Trying to evaluate $\int_{0}^{\infty}\frac{\ln(1+x^3)}{1+x^3}\frac{dx}{1+x^3}$

Question

I would like to work this out:

$$I=\large\int_{0}^{\infty}\frac{\ln(1+x^3)}{1+x^3}\frac{\mathrm dx}{1+x^3}$$

Making a sub: $u=x^3$, $dx=\frac{du}{3x^2}$

$$I=\frac{1}{3}\int_{0}^{\infty}\frac{\ln(1+u)}{u^{2/3}(1+u)^2}\mathrm du$$

Making a sub: $u=\tan^2(y)$, $du=\sec^2(y)dy$

$$I=\frac{2}{3}\int_{0}^{\pi/2}\frac{\ln\sec(y)}{\sec^2(y)\sqrt[3]{\tan^4(y)}}\mathrm dy$$

$$I=\frac{2}{3}\int_{0}^{\pi/2}\frac{\cot(y)\cos^2(y)\ln\sec(y)}{\sqrt[3]{\tan(y)}}\mathrm dy$$

I can't continue.

Maybe there is another alternative way to simplify $I$

Answer 1

The substitution $t = (1 + x^3)^{-1}$ yields $$ I = \frac{1}{3} \int \limits_0^1 - \ln(t) \left(\frac{t}{1-t}\right)^{2/3} \, \mathrm{d} t = f'\left(\frac{2}{3}\right) \, , $$ where $$ f(\alpha) \equiv - \frac{1}{3} \int \limits_0^1 \frac{t^\alpha}{(1-t)^{2/3}} \, \mathrm{d} t = - \frac{1}{3} \operatorname{B} \left(\frac{1}{3}, \alpha +1 \right) = - \frac{\operatorname{\Gamma} \left(\frac{1}{3}\right)}{3} \frac{\operatorname{\Gamma}(\alpha + 1)}{\operatorname{\Gamma}\left(\alpha + \frac{4}{3} \right)} ~~~ , \, \alpha > -1 \, .$$ The differentiation under the integral sign can be justified using the dominated convergence theorem.

In terms of the digamma function $\psi$ we now have $$ I = f'\left(\frac{2}{3}\right) = \frac{2}{9} \operatorname{\Gamma} \left(\frac{1}{3}\right) \operatorname{\Gamma} \left(\frac{2}{3}\right) \left[\operatorname{\psi} (2) - \operatorname{\psi} \left(\frac{5}{3}\right)\right] \, . $$ We can use the reflection formulas for $\Gamma$ and $\psi$ and the recurrence relation $\operatorname{\psi}(x + 1) = \operatorname{\psi}(x) + \frac{1}{x}$ to find $$ I = \frac{2}{9} \frac{\pi}{\sin\left(\frac{\pi}{3}\right)} \left[\operatorname{\psi}(1) - \operatorname{\psi} \left(\frac{1}{3}\right) - \pi \cot\left(\frac{\pi}{3}\right) - \frac{1}{2} \right] \, .$$ The special values $\operatorname{\psi}(1) = - \gamma$ and $\operatorname{\psi} \left(\frac{1}{3}\right) = - \frac{\pi}{2 \sqrt{3}} - \frac{3}{2} \ln (3) - \gamma$ then lead to the final result $$ I = \frac{2 \pi}{27} [\sqrt{3} (3 \ln(3) - 1) - \pi] \, .$$

Answer 2

In this solution we will not use the beta function or the digamma function.

Put \begin{equation*} f(s)=\int_{0}^{\infty}\dfrac{\ln(1+s^3x^3)}{(1+x^3)^2}\, \mathrm{d}x. \end{equation*} We want to calculate $f(1)$. However, \begin{equation*} f(1) = f(1)-f(0) = \int_{0}^{1}f'(s)\, \mathrm{d}s\tag{1} \end{equation*} and \begin{equation*} f'(s)=\int_{0}^{\infty}\dfrac{3s^2x^3}{(1+s^3x^3)(1+x^3)^2}\, \mathrm{d}x =[x^3=z] = \int_{0}^{\infty}\dfrac{s^2z^{1/3}}{(1+s^3z)(1+z)^2}\, \mathrm{d}z . \end{equation*} For $0<s<1$ we use keyhole contour integration and get \begin{equation*} f'(s) = \dfrac{2\pi\sqrt{3}}{9}\dfrac{2s+1}{(s^2+s+1)^2}\cdot s^2 . \end{equation*} Integration by parts in (1) will finally give us \begin{equation*} f(1) = \dfrac{2\pi}{27}\left(\sqrt{3}(3\ln 3 -1)-\pi\right). \end{equation*}

Answer 3

Let’s consider the parameterized integral $$ I(a)=\int_0^{\infty}\left(1+x^3\right)^a d x \stackrel{\frac{1}{1+x^3} \mapsto x}{=} \frac{\Gamma\left(\frac{4}{3}\right) \Gamma\left(-a-\frac{1}{3}\right)}{\Gamma(-a)} $$ where $a<0$.

Logarithmic differentiation w.r.t. $a$ yields $$ \frac{I^{\prime}(a)}{I(a)}=-\psi\left(-a-\frac{1}{3}\right)+\psi(-a) $$ Putting $a=-2$ gives our integral $$ \begin{aligned} \int_{0}^{\infty}\frac{\ln(1+x^3)}{1+x^3}\frac{dx}{1+x^3} &=I^{\prime}(-2)\\ & =I(-2)\left[\psi(2)-\psi\left(\frac{5}{3}\right)\right] \\ & =\frac{\Gamma\left(\frac{4}{3}\right) \Gamma\left(\frac{5}{3}\right)}{\Gamma(2)}\left[1-\gamma-\frac{3}{2}+\gamma-\frac{\pi}{2 \sqrt{3}}+\frac{3}{2} \ln 3\right] \\ & =\frac{4 \pi}{9 \sqrt{3}}\left(-\frac{1}{2}-\frac{\pi}{2 \sqrt{3}}+\frac{3}{2} \ln 3\right) \\ & =\frac{2 \pi}{27}[\sqrt{3}(3 \ln 3-1)-\pi] \end{aligned} $$

Answer 4

For an elementary integration, first evaluate \begin{align} J=\int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{1+x^{3}}dx =&\int_{0}^{\infty}\int_0^1 \frac{3y^2x^3}{(1+x^{3})(1+y^3x^3)}dy dx\\ =& \ \frac{2\pi}{\sqrt3}\int_0^1 \frac y{1+y+y^2}dy= \frac\pi{\sqrt3}\ln3-\frac{\pi^2}9 \end{align} Then, per integration-by-parts \begin{aligned} \int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{\left(1+x^{3}\right)^{2}} d x &=\int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{3 x^2} \ d\left(-\frac{1}{1+x^{3}}\right)\\ &=\int_{0}^{\infty} \frac{1}{\left(1+x^{3}\right)^2} - \frac{2\ln \left(1+x^{3}\right)}{3x^{3}} + \frac{2\ln \left(1+x^{3}\right)}{3\left(1+x^{3}\right)}\ d x\\&= \frac{4 \pi}{9\sqrt3}-\frac{2\pi}{3\sqrt3} +\frac{2}{3} J = \frac{2\pi}{9\sqrt3}(\ln 27 -1)-\frac{2\pi^2}{27} \end{aligned}