the integration23

The integral can be solved using differentiation under the integral sign .

$\begin{aligned} I(a) & = \int_0^\infty \frac {\tan^{-1}(ax^3)-\tan^{-1}(x^3)}x dx \\ \frac {\partial I(a)}{\partial a} & = \int_0^\infty \frac {x^2}{1+a^2x^6} dx & \small \blue{\text{Let }u = a^2x^6 \implies du = 6a^2 x^5 \ dx} \\ & = \frac 1{6a} \int_0^\infty \frac {u^{-\frac 12}}{1+u} du & \small \blue{\text{Beta function B }(m,n) = \int_0^\infty \frac {u^{m-1}}{(1+u)^{m+n}} du} \\ & = \frac 1{6a} \text{ B }\left(\frac 12, \frac 12 \right) & \small \blue{\text{and B }(m,n) = \frac {\Gamma(m)\Gamma(n)}{\Gamma(m+n)}} \\ & = \frac {\Gamma^2 \left(\frac 12 \right)}{6a \Gamma (1)} & \small \blue{\text{and }\Gamma \left(\frac 12 \right) = \sqrt \pi} \\ & = \frac \pi {6a} \\ \implies I(a) & = \int \frac \pi {6a} da = \frac {\pi \ln a}6 + \blue C & \small \blue{\text{where }C \text{ is the constant of integration.}} \\ I(1) & = C = 0 \\ \implies I(a) & = \frac {\pi \ln a}6 \end{aligned}$

Therefore $\displaystyle \int_0^\infty \frac {\tan^{-1}(3x^3) - \tan^{-1}(x^3)}x dx = I(3) = \frac \pi 6 \ln 3 \approx \boxed{0.575}$ .

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References:

• Beta function
• Gamma function