On the $\int_0^\infty \frac{x \ln(x)}{1+x^3} \, dx$

Question

While trying to evaluate, \begin{align*} \mathcal{I} &= \int_0^\infty \frac{x \ln(x)}{1+x^3} \, dx \tag{1} \\ &= \left( \int_0^1 + \int_1^\infty \right) \frac{x \ln(x)}{1+x^3} \, dx \\ &= \int_0^1 \frac{(x-1) \ln(x)}{1+x^3} \, dx \\ &= \int_0^1 (x-1) \ln(x) \left( \sum_{n \geq 0} (-1)^n x^{3n} \right) \, dx \\ &= \sum_{n \geq 0} (-1)^n \int_0^1 (x^{3n+1} - x^{3n}) \ln(x) \, dx \tag{2} \\ &= \sum_{n \geq 0} (-1)^{n-1} \left( \frac{1}{(3n+2)^2} - \frac{1}{(3n+1)^2} \right) \tag{3} \end{align*}

Two integrals in $(2)$ are well known and closely related to gamma function.

And I don't know how to evaluate $(3)$, I searched but didn't get so I thought let's try to evaluate $(1)$ using beta function,

\begin{align*} J(a) &= \int_0^\infty \frac{x^a}{1+x^3} \, dx \\ &= \int_0^\infty \frac{y^{(a-2)/3}}{3(1+y)} \, dy && (x^3 = y) \\ &= \int_0^\infty \frac{y^{((a+1)/3) - 1}}{3(1+y)} \, dy \\ &= \frac{1}{3} \, \beta \left( \frac{a+1}{3}, 1 - \frac{a+1}{3} \right) \\ J(a) &= \frac{1}{3} \frac{\pi}{\sin \left( \frac{\pi}{3} (a+1) \right)} \tag{4} \end{align*}

As we can see that, $$J'(1) = \mathcal{I}$$ And, $$J'(a) = \frac{-\pi^2}{9} \frac{\cos \left( \frac{\pi}{3} (a+1) \right)}{\sin^2 \left( \frac{\pi}{3} (a+1) \right)} \tag{5}$$ And, $$J'(1) = \frac{2\pi^2}{27} = 0.73108... $$

So can I conclude that, $$\sum_{n \geq 0} (-1)^{n-1} \left( \frac{1}{(3n+2)^2} - \frac{1}{(3n+1)^2} \right) = \frac{2\pi^2}{27} \qquad ? $$

I checked it on Wolfram Alpha and got $(1)$ is approximately equal to $0.731082$,
And $(3)$ is approximately equal to $0.7311..$

Or is there any other way to find the sum in $(3) \, ?$

Answer 1

One way to evaluate the sum is to use contour integration. Let $f(z)=\frac{\csc(\pi z)}{(3z+2)^2}-\frac{\csc(\pi z)}{(3z+1)^2}$

Then, the residue theorem guarantees that

$$\begin{align} \oint_{|z|=N+1/2}f(z)\,dz&=2\pi i\sum_{n=-N}^N\text{Res}(f(z), n)\\\\ &+2\pi i \text{Res}\left(\frac{\csc(\pi z)}{(3z+2)^2},z=-2/3\right)\\\\ &-2\pi i \text{Res}\left(\frac{\csc(\pi z)}{(3z+2)^2},z=-1/3\right)\\\\ &=2\pi i \sum_{n=-N}^N \frac{(-1)^n}{\pi}\left(\frac1{(3n+2)^2}-\frac1{(3n+1)^2}\right)\\\\ &+2\pi i (4\pi/27)\tag1 \end{align}$$

Letting $N\to \infty$, the left-hand side of $(1)$ goes to zero and we find that

$$\sum_{n=-\infty}^\infty \frac{(-1)^{n-1}}{\pi}\left(\frac1{(3n+2)^2}-\frac1{(3n+1)^2}\right)=\frac{4\pi}{27}$$

from which we obtain

$$\sum_{n=0}^\infty (-1)^{n-1}\left(\frac1{(3n+2)^2}-\frac1{(3n+1)^2}\right)=\frac{2\pi^2}{27}$$

as was to be shown!

Answer 2

Typing out the series we get $$1-\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{7^2}-\cdots$$ We may then write $$\begin{align} \sum_{n=0}^\infty (-1)^{n-1}\left(\frac{1}{(3n+2)^2}-\frac{1}{(3n+1)^2}\right) &= \sum_{n=1}^\infty (-1)^{n-1}\left(\frac{1}{n^2}-\frac{1}{(3n)^2}\right)\\\\ & = (1-3^{-2})\eta(2) \\\\ &= \frac{2\pi^2}{27} \end{align}$$ and we are done!

Answer 3

The story: Here is yet an other solution, it was started some hours ago, but could not be finalized, so just for having various ideas to attack similar integrals, i will complete and submit. The idea is to get a relation of the given integral with the logarithm and the dilogarithm, then use functional equations of these special functions. We associate the integrals having in place of the power $x^1$ in the numerator $1,x,x^2$, and observe that the linear combination $1-x+x^2$ reduces the algebraic complexity, and $(1-x+x^2)/(1+x^3)=1/(1+x)$ (and it appears times $\log x$).

Unfortunately, using $x^2$ leads to a divergent integral. So we morally isolate the "principal value", but to have a rigorous computation let us integrate on $(\epsilon,M)$ instead of $(0,\infty)$.

Also, the integral involving $x^2$ (instead of $x^1$) can be related by the substitution $y=x^3$ with an integral of the same shape, $1/(1+y)$ (and it appears times $\log y$, some substitution scalar also shows up).

The two sources of divergence cancel each other, up to some "monodromy contribution" of dilogarithms at related $\infty$-values, and exactly this contribution is the value of the integral to be computed.

The algebraic idea was easy when started, but putting things on paper with well defined notations is not so easy...

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Notations and facts: It is useful to have the notation $J_k(A)=\displaystyle\int_A\frac{x^k\log x}{1+x^3}\; dx$, where $A$ is an interval of the shape $(\epsilon, M)$, included in $(0,\infty)$. For $A=(0,\infty)$ we simply write $J_k(A)=\Bbb J_k$. We use below only intervals $A$ invariated by the map $x\to 1/x$, so $\epsilon M=1$.

Recall the primitives of $(\log x)/(x-a)$, and we introduce $F$ to be used below, $$ \int \frac{\log x}{x-a}\; dx = \operatorname{Li}_2\left(\frac xa\right) +\log x\log\left(1-\frac xa\right) +\text{Constant ,} $$ and we need only $$ \int \frac{\log x}{x+1}\; dx = \underbrace{\operatorname{Li}_2(-x) +\log x\log(1+x)}_{:=F(x)} +\text{Constant .} $$ involving the dilogarithm function $\operatorname{Li}_2$, we show the formula first for a real $a>0$ (substitution $x=ay$), and we may extend it for comlex values of $a$, if we want so, but then monodromy aspects have to be considered. In our case, we expect a real answer, so we may adjust in case of choices.

The dilogarithm satisfies the following functional equation, page 8, §2 of an article of Don Zagier: $$ \tag{*} \operatorname{Li}_2\left(\frac1z\right) = -\operatorname{Li}_2(z) -\frac 12\log^2(-z) -\frac{\pi^2}6\ . $$

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Computation: The substitution $x=\frac 1y$ shows the useful relation $$ J_1(A)=\int_A\frac{x\log x}{1+x^3}\; dx =\int_A\frac{-\frac 1y\log y}{\frac{1+y^3}{y^3}}\; \left(-\frac{dy}{y^2}\right) =-\int_A\frac{\log y}{1+y^3}\; dy=-J_0(A)\ . $$ The integral $J_2(A)$ can be rewritten, if convergent ($\epsilon>0$), since we can substitute $y=x^3$, so $x\in A$ implies $y\in A^3$, $$ J_2(A) =\int_A\frac{x^2\;\frac 13\log (x^3)}{1+x^3}\; dx =\frac 19\int_A\frac{\log x^3}{1+x^3}\; d(x^3) =\frac 19\int_{A^3}\frac{\log y}{y+1}\; dy\ . $$

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We can now compute: $$ \begin{aligned} 2J_1(A) &=J_2(A)-(J_0(A)-J_1(A)+J_2(A))\\ &=J_2(A) - \int_A\frac{(1-x+x^2)\log x}{1+x^3}\; dx\\ &=\frac 19\int_{A^3}\frac{\log x}{1+x}\; dx - \int_A\frac{(1-x+x^2)\log x}{1+x^3}\; dx =\left(\frac 19\int_{A^3}-\int_A\right)\frac{\log x}{1+x}\; dx\\ &=\left(\frac 19\int_{A^3}-\int_A\right)F'(x)\; dx \\ &=\frac 19(F(M^3)-F(\epsilon^3))\ -\ (F(M)-F(\epsilon))\ ,\\[2mm] &\qquad\text{ and we pass now to the limit $\epsilon\searrow 0$.}\\ &\qquad\text{ The term $\log x\log(1+x)$ does not contribute.}\\ &\qquad\text{ The dilog terms in $\epsilon$, $\epsilon^3$ have zero limit.}\\ &\qquad\text{ So we obtain:}\\ 2\Bbb J_1 &=\lim_{M\to\infty}\left(\frac 19 \operatorname{Li}_2(-M^3) -\operatorname{Li}_2(-M) \right)\\ &\qquad\text{ and with the functional equation $(*)$}\\ &=\lim_{\epsilon\searrow 0} \left(\ \frac 19\left( -\operatorname{Li}_2(-\epsilon^3) -\frac 12\log^2(\epsilon^3) -\frac{\pi^2}6\right) \ -\ \left( -\operatorname{Li}_2(-\epsilon) -\frac 12\log^2(\epsilon) -\frac{\pi^2}6\right) \ \right)\\ &=\frac {\pi^2}6\left(1-\frac 19\right)=\frac{4\pi^2}{27}\ .\text{ So} \\[3mm] \Bbb J_1 &=\bbox[yellow]{\ \frac {2\pi^2}{27}\ }\ . \end{aligned} $$ $\square$