I need help with this integral $$\int_0^\infty \frac{(\log x)^3}{x^2 +2x +2} dx$$
What I have done so far:
\begin{align*} I &= \int_0^1 \frac{(\log x)^3}{x^2 + 2x + 2} dx + \int_1^\infty \frac{(\log x)^3}{x^2 +2x +2} dx\\ &=−(\int_1^\infty \frac{(\log x)^3}{2x^2 + 2x + 1} dx) + \int_1^\infty \frac{(\log x)^3}{x^2 +2x +2} dx \end{align*}
After this I am not sure how to proceed. According to wolframalpha the approximate value of the integral is $2.55128$
For $-1 < s < 1$, define \begin{align*} I(s) &= \int_0^\infty \frac{x^s}{x^2 + 2x + 2} \ \mathrm{d}x \end{align*} Then, we see the integral required is $I'''(0)$. Hence, it suffices to evaluate the above integral.
Let $f(z) = \frac{z^s}{z^2 + 2z + 2}$ for $z \in \mathbb{C}$. Consider an anticlockwise keyhole contour $\Gamma$ of outer radius $R$ and inner radius $\varepsilon$ about the positive real axis. Let $\gamma_R$ denote the outer arc, $\gamma_\varepsilon$ denote the inner arc, and $\gamma_{\pm}$ denote the segments going away from and towards the origin respectively.
Note that this contour contains two singularities of $f$, namely $z = \sqrt{2}\exp\left (\frac{3\pi i}{4}\right ), \sqrt{2}\exp\left (\frac{5\pi i}{4}\right )$, both of order $1$. Hence, by Residue Theorem, we have \begin{align*} \oint_\Gamma \frac{z^s}{z^2 + 2z + 2} \ \mathrm{d}z &= 2\pi i \left (\operatorname{Res}\left (f(z), z = \sqrt{2}\exp\left (\frac{3\pi i}{4}\right )\right ) + \operatorname{Res}\left (f(z), z = \sqrt{2}\exp\left (\frac{5\pi i}{4}\right )\right ) \right )\\ &= 2^{\frac{s}{2}} \pi \left (\exp\left (\frac{3s\pi i}{4}\right ) - \exp\left (\frac{5s\pi i}{4}\right ) \right ) \end{align*} On the other hand, by bounding (using Estimation Lemma) we see \begin{align*} \left |\int_{\gamma_R} \frac{z^s}{z^2 + 2z + 2} \ \mathrm{d}z \right | &\leq \frac{2\pi R^{1+s}}{R^2 - 2R - 2} \to 0 \text{ as } R \to \infty\\ \left |\int_{\gamma_\varepsilon} \frac{z^s}{z^2 + 2z + 2} \ \mathrm{d}z \right | &\leq \frac{2\pi \varepsilon^{1+s}}{2 - 2\varepsilon - 2\varepsilon^2} \to 0 \text{ as } \varepsilon \to 0\\ \end{align*} We see \begin{align*} \int_{\gamma_-} \frac{z^s}{z^2 + 2z + 2} \ \mathrm{d}z &= \int_R^\varepsilon \frac{x^s \exp(2s\pi i)}{x^2 + 2x + 2} \ \mathrm{d}x\\ &= -\exp(2s\pi i) \int_\varepsilon^R \frac{x^s}{x^2 + 2x + 2} \ \mathrm{d}x \end{align*} Taking $\varepsilon \to 0$ and $R \to \infty$, we have \begin{align*} \oint_\Gamma \frac{z^s}{z^2 + 2z + 2} \ \mathrm{d}z &= (1 - \exp(2s\pi i))\int_0^\infty \frac{x^s}{x^2 + 2x + 2} \ \mathrm{d}x \end{align*} We therefore get \begin{align*} \int_0^\infty \frac{x^s}{x^2 + 2x + 2} \ \mathrm{d}x &= \frac{2^{\frac{s}{2}} \pi \left (\exp\left (\frac{3s\pi i}{4}\right ) - \exp\left (\frac{5s\pi i}{4}\right ) \right )}{1 - \exp(2s\pi i)}\\ &= \frac{2^{\frac{s}{2}} \pi (\omega^3 - \omega^5)}{1 - \omega^8} && \text{letting } \omega = \exp\left (\frac{s\pi i}{4}\right )\\ &= \frac{2^{\frac{s}{2}} \pi}{(\omega + \omega^{-1})(\omega^2 + \omega^{-2})}\\ &= \frac{\pi}{4} 2^{\frac{s}{2}} \sec \left (\frac{s\pi}{2} \right ) \sec \left (\frac{s\pi}{4} \right ) \end{align*} To finish off, we can use the Taylor series of these terms to quickly compute $I'''(0)$: \begin{align*} 2^{\frac{s}{2}} &= 1 + \frac{\log(2)}{2} s + \frac{\log(2)^2}{8} s^2 + \frac{\log(2)^3}{48} s^3 + O(s^4)\\ \sec \left (\frac{s\pi}{2} \right ) &= 1 + \frac{\pi^2}{8} s^2 + O(s^4)\\ \sec \left (\frac{s\pi}{4} \right ) &= 1 + \frac{\pi^2}{32} s^2 + O(s^4)\\ \implies \text{the $s^3$ coefficient} &= \frac{\pi \log(2)^3}{192} + \frac{\pi^3 \log(2)}{64} + \frac{\pi^3 \log(2)}{256}\\ &= \frac{\pi \log(2)^3}{192} + \frac{5\pi^3 \log(2)}{256} \end{align*}
Hence, we get \begin{align*} \boxed{\int_0^\infty \frac{\log(x)^3}{x^2 + 2x + 2} \ \mathrm{d}x = \frac{\pi \log(2)^3}{32} + \frac{15\pi^3 \log(2)}{128}} \end{align*} which agrees numerically with your answer.
Let $J_n= \int_0^\infty \frac{\ln^n x}{x^2 +\sqrt2 x +1}{dx}$ and express the integral as \begin{align} I=& \int_0^\infty \frac{\ln^3 x}{x^2 +2x +2} \overset{x\to \sqrt2 x}{dx} = \frac1{\sqrt2}\int_0^\infty \frac{\ln^3 (\sqrt2 x)}{x^2 +\sqrt2 x +1}{dx}\\ =&\ \frac1{\sqrt2}\left( J_3+\frac32\ln2 \ J_2 +\frac34\ln^22\ J_1 + \frac18\ln^32 \ J_0\right) \end{align} Note that $J_1\overset{x\to \frac1x}= -J_1=0$ and likewise $J_3= -J_3=0$. Furthermore \begin{align} J_0= &\int_0^\infty \frac{1}{x^2 +\sqrt2 x +1}{dx}=\frac\pi{2\sqrt2}\\ J_2= &\int_0^\infty \frac{\ln^2x}{x^2 +\sqrt2 x +1}{dx} =\sqrt2\ \Im\int_0^\infty\frac{\ln^2x}{x+e^{-i\frac\pi4}}dx\\ = & \ \sqrt2\ \Im\frac{d^2}{da^2}\bigg(\int_0^\infty\frac{x^a}{x+e^{-i\frac\pi4}}dx\bigg)_{a=0}\\ = & \ \sqrt2\ \Im\frac{d^2}{da^2}\bigg(-\pi e^{-i \frac{\pi a}4}\csc \pi a\bigg)_{a=0}=\frac{5\pi^3}{32\sqrt2} \end{align} Plug above results into $I$ to arrive at $$I= \frac{15\pi^3 }{128}\ln2+ \frac{\pi}{32} \ln^32 $$
\begin{align}J&=\int_0^\infty \frac{(\log x)^3}{x^2 +2x +2} dx\\ &\overset{u=\frac{x}{\sqrt{2}}}=\frac{1}{\sqrt{2}}\int_0^\infty \frac{\ln^3\left(\sqrt{2}u\right)}{u^2+\sqrt{2}u+1}du\\ &=\frac{3\ln 2}{2\sqrt{2}}\underbrace{\int_0^\infty\frac{\ln^2 u}{u^2+\sqrt{2}u+1}du}_{=K}+\frac{\pi\ln^3 2}{32}\\ S&=\int_0^\infty\int_0^\infty\frac{\ln^2(tu)}{(u^2+\sqrt{2}u+1)(t^2+\sqrt{2}t+1)}du\\ &\overset{z(t)=tu}=\int_0^\infty\int_0^\infty \frac{u\ln^2 z}{(u^2+\sqrt{2}u+1)(z^2+\sqrt{2}uz+u^2)}dudz\\ &=\int_0^\infty \left(-\frac{\pi\ln^2 z}{1+z^2}+\frac{z\ln^3 z}{z^3-z^2+z-1}+\frac{\ln^3 z}{z^3-z^2+z-1}+\frac{\frac{\pi\ln^2 z}{2}}{1+z^2}\right)dz\\ &=-\int_0^\infty \left(\frac{\pi\ln^2 z}{2(1+z^2)}+\frac{(1+z)\ln^3 z}{(1-z)(1+z^2)}\right)dz\\ &=-\frac{\pi^4}{16}-2\int_0^1 \frac{(1+z)\ln^3 z}{(1-z)(1+z^2)}dz\\ &=-\frac{\pi^4}{16}-2\int_0^1 \frac{\ln^3 z}{1-z}dz-\underbrace{\int_0^1 \frac{2z\ln^3 z}{1+z^2}dz}_{w=z^2}\\ &=-\frac{\pi^4}{16}+12\zeta(4)+\frac{21}{32}\zeta(4)=\frac{5\pi^4}{64} \end{align} On the other hand, \begin{align}S&=\frac{\pi K}{\sqrt{2}}\end{align} Therefore, \begin{align}K&=\frac{5\pi^3}{32\sqrt{2}}\\ J&=\frac{3\ln 2}{2\sqrt{2}}\times \frac{5\pi^3}{32\sqrt{2}}+\frac{\pi\ln^3 2}{32}=\boxed{\frac{\pi\ln^3 2}{32}+\frac{15\pi^3\ln 2}{128}} \end{align}
NB: \begin{align}\int_0^\infty \frac{\ln u}{u^2+\sqrt{2}u+1}du=\int_0^\infty \frac{\ln^3 u}{u^2+\sqrt{2}u+1}du&=0\\ \int_0^\infty \frac{1}{u^2+\sqrt{2}u+1}du&=\frac{\pi}{2\sqrt{2}}\\ \int_0^\infty\frac{\ln^2 u}{1+u^2}du&=\frac{\pi^3}{8}\\ \int_0^1 \frac{\ln^3x}{1-x}dx&=-6\zeta(4)\\ \int_0^1 \frac{\ln^3x }{1+x}dx&=-\frac{21}{4}\zeta(4) \end{align}
