Prove $H_{\frac{1}{8}}=\int_{0}^{1}\frac{x^{\frac{1}{8}}-1}{x-1}dx$

Question

I want to prove :

$$H_{\frac{1}{8}}=8-\frac{\pi}{2}-4\ln\left(2\right)-\frac{1}{\sqrt{2}}\left(\pi+\ln\left(2+\sqrt{2}\right)-\ln\left(2-\sqrt{2}\right)\right)$$

Where $H_n$ is the n-th harmonic number.

Using the integral representation of harmonic numbers it follows:

$$H_{\frac{1}{8}}=\int_{0}^{1}\frac{x^{\frac{1}{8}}-1}{x-1}dx$$

Setting $x^{\frac{1}{8}}\ \mapsto u$ we have: $$=8\int_{0}^{1}\frac{\left(u-1\right)u^{7}}{u^{8}-1}du=8\int_{0}^{1}\frac{u^{7}}{\sum_{i=0}^{7}u^{i}}dx$$

I know how to continue, but the way I use takes much time, so what would be the best way to prove the relation?

Answer 1

Let $t=x^{\frac{1}{8}}$

$$H_{\frac18}=\int_{0}^{1}\frac{x^{\frac{1}{8}}-1}{x-1}dx =8\int_{0}^{1}\frac{t^8-t^7}{t^8-1}dt =8-8\int_{0}^{1}\frac{t^7-1}{t^8-1}dt $$

Decompose the integrand

$$\frac{t^7-1}{t^8-1} = \frac14\frac{1}{t+1}+\frac14\frac{t+1}{t^2+1}+\frac12\frac{t^3+1}{t^4+1}$$

and the integral

$$H_{\frac18}=8-2\ln 2 - 2I_1 - 4I_2\tag 1$$

where

$$I_1 = \int_{0}^{1} \frac{t+1}{t^2+1}dt =\frac12\ln2+\frac\pi4\tag 2$$

$$I_2 = \int_{0}^{1} \frac{t^3+1}{t^4+1}dt=\frac14\ln2+\int_{0}^{1} \frac{1}{t^4+1}dt$$

Integrate

\begin{align} &\int_{0}^{1} \frac{2}{t^4+1}dt= \int_0^1\frac{1+x^2}{x^4+1} dx + \int_0^1\frac{1-x^2}{x^4+1} dx\\ =&\int_0^1\frac{d(x-\frac1{x})}{(x-\frac1{x})^2+2} - \int_0^1\frac{d(x+\frac1{x})}{(x+\frac1{x})^2-2} =\frac\pi{2\sqrt2}+\frac1{2\sqrt2}\ln\frac{\sqrt2+1}{\sqrt2-1} \end{align}

Then

$$I_2 =\frac14\ln2+\frac\pi{4\sqrt2}+\frac1{4\sqrt2}\ln\frac{\sqrt2+1}{\sqrt2-1}\tag 3$$

Plug (2) and (3) into (1), $$H_{\frac{1}{8}}=8-\frac{\pi}{2}-4\ln\left(2\right)-\frac{1}{\sqrt{2}}\left(\pi+\ln\left(2+\sqrt{2}\right)-\ln\left(2-\sqrt{2}\right)\right)$$

Answer 2

Using partial fraction decomposition $$8\frac{\left(u-1\right)u^{7}}{u^{8}-1}=8-2\frac{u+1}{u^2+1}-4\frac{u^3+1}{u^4+1}-2\frac{1}{u+1}$$ seems to be workable.

Answer 3

Here is a non-elementary approach to finding $H_{\frac{1}{8}}$ that makes use of the digamma function.

Making use of the result $$\frac{H_n}{n} = - \int_0^1 x^{n - 1} \ln (1 - x) \, dx,$$ a result that can be analytically extended to all $n \in \mathbb{R}$ such that $n > -1$. Setting $n = 1/8$ one obtains \begin{align} H_{\frac{1}{8}} &= -\frac{1}{8} \int_0^1 x^{-\frac{7}{8}} \ln (1 - x) \, dx\\ &= \frac{1}{8} \sum_{n = 1}^\infty \frac{1}{n} \int_0^1 x^{n - \frac{7}{8}} \, dx\\ &= \sum_{n = 1}^\infty \frac{1}{n(8n + 1)}\\ &= 8 \sum_{n = 1}^\infty \left (\frac{1}{8n} - \frac{1}{8n + 1} \right ). \end{align}

From one of the series representations for the digamma function, namely $$\psi(z + 1) = - \gamma + \sum_{n = 1}^\infty \left (\frac{1}{n} - \frac{1}{n + z} \right ),$$ where $\gamma$ denotes the Euler–Mascheroni constant, setting $z = 1/8$ we find $$\psi \left (\frac{9}{8} \right ) = -\gamma + 8 \sum_{n = 1}^\infty \left (\frac{1}{8n} - \frac{1}{8n + 1} \right ).$$ Thus $$H_{\frac{1}{8}} = \gamma + \psi \left (\frac{9}{8} \right ).$$ Now it is just a matter of finding the value for the digamma function.

It is known that for positive integers $r$ and $m$ ($r < m$), the digamma function may be expressed in terms of the Euler–Mascheroni constant and a finite number of elementary functions. Making use of the following recurrence formula for the digamma function, namely $$\psi (x + 1) = \psi(x) + \frac{1}{x},$$ we see that $$\psi \left (\frac{9}{8} \right ) = \psi \left (1 + \frac{1}{8} \right ) = \psi \left (\frac{1}{8} \right ) + 8.$$ The value for $\psi (1/8)$ can now be found from Gauss' digamma theorem which states that $$\psi \left (\frac{r}{m} \right ) = -\gamma - \ln (2m) - \frac{\pi}{2} \cot \left (\frac{r\pi}{m} \right ) + 2 \sum_{n = 1}^{\lfloor \frac{m - 1}{2} \rfloor} \cos \left (\frac{2 \pi n r}{m} \right ) \ln \left (\sin \left (\frac{\pi n}{m} \right ) \right ).$$ Setting $r = 1, m = 8$ gives \begin{align} \psi \left (\frac{1}{8} \right ) &= -\gamma - \ln (16) - \frac{\pi}{2} \cot \left (\frac{\pi}{8} \right ) + 2 \sum_{n = 1}^3 \cos \left (\frac{\pi n}{4} \right ) \ln \left (\sin \left (\frac{\pi n}{8} \right ) \right )\\ &= -\gamma - 4 \ln 2 - \frac{\pi}{2} (1 + \sqrt{2}) - \sqrt{2} \ln (1 + \sqrt{2}). \end{align} So finally we arrive at the following value for $H_{\frac{1}{8}}$ of $$H_{\frac{1}{8}} = 8 - 4 \ln 2 - \frac{\pi}{2}(1 + \sqrt{2}) - \sqrt{2} \ln (1 + \sqrt{2}).$$