frac{19}{\sqrt{a}}+9}{3a+10\sqrt{a}+3} \; \mathrm{d}a \\ &=\frac{\pi}{6} \int_0^4 \frac{\frac{1}{\sqrt{a}}}{\sqrt{a}+3} + \frac{\frac{6}{\sqrt{a}}}{3\sqrt{a}+1} \; \mathrm{d} a \\ &=\frac{\pi}{6} \left(2 \int_0^4 \frac{\mathrm{d}\left(\sqrt{a}+3\right)}{\sqrt{a}+3} + 4 \int_0^4 \frac{\mathrm{d}\left(3\sqrt{a}+1\right)}{3\sqrt{a}+1}\right) \\ &=\frac{\pi}{3} \left(\ln{\left(\sqrt{a}+3\right)}+2\ln{\left(3\sqrt{a}+1\right)}\right) \bigg \rvert_0^4 \\ I(4) &= \int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^2)} \mathrm{d}x =\boxed{\frac{\pi \ln{\left(\frac{245}{3}\right)}}{3}}\\ \end{align*}

Answer 2

First, the integral:

$$ I := \int_0^{\infty} \frac{\ln\left(1+4\,x^2\right)\left(1+9\,x^2\right)\left(9+\,x^2\right)+\left(9+x^2\right)\ln\left(4+x^2\right)\left(10+10\,x^2\right)}{\left(9+x^2\right)^2\left(1+9\,x^2\right)}\,\text{d}x $$

thanks to the linearity of the integrals it can be written as the algebraic sum of two integrals:

$$ I = \int_0^{\infty} \frac{\ln\left(1+4\,x^2\right)}{9+x^2}\,\text{d}x + \int_0^{\infty} \frac{10\left(1+x^2\right)\ln\left(4 + x^2\right)}{\left(9+\,x^2\right)\left(1+9\,x^2\right)}\,\text{d}x\,. $$

At this point, to calculate the first, it's sufficient to consider the function $J : [0,\,+\infty) \to \mathbb{R}$ of the law:

$$ J(a) := \int_0^{\infty} \frac{\ln\left(1+a\,x^2\right)}{9+x^2}\,\text{d}x $$

whose first derivative is equal to:

$$ J'(a) = \int_0^{\infty} \frac{x^2}{\left(9+x^2\right)\left(1+a\,x^2\right)}\,\text{d}x = \frac{\pi}{2\left(\sqrt{a}+3\,a\right)} $$

and therefore, going back, we get:

$$ J(a) = \int \frac{\pi}{2\left(\sqrt{a}+3\,a\right)}\,\text{d}a = \frac{\pi}{3}\,\ln\left(1 + 3\sqrt{a}\right) + c $$

where equaling the two expressions of $J(0)$ must be $c = 0$, from which:

$$ \int_0^{\infty} \frac{\ln\left(1+a\,x^2\right)}{9+x^2}\,\text{d}x = \frac{\pi}{3}\,\ln\left(1 + 3\sqrt{a}\right). $$

Similarly, to calculate the second, it's sufficient to consider the function $K : [0,\,+\infty) \to \mathbb{R}$ of the law:

$$ K(b) := \int_0^{\infty} \frac{10\left(1+x^2\right)\ln\left(4 + b\,x^2\right)}{\left(9+\,x^2\right)\left(1+9\,x^2\right)}\,\text{d}x $$

whose first derivative is equal to:

$$ K'(b) = \int_0^{\infty} \frac{10\,x^2\left(1+x^2\right)}{\left(9+x^2\right)\left(1+9\,x^2\right)\left(4+b\,x^2\right)}\,\text{d}x = \frac{\left(3+\frac{10}{\sqrt{b}}\right)\pi}{3\left(12+20\sqrt{b}+3\,b\right)} $$

and therefore, going back, we get:

$$ K(b) = \int \frac{\left(3+\frac{10}{\sqrt{b}}\right)\pi}{3\left(12+20\sqrt{b}+3\,b\right)}\,\text{d}b = \frac{\pi}{3}\,\ln\left(12+20\sqrt{b}+3\,b\right) + c $$

where equaling the two expressions of $K(0)$ must be $c = \frac{2\,\pi}{3}\,\ln(2) - \frac{\pi}{3}\,\ln(12)$, from which:

$$ \int_0^{\infty} \frac{10\left(1+x^2\right)\ln\left(4 + b\,x^2\right)}{\left(9+\,x^2\right)\left(1+9\,x^2\right)}\,\text{d}x = \frac{\pi}{3}\,\ln\left(12+20\sqrt{b}+3\,b\right) + \frac{2\,\pi}{3}\,\ln(2) - \frac{\pi}{3}\,\ln(12)\,. $$

In conclusion, setting $a = 4$ and $b = 1$, we get:

$$ I = \frac{\pi}{3}\,\ln\left(1 + 3\sqrt{4}\right) + \frac{\pi}{3}\,\ln\left(12+20\sqrt{1}+3\cdot 1\right) + \frac{2\,\pi}{3}\,\ln(2) - \frac{\pi}{3}\,\ln(12) = \frac{\pi}{3}\,\ln\left(\frac{245}{3}\right), $$

which is what is desired.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ With $\ds{a,b,c \in \mathbb{R}_{\ >\ 0}}$: \begin{align} \mrm{f}\pars{a,b,c} & \equiv \int_{0}^{\infty}{\ln\pars{a^{2} + b^{2}x^{2}} \over c^{2} + x^{2}}\,\dd x = \Re\int_{-\infty}^{\infty}{\ln\pars{a + bx\ic} \over \pars{x + c\ic}\pars{x - c\ic}}\,\dd x \end{align} With the change $\ds{\pars{x = -\,{s - a \over b}\,\ic \implies s = a + bx\ic}}$: \begin{align} \mrm{f}\pars{a,b,c} & = -\,\Im\int_{a - \infty\ic}^{a + \infty\ic}{b\ln\pars{s} \over \bracks{s - \pars{a - bc}}\bracks{s - \pars{a + bc}}}\,\dd s \\[5mm] & = -\,\Im\bracks{-2\pi\ic\,{b\ln\pars{a + bc} \over 2bc}} = \bbx{\pi\,{\ln\pars{a + bc} \over c}} \\ & \end{align}

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$$ \left\{\begin{array}{rcccccl} \ds{j_{1}} & \ds{=} & \ds{\int_{0}^{\infty}{\ln\pars{1 + 4x^{2}} \over 9 + x^{2}}\,\dd x} & \ds{=} & \ds{\mrm{f}\pars{1,2,3}} & \ds{=} & \ds{{\pi \over 3}\ln\pars{7}} \\[2mm] \ds{j_{2}} & \ds{=} & \ds{\int_{0}^{\infty}{\ln\pars{4 + x^{2}} \over 1 + 9x^{2}}\,\dd x} & \ds{=} & \ds{{1 \over 9}\,\mrm{f}\pars{2,1,{1 \over 3}}} & \ds{=} & \ds{{\pi \over 3}\,\ln\pars{7 \over 3}} \\[2mm] \ds{j_{3}} & \ds{=} & \ds{\int_{0}^{\infty}{\ln\pars{4 + x^{2}} \over 9 + x^{2}}\,\dd x} & \ds{=} & \ds{\mrm{f}\pars{2,1,3}} & \ds{=} & \ds{{\pi \over 3}\,\ln\pars{5}} \end{array}\right. $$ \begin{align} \mbox{} & \\ j_{1} + j_{2} + j{3} & = \bbx{{1 \over 3}\,\pi\ln\pars{245 \over 3}} \approx 4.6104 \\ & \end{align}