why does $\pi$ always show up in $\int_0 ^1 \frac{x^c}{1+x^k} dx$ if $c\neq mk-1$ for all $m \in \mathbb{N}$

Question

when I posted this question I was interested in the sum $ \sum_{n=0} ^{\infty} \frac{(-1)^n}{4n+3}$ but when I thought about the generalised sum $ \sum_{n=0} ^{\infty} \frac{(-1)^n}{kn+c +1}$ for all $k \in \mathbb{N} -\{ 1 \} , c \in \mathbb{N} \cup \{0 \} $ I want to know what values of $k,c$ make $\pi $ show up in as a result of this infinite sum.

It is convenient to turn the sum into integrals
$$ \sum_{n=0} ^{\infty} \frac{(-1)^n}{kn+c +1} = \sum_{n=0} ^{\infty} \int_0 ^1 (-1)^n x^{kn+c}dx = \int_0 ^1 \sum_{n=0} ^{\infty} (-1)^n x^{kn+c}dx = \int_0 ^1 x^c\sum_{n=0} ^{\infty} (-1)^n x^{kn}dx = \int \frac{x^c}{1+x^k} dx$$

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I noticed the following pattern

for k=2 $$\int_0 ^1 \frac{1}{1+x^2}dx =\frac{\pi}{4}$$ $$\color{red}{\int_0 ^1 \frac{x}{1+x^2}dx =\frac{\ln2}{2}}$$ $$\int_0 ^1 \frac{x^2}{1+x^2}dx =1- \frac{\pi}{4}$$ $$\color{red}{\int_0 ^1 \frac{x^3}{1+x}dx = \frac{1}{2} - \frac{\ln2}{2} }$$

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for $k=3$ $$\int_0 ^1 \frac{1}{1+x^3}dx =\frac{1}{18} (2\sqrt{3}\pi +\ln(64))$$ $$\int_0 ^1 \frac{x}{1+x^3}dx = \frac{1}{18} (2\sqrt{3}\pi -\ln(8))$$ $$\color{red}{\int_0 ^1 \frac{x^2}{1+x^3}dx =\frac{\ln2}{3}}$$ $$\int_0 ^1 \frac{x^3}{1+x^3}dx =1- (\frac{1}{9} (\sqrt{3}\pi +\ln(8)))$$ $$\int_0 ^1 \frac{x^4}{1+x^3}dx =\frac{1}{2} -\frac{1}{18} (-2\sqrt{3}\pi +\ln(64))$$ $$\color{red}{\int_0 ^1 \frac{x^5}{1+x^3}dx =\frac{1}{3} (1- \ln(2))}$$

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it seems that $\pi$ must show up in every case where $c \neq mk-1$ for all $m \in \mathbb{N}$ put I couldn't prove why.

Answer 1

If you ar familiar with the digamma function $$\int_0 ^1 \frac{x^c}{1+x^k} dx=\frac 1{2k}\left(\psi \left(\frac{c+k+1}{2 k}\right)-\psi\left(\frac{c+1}{2 k}\right)\right)$$ which, as you did, reduce to generalized harmonic numbers since $$\psi(p)=H_{p-1}- \gamma$$

If $c=(m k-1)$ $$\int_0 ^1 \frac{x^c}{1+x^k} dx=\frac 1{2k}\left(\psi \left(\frac{m+1}{2}\right)-\psi \left(\frac{m}{2}\right)\right)$$

$$\psi \left(\frac{m+1}{2}\right)-\psi \left(\frac{m}{2}\right)= (-1)^{m+1}\, 2\log (2)+a_m$$ where the first $a_m$ are $$\left\{0,2,-1,\frac{5}{3},-\frac{7}{6},\frac{47}{30},-\frac{37}{30}, \frac{319}{210},-\frac{533}{420},\frac{1879}{1260},-\frac{1627}{1 260},\frac{20417}{13860}\right\}$$

For the coefficient of $\pi$ in the most general case, use what @Miracle Invoker gave in comments.

Answer 2

By the reflection property of Beta function

Letting $y=x^k$ transforms the integral into a Beta function as $$ I=\frac{1}{k} \int_0^1 \frac{y^{\frac{c+1}{k}-1}}{1+y} d y $$ So if $c \neq m k-1, \forall m \in N$, then $ \frac{c+1}{k} \in \mathbb{N}.$ By the reflection property of Beta function, we have $$ B(x, 1-x)=\pi \csc (\pi x) \quad \textrm{ for all }x \notin \mathbb{Z}. $$ Hence $$ I=\frac{\pi}{k} \csc \frac{(k+1) \pi}{k} $$ which explains why $\pi$ always appears in the the integral if $c \neq m k-1, \forall m \in N$.