# Different method of proving $\int_{0}^{1}{ dx\over x}\ln\left({1-\sqrt{x}\over 1+\sqrt{x}}\cdot{1+\sqrt[3]{x}\over 1-\sqrt[3]{x}}\cdots\right)=-\pi^2$

Given the integral $(1)$

$$\text{Prove that}\ \int_{0}^{1}{\mathrm dx\over x}\ln\left({1-\sqrt{x}\over 1+\sqrt{x}}\cdot{1+\sqrt[3]{x}\over 1-\sqrt[3]{x}}\cdot{1-\sqrt[5]{x}\over 1+\sqrt[5]{x}}\right)=-\pi^2\tag1$$

An attempt:

$u=x^2,x^3 \text{and}\ x^5$, then $(1)$ becomes

$$\int_{0}^{1}{\mathrm du\over u}\ln\left[\left({1-u\over 1+u}\right)^2\left({1+u\over 1-u}\right)^3\left({1-u\over 1+u}\right)^5\right]\tag2$$

Simplify to

$$\int_{0}^{1}{\mathrm du\over u}\ln\left({1-u\over 1+u}\right)^4\tag3$$

Apply $\ln\left({1+u\over 1-u}\right)$ series, then we have

$$-\sum_{n=0}^{\infty}{1\over (2n+1)}\int_{0}^{1}u^{2n}\mathrm du\tag4$$

$$-8\sum_{n=0}^{\infty}{1\over (2n+1)^2}=-\pi^2\tag5$$

Looking for another method of proving $(1)$

• How exactly do you define such a substitution? That is how have you made the variable u simultaneously equal to three different expressions in x – Triatticus Mar 25 '17 at 18:36
• @Triatticus, use the logarithm to separate the integral into a sum of integrals. Then use a substitution in each one. – Zaid Alyafeai Mar 25 '17 at 18:40
• A shorter approach would be to use the log to separate the integral into three integrals. Then use the summation in (4) directly. – Zaid Alyafeai Mar 25 '17 at 18:44
• What does this integral mean? Is there some physical interpretation? It's interesting...Thanks – PiE Mar 25 '17 at 22:57

Let us generalize to

$$\int^1_0 \frac{\mathrm d x}{x} \log \left(\frac{1-x^a}{1+x^a}\right)$$

Let $x^a = y$ which implies $\mathrm d x = \frac{1}{a}y^{\frac{1}{a}-1} \mathrm d y$

\begin{align}\int^1_0 \frac{\frac{1}{a}y^{\frac{1}{a}-1}\mathrm dy}{y^{\frac{1}{a}}} \log \left(\frac{1-y}{1+y}\right) &= \frac{1}{a}\int^1_0 \frac{\mathrm dy}{y} \log \left(\frac{1-y}{1+y}\right) \\&=\frac{1}{a}\int^1_0 \frac{\log(1-y)}{y} \mathrm dy- \frac{1}{a}\int^1_0 \frac{\log(1+y)}{y} \mathrm dy \\&= \frac{\mathrm{Li}_2(-1)-\mathrm{Li}_2(1)}{a} \\ &= -\frac{\pi^2}{4a}\end{align}

Now consider the general case

\begin{align}\int^1_0 \frac{\mathrm d x}{x} \log \left(\prod_{k=1}^{n}\frac{1-x^{a_k}}{1+x^{a_k}}\right) &=\sum_{k=1}^n\int^1_0 \frac{\mathrm d x}{x} \log \left(\frac{1-x^{a_k}}{1+x^{a_k}}\right) \\&=-\frac{\pi^2}{4}\sum_{k=1}^n\frac{1}{a_k} \end{align}

So we get the result for $a_k \geq 1$

$$\int^1_0 \frac{\mathrm d x}{x} \log \left[\prod_{k=1}^{n}\frac{1-x^{1/a_k}}{1+x^{1/a_k}}\right] =-\frac{\pi^2}{4}\sum_{k=1}^n a_k$$

Similarly

$$\int^1_0 \frac{\mathrm d x}{x} \log \left[\prod_{k=1}^{n}(-1)^{k-1}\frac{1-x^{1/a_k}}{1+x^{1/a_k}}\right] =\frac{\pi^2}{4}\sum_{k=1}^n a_k (-1)^k$$

In your case we have the sequence

$$a_k = 2, 3 , 5$$

Hence

\begin{align}\int_{0}^{1}{\mathrm dx\over x}\ln\left({1-\sqrt{x}\over 1+\sqrt{x}}\cdot{1+\sqrt[3]{x}\over 1-\sqrt[3]{x}}\cdot{1-\sqrt[5]{x}\over 1+\sqrt[5]{x}}\right)&= \int_{0}^{1}{\mathrm dx\over x}\log \prod^3_{k=1}(-1)^{k-1}\left(\frac{1-x^{1/a_k}}{1+x^{1/a_k}}\right)\\&=\frac{\pi^2}{4}\sum_{k=1}^3 (-1)^k a_k \\&= \frac{\pi^2(-2+3-5)}{4} = -\pi^2\end{align}

Perhaps if we are able to interchange the limit and the integral and under some conditions on $a_k$

$$\int^1_0 \frac{\mathrm d x}{x} \log \left[\prod_{k=1}^{\infty}\frac{1-x^{1/a_k}}{1+x^{1/a_k}}\right] =-\frac{\pi^2}{4}\sum_{k=1}^{\infty} a_k$$

• Thank you @Zaid Alyafeai for the generalised case. I was trying to but couldn't. – gymbvghjkgkjkhgfkl Mar 25 '17 at 23:04
• @Bui, you are welcome. You were very close to the general form. – Zaid Alyafeai Mar 25 '17 at 23:07