How to calculate $\int_{0}^{\infty}\frac{x^{2}\ln x}{(1+x^{3})^4}dx$

Question

I have to calculate this definite integral: $$ \int_{0}^{\infty}\frac{x^{2}\ln x}{(1+x^{3})^4}dx$$

I calculatated as a indefinite integral and the answer is $$-\dfrac{\ln\left(x^3+1\right)}{27}-\dfrac{\ln\left(x\right)}{9\left(x^3+1\right)^3}+\dfrac{\ln\left(x\right)}{9}+\dfrac{1}{27\left(x^3+1\right)}+\dfrac{1}{54\left(x^3+1\right)^2}$$

But what I have to do if I want to calculate the definite integral? How can I apply here the fundamental theorem of calculus

Answer 1

Consider the Mellin transform $$\int_{0}^{+\infty}\frac{x^{s}}{\left(1+x^{3}\right)^{4}}dx.$$ Now, it is simple to observe that we have the formal power series $$\frac{x^{s}}{\left(1+x^{3}\right)^{4}}=-\frac{1}{6}\sum_{n\geq0}\left(-1\right)^{n}n\left(n-1\right)\left(n-2\right)x^{3n-9+s}$$ hence, by the bracket's method, we get $$\int_{0}^{+\infty}\frac{x^{s}}{\left(1+x^{3}\right)^{4}}dx=-\frac{1}{6}\sum_{n\geq0}\left(-1\right)^{n}n\left(n-1\right)\left(n-2\right)\left\langle 3n-8+s\right\rangle $$ $$=-\frac{1}{18}\Gamma\left(\frac{8-s}{3}+1\right)\frac{8-s}{3}\left(\frac{8-s}{3}-1\right)\left(\frac{8-s}{3}-2\right)\Gamma\left(\frac{s-8}{3}\right).$$ Then $$\int_{0}^{+\infty}\frac{x^{2}\log\left(x\right)}{\left(1+x^{3}\right)^{4}}dx=-\frac{1}{18}\lim_{s\rightarrow2}\frac{d}{ds}\Gamma\left(\frac{8-s}{3}+1\right)\frac{8-s}{3}\left(\frac{8-s}{3}-1\right)\left(\frac{8-s}{3}-2\right)\Gamma\left(\frac{s-8}{3}\right)$$ $$=\color{red}{-\frac{1}{18}}.$$

Answer 2

Well, that's the opposite of what typically happens: you've done pretty much all of the job, and then are asking... for what, really? By looking at your expression, I would doubt that it has an elementary antiderivative. But it does, and you've found it. That was the hardest part; what remains is just plugging the limits. So what if there is an indeterminacy or two? They can be treated with well-known methods.

$$\lim\limits_{x\to0}\left(\underbrace{-\dfrac{\ln(x^3+1)}{27}}_{\color{red}0}-\dfrac{\ln(x)}{9(x^3+1)^3}+\dfrac{\ln(x)}{9}\underbrace{+\dfrac{1}{27\left(x^3+1\right)}}_{\color{red}{1/27}} \underbrace{+\dfrac{1}{54\left(x^3+1\right)^2}}_{\color{red}{1/54}}\right)=\\= \lim\limits_{x\to0}\dfrac{\ln(x)\Big((x^3+1)^3-1\Big)}{9(x^3+1)^3}+\dfrac{1}{18} = \lim\limits_{x\to0}\dfrac{x^3\ln(x)}{3}+\dfrac{1}{18}=\dfrac{1}{18}$$

Now the other one. $$ \lim\limits_{x\to\infty}\left(-\dfrac{\ln(x^3+1)}{27} \underbrace{-\dfrac{\ln(x)}{9(x^3+1)^3}}_{\color{red}0}+\dfrac{\ln(x)}{9}\underbrace{+\dfrac{1}{27(x^3+1)}}_{\color{red}0} \underbrace{+\dfrac{1}{54(x^3+1)^2}}_{\color{red}0}\right) = \\ = \lim\limits_{x\to\infty}\left(-\dfrac{\ln(x^3+1)}{27}+\dfrac{\ln(x)}{9}\right)=\dfrac1{27}\lim\limits_{x\to\infty}\ln{x^3\over x^3+1}=0$$

Answer 3

If you are open to engineering logic: Let $M\gg1$ and $0<\epsilon\ll1$. Then, evaluation gives: $$-\frac{\ln(M^3+1)}{27}+\frac{\ln(\epsilon^3+1)}{27}-\frac{\ln M}{9(M^3+1)^3}+\frac{\ln\epsilon}{9(\epsilon^3+1)^3}+\frac{\ln M}{9}-\frac{\ln\epsilon}{9}+\frac{1}{27(M^3+1)}-\frac{1}{27(\epsilon^3+1)}+\frac{1}{54(M^3+1)^2}-\frac{1}{54(\epsilon^3+1)^2}.$$ Now, let $M\rightarrow\infty$ and $\epsilon\rightarrow0$. Then, "in the limit": $$-\frac{\ln M}{9}+0-0+\frac{\ln\epsilon}{9}+\frac{\ln M}{9}-\frac{\ln\epsilon}{9}+0-\frac1{27}+0-\frac1{54}=-\frac1{27}-\frac1{54}=\color{red}{-\frac1{18}}.$$

Answer 4

Integration by parts and by substitution will work for this integral. In fact \begin{eqnarray} &&\int_{0}^{\infty}\frac{x^{2}\ln (x)}{(1+x^{3})^4}\mathrm{d}x=\frac19\int_{0}^{\infty}\frac{\ln(x^3)}{(1+x^3)^4}\mathrm{d}(x^3)=\frac19\int_{0}^{\infty}\frac{\ln(x)}{(1+x)^4}\mathrm{d}x\\ &=&\frac19\bigg[\int_{0}^{1}\frac{\ln(x)}{(1+x)^4}\mathrm{d}x+\overset{x\to\frac1x}{\int_{1}^{\infty}\frac{\ln(x)}{(1+x)^4}\mathrm{d}x}\bigg]\\ &=&\frac19\bigg[\int_{0}^{1}\frac{\ln(x)}{(1+x)^4}\mathrm{d}x-{\int_{0}^{1}\frac{x^2\ln(x)}{(1+x)^4}\mathrm{d}x}\bigg]\\ &=&\frac19\int_{0}^{1}\frac{(1-x^2)\ln(x)}{(1+x)^4}\mathrm{d}x=\frac19\int_{0}^{1}\ln(x)\mathrm{d}\bigg[\frac{x}{(1+x)^3}\bigg]\\ &=&\frac19\ln(x)\frac{x}{(1+x)^3}\bigg|_0^1-\frac19\int_0^1\frac1{(1+x)^3}\mathrm{d}x=-\frac1{18}.\\ \end{eqnarray}

Answer 5

Your integral is also the derivative of the Beta function. Given that $\ln(x) = \displaystyle \lim_{t\to 0+} \frac{d}{dt} x^t$

\begin{align*} \int_{0}^{\infty} \frac{x^{2}\ln(x)}{(1+x^3)^4}dx=&\lim_{t \to 0} \frac{d}{dt} \int_{0}^{\infty} \frac{x^{2+t}}{(1+x^3)^{4}} \\ =& \lim_{t\to 0+} \frac{d}{dt} \frac{1}{3}\int_{0}^{1} (1-w)^{\frac{t}{3}}w^{2-\frac{t}{3}}dw \quad \left(w \mapsto \frac{1}{1+x^3}\right)\\ =& \lim_{t\to 0+} \frac{d}{dt} \frac{1}{3}B\left(\frac{t}{3}+1,3-\frac{t}{3}\right)\\ =& \lim_{t\to 0+} \frac{d}{dt} \frac{1}{3}\frac{\Gamma\left(\frac{t}{3}+1\right)\Gamma\left(3-\frac{t}{3}\right)}{\Gamma(4)}\\ =& \lim_{t\to 0+} \frac{d}{dt} \frac{1}{486}t(6-t)(3-t)\Gamma\left(\frac{t}{3}\right)\Gamma\left(1-\frac{t}{3}\right) \quad \left(\Gamma(z+1) = z\Gamma(z) \right)\\ =& \lim_{t\to 0+} \frac{d}{dt} \frac{\pi }{486}(t^3-9t^2+18t)\csc\left(\frac{\pi t}{3}\right) \quad \left(\Gamma(1-z)\Gamma(z) = \frac{\pi}{\sin(\pi z)} \right)\\ =& -\frac{1}{18} \end{align*}