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I started like this: $$\int _0^{\infty }\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx\: =\int _0^1\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx+\overset {x=\frac{1}{x}}{\int _1^{\infty }\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx}$$ $$=2\int _0^1\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx+5G$$ where $G$ is Catalan's constant. But the integral left is very hard to calculate. As suggested by Zacky one can use the sub $x=\frac{1-t}{1+t}$ and get these integrals. $$\ln 2\int _0^1\frac{1}{1+t^2}\:dt+\ln 5\int _0^1\frac{1}{1+t^2}\:dt+\int _0^1\frac{\ln \left(t^2+1-\frac{2}{\sqrt{5}}\right)}{1+t^2}\:dt$$ $$+\int _0^1\frac{\ln \left(t^2+1+\frac{2}{\sqrt{5}}\right)}{1+t^2}\:dt-5\int _0^1\frac{\ln \left(1+t\right)}{1+t^2}\:dt$$ To evaluate those one can use the identity $$\int _0^1\frac{\ln \left(b+ax^2\right)}{1+x^2}\:dx=\frac{\pi }{2}\ln \left(\sqrt{a}+\sqrt{b}\right)+\text{Ti}_2\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right)-G$$

In the end, the integral evaluates to \begin{align} \int _0^1\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx &= -\frac{3\pi }{8}\ln 2+\frac{\pi }{4}\ln 5-2G\\ &+\frac{\pi }{2}\ln \left(1+\sqrt{1-\frac{2}{\sqrt{5}}}\right)+\text{Ti}_2\left(\frac{\sqrt{1-\frac{2}{\sqrt{5}}}-1}{\sqrt{1-\frac{2}{\sqrt{5}}}+1}\right)\\ &+\frac{\pi }{2}\ln \left(1+\sqrt{1+\frac{2}{\sqrt{5}}}\right)+\text{Ti}_2\left(\frac{\sqrt{1+\frac{2}{\sqrt{5}}}-1}{\sqrt{1+\frac{2}{\sqrt{5}}}+1}\right) \end{align}

But, I have no idea how to simplify the $\text{Ti}_2\left(z\right)$ terms, which seems possible because I found that a similar version can be expressed without these $$\int _0^{\infty }\frac{\ln \left(1+x^5\right)}{\left(1+x^2\right)^2}\:dx=-\frac{5\pi }{8}-\frac{7\pi }{40}\ln \left(2\right)+\frac{\pi }{5}\ln \left(4+\sqrt{10-2\sqrt{5}}\right) +\frac{\pi }{10}\ln \left(43+7\sqrt{5}+4\sqrt{130+38\sqrt{5}}\right)+\frac{G}{10}$$

NB:

$\displaystyle x\in\mathbb{R},\text{Ti}_2(x)=\int_0^x \frac{\arctan t}{t}dt$

FDP
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Dennis Orton
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    Well, after the substitution $x\to \frac{1-x}{1+x}$ there is $$\int_0^1 \frac{\ln(1+x^5)}{1+x^2}dx=\int_0^1 \frac{\ln 10 -5\ln(1+x)+\ln(1+\frac{2}{\sqrt 5}+x^2)+\ln(1-\frac{2}{\sqrt 5}+x^2)}{1+x^2}dx$$ And all there's left is to use this general result. So does this link answer your question? – Zacky May 15 '20 at 10:00
  • @ThreeSidedCoin Came back to doing this one but i have a question. Do you have any idea on how could one simplify the $\text{Ti}_2$ terms? and if you could, could you post it? – Dennis Orton Jul 29 '20 at 01:43
  • Maybe https://math.stackexchange.com/questions/2077867/how-to-justify-differentiation-under-the-integral-in-calculation-of-int-01-f helps, or maybe https://math.stackexchange.com/questions/1669992/how-to-evaluate-this-integral-int-0-infty-frac-ln-left-1x3-right or https://math.stackexchange.com/questions/3247341/evaluate-int-0-infty-frac-ln1x31x2dx or https://math.stackexchange.com/questions/1688024/how-to-evaluate-these-two-integrals – Gerry Myerson Jul 29 '20 at 02:43
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    @GerryMyerson im afraid your comment is no longer needed to the post, the integral in question is already evaluated, but the thing is how to simplify those last terms with the $\text{Ti}_2$ function – Dennis Orton Jul 29 '20 at 03:03
  • What is the ${\rm Ti}_2$ function, please? – Gerry Myerson Jul 29 '20 at 06:10
  • @GerryMyerson its the inverse tangent integral. – Dennis Orton Jul 29 '20 at 06:37

1 Answers1

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Continue to reduce the result in the post as follows

\begin{align} & \text{Ti}_2\left(\frac{\sqrt{1-\frac{2}{\sqrt{5}}}-1}{\sqrt{1-\frac{2}{\sqrt{5}}}+1}\right)+\text{Ti}_2\left(\frac{\sqrt{1+\frac{2}{\sqrt{5}}}-1}{\sqrt{1+\frac{2}{\sqrt{5}}}+1}\right)\\ =&\ \text{Ti}_2(\tan\frac\pi{20}) - \text{Ti}_2(\tan\frac{3\pi}{20}) =\int_{\tan\frac{3\pi}{20} }^{\tan\frac{\pi}{20} }\frac{\tan^{-1}x}xdx\\ \overset{ibp}=& \ \frac\pi{20}\ln \tan\frac\pi{20}- \frac{3\pi}{20}\ln \tan\frac{3\pi}{20}+\int^{\frac{3\pi}{20} }_{\frac{\pi}{20} }\ln(\tan t )dt\\ =& \ \frac\pi{20}\ln \frac{\sqrt{1+\frac{2}{\sqrt{5}}}-1}{\sqrt{1+\frac{2}{\sqrt{5}}}+1}- \frac{3\pi}{20}\ln \frac{1-\sqrt{1-\frac{2}{\sqrt{5}}}}{1+\sqrt{1-\frac{2}{\sqrt{5}}}}-\frac25G\\ =& -\frac25G -\frac\pi5 \ln\frac2{\sqrt5 } -\frac\pi5\ln \left(1+\sqrt{1+\frac{2}{\sqrt{5}}}\right) +\frac{3\pi}5\ln \left(1+\sqrt{1-\frac{2}{\sqrt{5}}}\right) \end{align}

where $\int^{\frac{3\pi}{20} }_{\frac{\pi}{20} }\ln(\tan t )dt=-\frac25G$. Then, plug into the result to arrive at the elementary expression for the integral

\begin{align} \int_0^\infty\frac{\ln(1+x^5)}{1+x^2} &= \frac15G -\frac{19\pi }{20}\ln 2+\frac{3\pi }{5}\ln 5\\ &\hspace{5mm}+\frac{4\pi}5 \ln \left(1+\sqrt{1+\frac{2}{\sqrt{5}}}\right)+\frac{8\pi}5\ln \left(1+\sqrt{1-\frac{2}{\sqrt{5}}}\right)\\ \end{align}

Quanto
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