How to find the closed form for the integral $I_n:=\int_{0}^{1} \frac{x^{n} \ln x}{1+x^{2}} d x, {} $ where $ n\in N$?

Question

Inspired by the post, I started to investigate the integral in general, $$ I_n:=\int_{0}^{1} \frac{x^{n} \ln x}{1+x^{2}} d x, $$ where $ n\in N.$

Using the infinite geometric series, $$\frac{1}{1+t}=\sum_{k=0}^{\infty}(-1)^{k} t^{k} \text { for }|t|<1, $$ we can convert the integral $I_n $ into \begin{aligned}I_{n} &=\int_{0}^{1} x^{n} \ln x \sum_{k=0}^{\infty}(-1)^{k} x^{2 k} d x \\&=\sum_{k=0}^{\infty}(-1)^{k} \underbrace{\int_{0}^{1} x^{n+2 k} \ln x d x}_{J_{k}}\end{aligned} Using Integration by parts yields \begin{aligned}J_{k} &=\int_{0}^{1} \ln x d\left(\frac{x^{n+2 k+1}}{n+2 k+1}\right) \\&=-\frac{1}{n+2 k+1} \int_{0}^{1} x^{n+2 k} d x \\&=-\frac{1}{(x+2 k+1)^{2}}\end{aligned} We can now conclude that $$ \boxed{I_{n}=-\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(n+2 k+1)^{2}}} $$ Then we shall study and simplify the sum case by case.

A. When $ n=2m+1$ is odd, \begin{aligned} I_{n}&=-\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 m+2+2 k)^{2}}\\&=\frac{(-1)^{m}}{4} \sum_{k=1}^{\infty} \frac{(-1)^{m+k}}{(m+k)^{2}}\\&=\frac{(-1)^{m}}{4}\left(\sum_{k=m+1}^{\infty} \frac{(-1)^{k}}{k^{2}}\right)\\&=\frac{(-1)^{m}}{4}\left[\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k^{2}}-\sum_{k=1}^{m} \frac{(-1)^{k}}{k^{2}}\right]\end{aligned} $$\boxed{I_{2m+1}=\frac{(-1)^{m+1}}{4}\left(\frac{\pi^{2}}{12}+\sum_{k=1}^{m} \frac{(-1)^k}{k^{2}}\right)}$$ For examples, $$I_{1}=-\frac{\pi^{4}}{48}; \quad I_{3}=-\frac{1}{4}\left[-\frac{\pi^{2}}{12}-(-1)\right]=\frac{\pi^{2}}{48}-\frac{1}{4}; \quad I_{5}=\frac{1}{4}\left[-\frac{\pi^{2}}{12}-\left(-1+\frac{1}{4}\right)\right]=-\frac{\pi^{2}}{48}+\frac{3}{16}$$ B. When $ n=2m$ is even, \begin{aligned}I_{n} &=-\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 m+2 k+1)^{2}} \\&=(-1)^{m+1} \sum_{k=0}^{\infty} \frac{(-1)^{m+k}}{(2(m+k)+1)^{2}} \\&=(-1)^{m+1}\left[\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{2}}-\sum_{k=0}^{m-1} \frac{(-1)^{k}}{(2 k+1)^{2}}\right] \\ \end{aligned} $$\boxed{I_{2m}=(-1)^{m+1}\left(G-\sum_{k=0}^{m-1} \frac{(-1)^{k}}{(2 k+1)^{2}}\right) } $$ For examples, $$ I_{2} =G-1; \quad I_{4} =-\left[G-\left(1-\frac{1}{9}\right)\right]=\frac{8}{9}-G; \quad I_6=G-\frac{209}{225}, $$ where $G$ is the Catalan’s Constant.

My Question

Is there any other simpler method? Your opinions and alternative solutions are warmly welcome.

Answer 1

Alternative way of obtaining the identity $$ I_{n}=-\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(n+2 k+1)^{2}} $$ would be to take a derivative of integral with respect to $n$ $$ J(n) = \int_0^1 \frac{x^n}{x^2 + 1} \, dx = \int_0^1 \sum_{k=0}^\infty (-1)^k x^{n + 2k} \, dx = \sum_{k = 0}^\infty (-1)^k \frac{1}{n + 2k + 1}. $$ $$ I_n = J'(n) = - \sum_{k=0}^\infty \frac{(-1)^k}{(n + 2k + 1)^2}. $$

Seems a bit simpler than integration by parts, but anyway one needs to justify if it's really possible to differentiate / integrate infinite series term by term here.

Answer 2

Establish the recursive formula \begin{align} I_{n} =\int_0^1 x^{n-2} \ln x\>dx -I_{n-2} =-\frac{1}{(n-1)^2}-I_{n-2} \end{align} along with $ I_0=\int_0^1 \frac{\ln x}{1+x^2}dt=-G$ and $I_1=\int_0^1 \frac{x\ln x}{1+x^2}dt=-\frac{\pi^2}{48}$. Then $$I_{2m}=\sum_{k=1}^{m}\frac{(-1)^{m+k-1}}{(2k-1)^2}+(-1)^m I_0 = \sum_{k=1}^{m}\frac{(-1)^{m+k-1}}{(2k-1)^2}-(-1)^{m}G $$ \begin{align}I_{2m+1} = \sum_{k=1}^{m} \frac{(-1)^{m+k-1}}{(2k)^{2}} +(-1)^m I_1 =\sum_{k=1}^{m} \frac{(-1)^{m+k-1}}{4k^{2}} -(-1)^{m}\frac{\pi^{2}}{48} \end{align}

Answer 3

Set $y=x^2$ in

$$\int_0^1\frac{y^{m}}{1+y}\mathrm{d}y=\frac12\left[\psi\left(\frac{m+2}{2}\right)-\psi\left(\frac{m+1}{2}\right)\right],\quad \mathfrak{R}(m)>-1$$

we have

$$2\int_0^1\frac{x^{2m+1}}{1+x^2}\mathrm{d}x=\frac12\left[\psi\left(\frac{m+2}{2}\right)-\psi\left(\frac{m+1}{2}\right)\right]$$

let $m=\frac{n-1}{2}$

$$\int_0^1\frac{x^{n}}{1+x^2}\mathrm{d}x=\frac14\left[\psi\left(\frac{n+3}{4}\right)-\psi\left(\frac{n+1}{4}\right)\right]$$

Now differentiate both sides w.r.t $n$ we get

$$\int_0^1\frac{x^{n}\ln(x)}{1+x^2}\mathrm{d}x=\frac1{16}\left[\psi^{(1)}\left(\frac{n+3}{4}\right)-\psi^{(1)}\left(\frac{n+1}{4}\right)\right]$$