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

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?

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

• I apologise for posting an incomplete answer, it would be very much appreciated if somebody could help me in evaluating the sum. Many thanks! – Pranav Arora Jun 4 '14 at 6:42
• to compute your sum: $I(0)=\sum_n\frac{1}{(2n)^2}-\sum_n\frac{1}{(6n)^2}-\sum_{gcd(n,6)=1}\frac{1}{n^2}=(2^{-2}-6^{-2}-(1-2^{-2})(1-3^{-2}))\sum_n\frac{1}{n^2}=-\frac{4}{9}\frac{\pi^2}{6}=-\frac{2\pi^2}{27}$ – user8268 Jun 4 '14 at 7:22
• @user8268: Thanks for your input but I have no idea about how you wrote the sum in that way and evaluated it. Should I create a new thread for this? :) – Pranav Arora Jun 4 '14 at 7:33
• is the first equality clear? If yes, the $2^{-2}$ and $3^{-2}$ terms in the second equality are I hope OK; is the $(1-2^{-2})(1-3^{-2})$ term unclear? (it comes from Euler's $\sum_n 1/n^2=\prod_p1/(1-1/p^2)$; when we sum only over $n$'s prime to $6$, we need to omit $2$ and $3$ from the product) – user8268 Jun 4 '14 at 8:25
• @user8268: I tried writing down a few terms of my summation and it looked similar to your first equality but how did it occur to you that we can write the sum that way? Even after writing down so many terms, it wasn't obvious to me that I can rewrite the sum in that way. Is this some kind of standard exercise found in books? I am interested in knowing more about this. Also, I haven't seen that Euler's sum before. Thanks a lot. :) – Pranav Arora Jun 4 '14 at 11:46

Substitute $e^{-x}=t$, then the integral can be written as: \begin{align*} \int_0^\infty \frac{x\,dx}{\sqrt[\large 3]{(e^{3x}-1)^2}}\, dx &= -\int_{0}^{1} \, \frac{t\, \log{t}}{\left(1-t^3\right)^{2/3}}\, dt\tag 1 \end{align*}

Consider:

\begin{align*} I(a) &= \int_{0}^{1} \, \frac{t^{a+1}}{\left(1-t^3\right)^{2/3}}\, dt \\ &= -\frac{1}{3} \, {\rm B}\left(\frac{1}{3}, \frac{a+2}{3}\right) \\ I'(0) &= \int_{0}^{1} \, \frac{t\, \log{t}}{\left(1-t^3\right)^{2/3}}\, dt \\ &= \frac{1}{9} \, {\left(\gamma + \psi\left(\frac{2}{3}\right)\right)} {\rm B}\left(\frac{1}{3}, \frac{2}{3}\right) \tag{2} \end{align*}

Simplifying $(2)$ by using Gauss's digamma theorem for $m<k$

$\displaystyle \psi\left(\frac{m}{k}\right) = -\gamma -\ln(2k) -\frac{\pi}{2}\cot\left(\frac{m\pi}{k}\right) +2\sum_{n=1}^{\lfloor \frac{k-1}{2} \rfloor} \cos\left(\frac{2\pi nm}{k} \right) \ln\left(\sin\left(\frac{n\pi}{k}\right)\right)$

and Euler's reflection formula for gamma functions,

$\displaystyle {\rm B}(1-z, z)= \Gamma(1-z)\Gamma(z) = \frac{\pi}{\sin(\pi z)}$

and from $(1)$,

\begin{align*} \int_0^\infty \frac{x\,dx}{\sqrt[\large 3]{(e^{3x}-1)^2}}\, dx &= \frac{\pi}{3\sqrt 3}\left(\log 3-\frac{\pi}{3\sqrt 3}\right) \end{align*}

• My slow typing allowed for not making the first answer. Only aspect I would add is that it is easier to use \begin{align} \psi(2/3) = \frac{1}{6}(\sqrt{3} \pi - 9 \ln 3) - \gamma \end{align} and \begin{align} \int_{0}^{1} t^{\mu-1} (1-t)^{\nu-1} \ln(t) \ dt = B(\mu, \nu) ( \psi(\mu) - \psi(\mu+\nu)) \end{align}. – Leucippus Jun 4 '14 at 6:20
• Hi, thanks for your input. I don't think I'm faster either, I just use Emacs+AUCTeX, which allows me to define my shortcuts for symbols, and copy it here after typing. – gar Jun 4 '14 at 6:31
• @Integrals : You are welcome! – gar Jun 20 '14 at 3:24

• Nice approach, but it looks like you have a typo (missing log) at the beginning. – David H Aug 7 '14 at 23:51
• @DavidH, thanks for pointing out the typo. – xpaul Aug 8 '14 at 0:49