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Polygamma identity [entire topic moved to new forum]

Apostolos J. Kos on Thursday, October 01 2015, 09:22 AM
0

Let \( \psi \) denote the digamma function. Evaluate:

$$\psi\left( \frac{7}{8} \right) - \psi\left(\frac{3}{8}\right)$$

I discovered this result by chance yesterday. I have a solution, not a complicated one.

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Accepted Answer

  • Replied by Apostolos J. Kos on Sunday, October 18 2015, 10:24 PM
    We are invoking the definition of the digamma, namely:

    $$\psi(z)=-\gamma + \sum_{n=0}^{\infty} \left[ \frac{1}{n+1} - \frac{1}{z+n} \right], \;\; z \notin \mathbb{Z}^{-} \cup \{0 \}$$

    So, we have successively:

    $$\begin{aligned}
    \psi\left( \frac{7}{8} \right) - \psi\left(\frac{3}{8}\right) &= \sum_{n=0}^{\infty}\left ( \frac{1}{n+3/8} - \frac{1}{n+ 7/8} \right ) \\
    &=8\sum_{n=0}^{\infty}\left ( \frac{1}{8n+3} - \frac{1}{8n+7} \right ) \\
    &= 8\sum_{n=0}^{\infty}\left ( \frac{1}{4\cdot 2n +3} - \frac{1}{4\cdot (2n+1)+3} \right )\\
    &= 8\sum_{n=0}^{\infty} \frac{(-1)^n}{4n+3}\\
    &= 8\sum_{n=0}^{\infty}(-1)^n \int_{0}^{1}x^{4n+2}\, {\rm d}x \\
    &= 8\int_{0}^{1}x^2 \sum_{n=0}^{\infty}(-1)^n x^{4n} \, {\rm d}x \\
    &= 8\int_{0}^{1}\frac{x^2}{1+x^4}\, {\rm d}x \\
    &= \sqrt{2}\left [ \pi+ 2\ln \left ( \sqrt{2}-1 \right ) \right ]
    \end{aligned}$$

    For the integral \( \int_0^1 \frac{x^2}{1+x^4}\, {\rm d}x \) unfortunately I don't have a neat solution.

    1 vote by Grigorios Kostakos

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  •  
    Replied by Grigorios Kostakos on Monday, October 19 2015, 05:31 AM · Hide · #1
    The evaluation of the integral follows :


    \begin{align*}
    \int_0^1{\frac{8x^2}{1+x^4}\,{\rm{d}}x}&=\int_0^1{\frac{2x\sqrt{2}}{x^2-x\sqrt{2}+1}\,{\rm{d}}x}-\int_0^1{\frac{2x\sqrt{2}}{x^2+x\sqrt{2}+1}\,{\rm{d}}x}\\ &=\sqrt{2}\int_0^1{\frac{2x-\sqrt{2}}{x^2-x\sqrt{2}+1}\,{\rm{d}}x}+2\int_0^1{\frac{2}{2x^2-2x\sqrt{2}+2}\,{\rm{d}}x}\,-\\
    &\qquad\qquad\sqrt{2}\int_0^1{\frac{2x+\sqrt{2}}{x^2+x\sqrt{2}+1}\,{\rm{d}}x}+2\int_0^1{\frac{2}{2x^2-2x\sqrt{2}+2}\,{\rm{d}}x}\\
    &=\sqrt{2}\int_0^1{\frac{\big(x^2-x\sqrt{2}+1\big)'}{x^2-x\sqrt{2}+1}\,{\rm{d}}x}+2\sqrt{2}\int_0^1{\frac{\sqrt{2}}{\big(x\sqrt{2}-1\big)^2+1}\,{\rm{d}}x}\,-\\
    &\qquad\qquad\sqrt{2}\int_0^1{\frac{\big(x^2+x\sqrt{2}+1\big)'}{x^2+x\sqrt{2}+1}\,{\rm{d}}x}+2\sqrt{2}\int_0^1{\frac{\sqrt{2}}{\big(x\sqrt{2}+1\big)^2+1}\,{\rm{d}}x}\\
    &=\sqrt{2}\,\Big[\log\big(x^2-x\sqrt{2}+1\big)\Big]_0^1+2\sqrt{2}\,\Big[\arctan\big(x\sqrt{2}-1\big)\Big]_0^1\,\\
    &\qquad\qquad-\sqrt{2}\,\Big[\log\big(x^2+x\sqrt{2}+1\big)\Big]_0^1+2\sqrt{2}\,\Big[\arctan\big(x\sqrt{2}-1\big)\Big]_0^1\\
    &=\sqrt{2}\,\log\big(2-\sqrt{2}\,\big)+\frac{3\pi\sqrt{2}}{4}-\sqrt{2}\,\log\big(2+\sqrt{2}\,\big)+\frac{\pi\sqrt{2}}{4}\\
    &=\sqrt{2}\,\big(2\log\big(\sqrt{2}-1\big)+\pi\big)\\
    &=\sqrt{2}\,\big(\log\big(3-2\sqrt{2}\,\big)+\pi\big)\,.
    \end{align*}
    1 vote by Apostolos J. Kos
    Grigorios Kostakos
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