calculus - The 3 Integral $int_0^infty {x,{rm d}xover sqrt[3]{,left(e^{3x}-1right)^2,}}=frac{pi}{3sqrt 3}big(log 3-frac{pi}{3sqrt 3} big)$

Hi I am trying evaluate this integral and obtain the closed form:$$
I:=\int_0^\infty \frac{x\,dx}{\sqrt[\large 3]{(e^{3x}-1)^2}}=\frac{\pi}{3\sqrt 3}\left(\log 3-\frac{\pi}{3\sqrt 3} \right).
$$
The integral and result has all 3's everywhere. I am not sure how to approach this on. The denominator seems to be a problem.

If $\displaystyle x=\frac{2in\pi}{3}$ we have a singularity but I am not sure how to use complex methods. We will have a branch cut because of the root function singularity.

Differentiating under the integral sign did not help either. I tried partial integration with $v=(e^{3x}-1)^{\frac{2}{3}}$ but this did not simplify since I get a power $x^n, \ (n>1)$ in the new integral.

Thanks, how can we evaluate the integral I?

Answer

Use the substitution $e^{3x}-1=t^3 \Rightarrow dx=\frac{t^2}{1+t^3}\,dt$ to obtain:
$$I=\frac{1}{3}\int_0^{\infty} \frac{\ln(1+t^3)}{1+t^3}\,dt$$
Consider
$$I(a)=\frac{1}{3}\int_0^{\infty} \frac{\ln(a^3+t^3)}{1+t^3}\,dt$$
$$\Rightarrow I'(a)=\int_0^{\infty} \frac{a^2}{(1+t^3)(a^3+t^3)}\,dt=\frac{a^2}{a^3-1}\left(\int_0^{\infty} \frac{dt}{1+t^3}-\int_0^{\infty} \frac{dt}{a^3+t^3} \right)$$

Both the integrals are easy to evaluate, hence
$$I'(a)=\frac{a^2}{a^3-1}\left(\frac{2\pi}{3\sqrt{3}}-\frac{2\pi}{3\sqrt{3}a^2}\right)=\frac{2\pi}{3\sqrt{3}}\frac{a^2-1}{a^3-1}=\frac{2\pi}{3\sqrt{3}}\frac{a+1}{a^2+a+1}$$
$$I(a)=\frac{2\pi}{3\sqrt{3}}\left(\frac{1}{2}\ln(a^2+a+1)+\frac{\arctan\left(\frac{2a+1}{\sqrt{3}}\right)}{\sqrt{3}}\right)+C\,\,\,\,\,\,\,(*)$$
To determine $C$, I evaluate $I(0)$ i.e
$$I(0)=\int_0^{\infty} \frac{\ln t}{1+t^3}\,dt=\int_0^1 \frac{\ln t}{1+t^3}\,dt+\int_1^{\infty} \frac{\ln t}{1+t^3}\,dt=J_1+J_2$$
In $J_1$, use the substitution $t^3=y$ to get:
$$J_1=\frac{1}{9}\int_0^1 \frac{\ln y}{1+y}\frac{dy}{y^{2/3}}=\frac{1}{9}\sum_{k=0}^{\infty} (-1)^k\int_0^1 y^{k-2/3}\ln y=-\sum_{k=0}^{\infty} \frac{(-1)^k}{(3k+1)^2}$$
For $J_2$, perform the transformation $t \mapsto 1/t$ and use the substitution $t^3=y$ to obtain:
$$J_2=\sum_{k=0}^{\infty} \frac{(-1)^k}{(3k+2)^2}$$
$$\Rightarrow I(0)=\sum_{k=0}^{\infty}(-1)^k \left(\frac{1}{(3k+2)^2}-\frac{1}{(3k+1)^2}\right)$$

I couldn't evaluate the above sum, W|A gives $-2\pi^2/27$. I know I shouldn't be posting this answer in the first place when I don't even know how to evaluate the sum but I wrote a lot already and didn't realise that it would be difficult to evaluate the sum. I hope the community doesn't downvote this and I hope somebody could show how to evaluate this sum.

I found an another method to evaluate the sum.

$$\sum_{k=0}^{\infty}(-1)^k \left(\frac{1}{(3k+2)^2}-\frac{1}{(3k+1)^2}\right)=\sum_{k=0}^{\infty} (-1)^k \int_0^1 \int_0^1 \left((xy)^{3k+1}-(xy)^{3k}\right)\,dx\,dy$$
$$=\int_0^1\int_0^1 \frac{xy-1}{1+x^3y^3}=-\frac{2\pi^2}{27}\,\,\,\,\,(**)$$
$$I(0)=\frac{2\pi}{3\sqrt{3}}\left(\frac{\pi}{6\sqrt{3}}\right)+C=\frac{-2\pi^2}{27} \Rightarrow C=\frac{-3\pi^2}{27}$$
$$\Rightarrow I(1)=\frac{2\pi}{3\sqrt{3}}\left(\frac{\ln 3}{2}+\frac{\pi}{3\sqrt{3}}\right)-\frac{3\pi^2}{27}=\frac{\pi}{3\sqrt{3}}\left(\ln 3-\frac{\pi}{3\sqrt{3}}\right)$$

$\blacksquare$

---

Proof of $(*)$:

$$\int \frac{a+1}{a^2+a+1}\,da=\frac{1}{2}\int \frac{2a+1}{a^2+a+1}\,da+\frac{1}{2}\int \frac{1}{a^2+a+1}\,da=C+D$$
$$C=\frac{1}{2}\int \frac{2a+1}{a^2+a+1}\,da \stackrel{a^2+a+1 \mapsto t}{=} \frac{1}{2}\int \frac{dt}{t}=\frac{1}{2}\ln t=\frac{1}{2}\ln(a^2+a+1)$$
$$D=\frac{1}{2}\int \frac{1}{a^2+a+1}\,da=\frac{1}{2}\int \frac{1}{\left(a+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\,da=\frac{\arctan\left(\frac{2a+1}{\sqrt{3}}\right)}{\sqrt{3}}$$

---

Proof of $(**)$

$$\int_0^1 \int_0^1 \frac{xy-1}{1+x^3y^3}\,dx\,dy=\int_0^1 \frac{\ln(1-y+y^2)-2\ln(1+y)}{3y}\,dy$$
$$=\frac{1}{3}\int_0^1 \frac{\ln(1-y+y^2)}{y}\,dy-\frac{2}{3}\int_0^1 \frac{\ln(1+y)}{y}\,dy=\frac{1}{3}X_1-\frac{2}{3}X_2$$

$$\begin{aligned}
X_1&=\int_0^1\frac{\ln(1-y+y^2)}{y}\\
&=-\sum_{k=1}^{1} \frac{1}{k} \int_0^1 y^{k-1}(1-y)^k=-\sum_{k=1}^{\infty} \frac{1}{2k}\int_0^{\infty} y^{k-1}(1-y)^{k-1} \\
&=-\frac{1}{2}\sum_{k=0}^{\infty} \frac{1}{k+1}\frac{k!^2}{(2k+1)!}\\

\end{aligned}$$
Ron Gordon has shown an excellent method to evaluate the above sum here: Definite Integral $\int_0^1\frac{\ln(x^2-x+1)}{x^2-x}\,\mathrm{d}x$

Hence, $X_1=-\frac{\pi^2}{18}$.

$$\begin{aligned}
X_2 &=\int_0^1 \frac{\ln(1+y)}{y}\,dy=-\sum_{k=1}^{\infty} \frac{(-1)^k}{k} \int_0^1 y^{k-1}\,dy\\
&=\sum_{k=1}^{\infty} -\frac{(-1)^k}{k^2}\\
&=\frac{\pi^2}{12}\\
\end{aligned}$$

Therefore,
$$\int_0^1 \int_0^1 \frac{xy-1}{1+x^3y^3}=-\frac{2\pi^2}{27}$$