An amazing integral with a typo

Question

This question Concerns a certain integral $$f(p)=\int_0^1\frac{x^p(1-x)^p}{1+x^2}\mathrm dx$$ Which, as proven by Calvin Khor, has the property that $f(4k)-4^{k-1}(-1)^k\pi\in\Bbb Q$ when $k\in\mathbb N$.

When I was investigating this question, I made a typo (switching the places of the "$)$" and the "${}^p$") when inputting the integral into Mathematica, and instead inputted the integral $$g(p)=\int_0^1\frac{x^p(1-x^p)}{1+x^2}\mathrm dx$$ However, rather unexpectedly, this appears to have the same property as well, with my Mathematica spitting out $$g(4\cdot 1)=\frac{2}{35} \\ g(4\cdot 2)=\frac{196}{6~435} \\ g(4\cdot 3)=\frac{208~786}{10~140~585} \\ g(4\cdot 4)=\frac{489~772~744}{31~556~720~475} \\ \text{etc}.$$

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My question is simple - can someone explain why?

Answer 1

Observe that

• $ \frac{ x^{4a} - 1 } { 1+x^2} = \frac{ x^{4a} - 1 } {x^4 - 1 } \times \frac{ x^4 - 1 } { x^2 + 1 } $ is a polynomial with rational coefficients.
• (In fact, the coefficients are $0, 1, -1$.)
• Hence it's integral is a polynomial with rational coefficients.
• Hence $ \int_{0}^1 \frac{ x^{4a} - 1 } { 1+x^2}\, dx = [f(x) ]_0^1$ is a rational number.
• Hence $ \int_{0}^1 \frac{ x^{4a} - x^{8a} } { 1+x^2}\, dx $ is a rational number.
• Bonus: Show that $\frac{ x^{4a} - x^{8a} } { 1+x^2} = x^{4a} - x^{4a+2} + x^{4a+4} -x^{4a+6} \ldots + x^{8a-4} - x^{8a-2}$. Hence, the integral from 0 to 1 is equal to the rational number $$ \frac{1}{4a+1} - \frac{1}{4a+3} + \frac{1}{4a+5} - \frac{1}{4a+7} \ldots + \frac{1}{8a-3} - \frac{1}{ 8a-1} .$$

Furthermore, to understand why this doesn't hold for $p \neq 4k$, we can study it in a similar manner.

• For $ p = 4k+1$, show that $\frac{x^{p} - x^{2p}} { x^2 + 1 } = f(x) + \frac{ 1 + x } { 1 + x^2 }$, where $f(x)$ is a polynomial with rational(integer) coefficients. Hence, the integral is of the form rational number $ +\frac{ \pi + \log 4 } { 4} $.
• For $ p = 4k+2$, show that $\frac{x^{p} - x^{2p}} { x^2 + 1 } = f(x) + \frac{ -2 } { 1 + x^2 }$, where $f(x)$ is a polynomial with rational(integer) coefficients. Hence, the integral is of the form rational number $ - \frac{ \pi} { 2} $.
• For $ p = 4k+3$, show that $\frac{x^{p} - x^{2p}} { x^2 + 1 } = f(x) + \frac{ 1-x } { 1 + x^2 }$, where $f(x)$ is a polynomial with rational(integer) coefficients. Hence, the integral is of the form rational number $ +\frac{ \pi - \log 4} { 4} $.
• Bonus: We can similarly hunt down the polynomial $f(x)$ and can state what the rational number is.

Answer 2

By considering the integral, then: \begin{align} g(p) &= \int_{0}^{1} \frac{x^p \, (1 - x^p)}{1+x^2} \, dx \\ &= \sum_{n=0}^{\infty} (-1)^n \, \int_{0}^{1} x^{p+2 n} \, (1 - x^p) \, dx \\ &= \sum_{n=0}^{\infty} (-1)^n \, \left(\frac{1}{2n+p+1} - \frac{1}{2n+2p+1} \right) \\ &= \frac{1}{4} \, \left( \psi\left(\frac{p+3}{4}\right) - \psi\left(\frac{p+1}{4}\right) - \psi\left(\frac{2p+3}{4}\right) + \psi\left(\frac{2p+1}{4}\right) \right), \end{align} where $\psi(x)$ is the digamma function. Letting $ p \to 4 p$ gives $$ g(4 p) = \frac{1}{4} \, \left( \psi\left(p + \frac{3}{4}\right) - \psi\left(p + \frac{1}{4}\right) - \psi\left(2 p +\frac{3}{4}\right) + \psi\left(2 p + \frac{1}{4}\right) \right). $$ Using $$ \psi(x+1) = \psi(x) + \frac{1}{x} $$ then for each $p$ the values of $g(4 p)$ can be calculated. For $p=0,1,2$ then \begin{align} g(4 \cdot 0) &= 0 \\ g(4 \cdot 1) &= \frac{1}{5} - \frac{1}{7} = \frac{2}{35} \\ g(4 \cdot 2) &= \frac{1}{9} - \frac{1}{11} + \frac{1}{13} - \frac{1}{15} = \frac{196}{6435} \end{align}

In another form $$ g(4 \, p) = \sum_{j=2p}^{4p-1} \frac{(-1)^j}{2j+1}. $$

For note(s): \begin{align} f(p) &= \int_{0}^{1} \frac{x^p \, (1-x)^p}{1+x^2} \, dx \\ &= \sum_{n=0}^{\infty} (-1)^n \, B(2n + p + 1, p + 1) \\ &= \left( \frac{\Gamma(p+1)}{\Gamma(2p+2)}\right)^2 \, {}_{3}F_{2}\left( \frac{p+1}{2}, \, \frac{p+3}{2}, \, 1; \, p+1, \, p + \frac{3}{2}; \, -1 \right), \end{align} where $B(x, y)$ is the Beta function and ${}_{3}F_{2}$ is a hypergeometric function.