While browsing the post Is there any integral for the golden ratio $\phi$?, I came across this nice answer, $$ \int_0^\infty \frac{1}{1+x^{10}}dx=\frac{\pi\,\phi}5$$ it seems the general form is just

$$p \int_0^\infty \frac{1}{1+x^{p}}dx=\color{blue}{\frac{\pi}{\sin\big(\tfrac{\pi}{p}\big)}}$$

I wondered about $$\int_0^\color{red}1 \frac{1}{1+x^p}dx=\,?$$ Mathematica could find messy closed-forms for $p=5,7$. After some laborious simplification, $$5\int_0^1 \frac{1}{1+x^5}dx=\frac{\pi\sqrt{\phi}}{5^{1/4}}+\ln2+\sqrt{5}\ln\phi$$

Question 1: In general, is it true that for any $p$ ,

$$2p\,\int_0^1 \frac{1}{1+x^p}dx=\color{blue}{\frac{\pi}{\sin\big(\tfrac{\pi}{p}\big)}}+2\ln2-\psi\big(\tfrac{1}{p}\big)+\psi\big(\tfrac{p-1}{2p}\big)+\psi\big(\tfrac{p+1}{2p}\big)-\psi\big(\tfrac{p-1}{p}\big)$$

where $\psi(z)$ is the digamma function?

Note: The four digammas, implemented in Mathematica as PolyGamma[z], can be expressed as a sum of cosines x logarithms for odd $p=2m+1$. Let $k=\frac{2n-1}{p}\pi$, then, $$-\psi\big(\tfrac{1}{p}\big)+\psi\big(\tfrac{p-1}{2p}\big)+\psi\big(\tfrac{p+1}{2p}\big)-\psi\big(\tfrac{p-1}{p}\big)=-4\sum_{n=1}^m \cos (k)\ln\big(\sin\tfrac{k}{2}\big)$$

Question 2: For even $p$, can we can also avoid the digamma by using cosines and logarithms?

  • $\begingroup$ Can taylor expansion help? $\endgroup$ – vidyarthi Nov 26 '16 at 4:11
  • $\begingroup$ When the integral is evaluated by Maxima, which is similar to Mathematica, it uses factorization of denominator followed by partial fraction and some substitutions to get a complicated expression involving series of logarithms and the arctan function $\endgroup$ – vidyarthi Nov 26 '16 at 4:19
  • $\begingroup$ @vidyarthi: Can you put a simplified version of Maxima's result? $\endgroup$ – Tito Piezas III Nov 26 '16 at 4:22
  • $\begingroup$ here is the image of the final result for $p=10$ taken from integral-calculator.com which uses maxima-imgur.com/iezjQ2u $\endgroup$ – vidyarthi Nov 26 '16 at 4:29
  • 2
    $\begingroup$ You are asking, slightly in disguise, for explicit forms of the sum of the series $$\sum_{n=0}^\infty\frac{(-1)^n}{1+np}=p\sum_{n=0}^\infty\frac{1}{(1+2np)(1+(2n+1)p)}$$ which converges at least for every positive $p$. The "solutions" based on hypergeometric or digamma functions are, if only one looks at the definition of these special functions, mere rewritings of the sum of this series. $\endgroup$ – Did Nov 26 '16 at 8:46

I'm only going to address Question 1.

The expression proposed in Question 1 is true. However, it is a little bit too complicated than necessary. A simpler version of the expression is

$$2p\int_0^1 \frac{dx}{1+x^p} = \psi\left(\frac{1}{2p} + \frac12\right) - \psi\left(\frac{1}{2p}\right)$$

From reflection formula, take logarithm and differentiate, we get $$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin\pi z} \implies \psi(z) - \psi(1-z) = \pi\cot\pi z$$ This leads to $$\frac{\pi}{\sin z} = \pi\cot\frac{\pi z}{2} - \pi\cot\pi z = \psi\left(\frac{z}{2}\right) - \psi\left(1-\frac{z}{2}\right) - \psi(z) + \psi(1-z) $$ From duplication formula, take logarithm and differentiate, we get $$\Gamma(z)\Gamma\left(z+\frac12\right) = 2^{1-2z}\sqrt{\pi}\Gamma(2z) \implies \psi(z) + \psi\left(z + \frac12\right) = -2\log 2 + 2\psi(2z) $$ Apply these to RHS of Question 1, we can expose RHS to following mess $$ \left[ \color{red}{\psi\left(\frac{1}{2p}\right)} - \psi\left(1 - \frac{1}{2p}\right) - \color{green}{\psi\left(\frac{1}{p}\right)} + \color{blue}{\psi\left(1 - \frac{1}{p}\right)} \right] + \left[ \color{green}{2\psi\left(\frac1p\right)} - \color{red}{\psi\left(\frac{1}{2p}\right)} - \color{magenta}{\psi\left(\frac{p+1}{2p}\right)}\right]\\ - \color{green}{\psi\left(\frac{1}{p}\right)} + \psi\left(\frac{p-1}{2p}\right) + \color{magenta}{\psi\left(\frac{p+1}{2p}\right)} - \color{blue}{\psi\left(\frac{p-1}{p}\right)} $$ After massive cancellation, we can simplify RHS to $$ \psi\left(\frac{p-1}{2p}\right) - \psi\left(1 - \frac{1}{2p}\right) = \psi\left( 1 - \left(\frac{1}{2p} + \frac12\right)\right) - \psi\left(1 - \frac{1}{2p}\right) = \psi\left(\frac{1}{2p} + \frac12\right) - \psi\left(\frac{1}{2p}\right) $$ Recall following expansion of digamma function

$$\psi(z) = \frac{1}{z} + \sum_{n=1}^\infty \left(\frac{1}{z+n} - \frac{1}{n}\right)$$ We find $$\begin{align} \text{RHS} &= \frac{1}{\frac{1}{2p}} - \frac{1}{\frac{1}{2p} + \frac12} + \sum_{n=1}^\infty\left(\frac{1}{\frac{1}{2p}+n} - \frac{1}{\frac{1}{2p} + n + \frac12}\right)\\ &= 2\sum_{n=0}^\infty\frac{(-1)^n}{\frac{1}{p}+n} = 2\sum_{n=0}^\infty\int_0^1 (-1)^n t^{\frac{1}{p}+n-1} dt = 2\int_0^1 \sum_{n=0}^\infty (-1)^n t^{\frac{1}{p}+n-1} dt\\ &= 2 \int_0^1 \frac{t^{\frac{1}{p}-1}}{1+t} dt = 2p\int_0^1 \frac{dx}{1+x^p} = \text{LHS} \end{align} $$

  • $\begingroup$ Thanks for the confirmation. I started with $p=5$ and arrived at the complicated version which shows that when $p$ is odd, the four digammas (hence the integral) is expressible in terms of elementary functions. However, when $p$ is even, perhaps one has to stay with the digammas. $\endgroup$ – Tito Piezas III Nov 26 '16 at 11:20

This answer is incomplete as it does not proof all the steps completely:

Let us start with the partial fraction expansion $$\tag{1}\frac{p}{1+x^p}=- \sum_{j=1}^p\frac{\omega^{2j-1}}{x-\omega^{2j-1}}$$ with $\omega= e^{i\pi/p}$.

Now, we can integrate and obtain $$\int_0^1\!dx \frac{p}{1+x^p} = -\sum_{j=1}^p \omega^{2j-1} \log\left(1-\omega^{1-2j}\right). $$

We can simplify this formula a bit taking the real part. This yields $$\int_0^1\!dx \frac{p}{1+x^p} = \pi\sum_{j=1}^{p} \frac{1-2j+p}{2p} \sin\left(\frac{\pi(1-2j)}{p}\right)+\sum_{j=1}^{p}\cos\left(\frac{\pi(1-2j)}{p}\right) \log\left[2 \sin\left(\frac{\pi(1-2j)}{2p}\right)\right].$$

The first term can be summed explicitly $$\int_0^1\!dx \frac{p}{1+x^p} =\frac{\pi}{2\sin(\pi/p)}+\sum_{j=1}^{p}\cos\left(\frac{\pi(1-2j)}{p}\right) \log\left[2 \sin\left(\frac{\pi(1-2j)}{2p}\right)\right].$$ The second term can be related to the digamma function.

Formula (1) can be proven by multiplying the expression by $1+x^p$. For the right hand side, we obtain $$- \sum_{j=1}^p\omega^{2j-1} \prod_{k\neq j} (x-\omega^{2k-1}) = -x^{p-1}\sum_{j=1}^p\omega^{2j-1}+ x^{p-2} \sum_{j=1}^p\omega^{2j-1} \sum_{k,k'\neq j} \omega^{2k-1}\omega^{2k'-1} - \cdots\\+(-1)^p \sum_{j=1}^p\omega^{2j-1} \prod_{k\neq j} \omega^{2k-1}$$ as $1+x^p=\prod_{k=1}^p (x- \omega^{2j-1})$. The result (1) follows from some not so straightforward combinatorics.


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