5

These integrals are closely related since $\frac{\pi^2}{8}=\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}$ and $G=\sum_{n\mathop=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^2}$. I'm not able to prove them though.

Show that $$\int_0^1 \frac{1}{\sqrt{x^2-1}}\log\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\right)\,\mathrm{d}x =\frac{\pi^2}{8}$$ $$\int_1^{\infty} \frac{1}{x\sqrt{x^2-1}}\log\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\right)\,\mathrm{d}x =2G$$

where $G$ is Catalan's constant.

user85798
  • 1

2 Answers2

8

The first integral is easier than it looks. Write

$$\begin{align}I &= \int_0^1 \frac{dx}{\sqrt{x^2-1}} \log{\left (\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} \right )} \\ &= -i \int_0^1 \frac{dx}{\sqrt{1-x^2}} \log{\left (x+i \sqrt{1-x^2}\right )}\\ &= \int_0^1 \frac{dx}{\sqrt{1-x^2}} \arccos{x}\\ &= -\frac12 \left [\arccos^2{x} \right ]_0^1\\ &= \frac{\pi^2}{8}\end{align}$$

The second integral is also straightforward, using the sub $x=\cosh{t}$:

$$\begin{align}J &= \int_1^{\infty} \frac{dx}{x \sqrt{x^2-1}} \log{\left (\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} \right )} \\ &= \int_0^{\infty} dt \frac{t}{\cosh{t}}\\ &=2 \sum_{k=0}^{\infty} (-1)^k \int_0^{\infty} dt \, t \, e^{-(2 k+1) t}\\ &= 2 \sum_{k=0}^{\infty} \frac{(-1)^k}{(2 k+1)^2} \\ &= 2 G\end{align}$$

Ron Gordon
  • 138,521
3

Note that $\cosh^{-1}{x}=\log\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\right)$ and $\frac{d}{dx}\cosh^{-1}{x}=\frac{1}{\sqrt{x^2-1}}$. Substituting $u=\cosh^{-1}{x}$ into the first integral, we get:

$$\int_0^1 \frac{1}{\sqrt{x^2-1}}\log\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\right)\,\mathrm{d}x=\int_{\frac{i\pi}{2}}^{0}udu=\frac12u^2\big |_{\frac{i\pi}{2}}^0=-\frac12(\frac{i\pi}{2})^2=\frac{\pi^2}{8}.$$

David H
  • 29,921