calculus - The integral $J(m,n):=int_0^1 frac{x^m}{x^n+1}dx$

Here

Is there a general formula for $I(m,n)$?

I asked for a general formula for $\ I(m,n):=\int_0^{\infty} \frac{x^m}{x^n+1}dx$.

Is the formula

$$I(m,n)=\frac{\pi}{n\sin((m+1)\frac{\pi}{n})}$$ true for every pair $\ (m,n)\ $ of real numbers with $\ 0\le m\le n-2\ $ and not just for non-negative integers with $\ 0\le m\le n-2\ $ ?

I wondered whether there is a similar closed form for

$$J(m,n):=\int_0^1 \frac{x^m}{x^n+1}dx$$

I figured out that

$$I(m,n)=J(m,n)+J(n-2-m,n)$$ holds for $0\le m\le n-2$.

For $\ m=0\ $, the first few values are

$$J(0,0)=\frac{1}{2}$$ $$J(0,1)=\ln(2)$$ $$J(0,2)=\frac{\pi}{4}$$ $$J(0,3)=\frac{2\sqrt{3}\pi+\ln(64)}{18}$$ $$J(0,4)=\frac{\pi+2\coth^{-1}(\sqrt{2})}{4\sqrt{2}}$$ $$J(0,5)=\frac{1}{5}\sqrt{\frac{1}{10}(5+\sqrt{5})}\pi+\frac{1}{20}(\ln(16)+2\sqrt{5}\coth^{-1}(\frac{3}{\sqrt{5}}))$$ $$J(0,6)=\frac{\pi+\sqrt{3}\ln(2+\sqrt{3})}{6}$$

Is there a general formula for $\ J(m,n)\ $ and does it hold for all pairs $\ (m,n)\ $ of real numbers with $\ 0\le m\le n-2$ ?

Answer

There does exist a closed form for

$$J(m,n):=\int_0^1 \frac{x^m}{x^n+1}dx \tag1$$ in terms of the digamma function $\psi(\cdot)$.

Proposition. Let $m=1,2,\cdots$ and $n=1,2,\cdots$. One has

$$J(m,n)=\frac1{2n} \psi\left(\frac{m+n+1}{2n}\right)-\frac1{2n}\psi\left(\frac{m+1}{2n} \right) \tag2$$

then using

$$\psi\left(r+1\right)-\psi\left(r \right)=\frac1r,\quad r \in \mathbb{Q}^*,\tag3$$ and $$\psi\left(\frac{m}{2n}\right) = -\gamma -\ln(4n) -\frac{\pi}{2}\cot\left(\frac{m\pi}{2n}\right) +2\sum_{k=1}^{n-1} \cos\left(\frac{\pi km}{n} \right) \ln\sin\left(\frac{k\pi}{2n}\right) \quad (m<2n)\tag4$$ one gets a closed form in terms of a finite number of elementary functions.

Hint. By the change of variable, $x=u^{1/n}$, $dx=\dfrac1n u^{1/n-1}du$, one may write

$$\begin{align} J(m,n)&=\int_0^1 \frac{x^m}{x^n+1}dx \\&=\frac1n \int_0^1 \frac{u^{\frac{m+1}{n}-1}}{1+u}du \\&=\frac1n \int_0^1 \frac{u^{\frac{m+1}{n}-1}(1-u)}{1-u^2}du \\&=\frac1{2n} \int_0^1 \frac{v^{\frac{m+n+1}{2n}-1}}{1-v}dv-\frac1{2n} \int_0^1 \frac{v^{\frac{m+1}{2n}-1}}{1-v}dv \\&=\frac1{2n} \psi\left(\frac{m+n+1}{2n}\right)-\frac1{2n}\psi\left(\frac{m+1}{2n} \right) \end{align}$$ then one may conclude with Gauss's digamma theorem.

Edit. For any real numbers $a, b$ such that $a>0$ and $b>0$ we have

$$\int_0^1 \frac{x^a}{x^b+1}\:dx=\frac1{2b} \psi\left(\frac{a+b+1}{2b}\right)-\frac1{2b}\psi\left(\frac{a+1}{2b} \right)$$