Is it possible to express explicitly the general term of the sequence $ I_n = \int_0^1 t^n \arctan t \, \mathrm{d} t \, $?

Question

In a dual integral equation situation, I have to deal with the following sequence:

$$ I_n = \int_0^1 t^n \arctan t \, \mathrm{d} t \quad\quad (n \ge 1) \, . $$

It is easy to compute the first terms e.g. $I_1= \pi/4-1/2$ or $I_2=\pi/12+(\ln2-1)/6$ etc.

But is it possible to express explicitly the general term of this sequence directly as function of $n$? Hints and ideas welcome.

Greetings,

R

Answer 1

$$I_n = \int_0^1 t^n \arctan t \, dt=\frac{t^{n+1}}{n+1}\arctan t\vert_0^1- \int_0^1\frac{t^{n+1}}{n+1}\frac{dt}{1+t^2}=\frac{1}{n+1}\frac{\pi}{4}-\frac{1}{n+1}J_{n+1}$$

where $J_n=\int_0^1 t^n(1+t^2)^{-1}\,dt$

Now, $J_n$ can be expressed as a finite sum, but it depends on the parity (even/odd-ness) of $n$. Note that:

$$J_n+J_{n-2}=\int_0^1t^{n-2}\,dt=\frac{1}{n-1}, J_0=\frac{\pi}{4},J_1=\frac{1}{2}\ln 2$$

Unpacking this a bit, one can see that:

$$J_{2n}+J_{2n-2}=\frac{1}{2n-1}\implies (-1)^{n}J_{2n}-(-1)^{n-1}J_{2(n-1)}=\frac{(-1)^n}{2n-1}$$

Telescoping brings this to:

$$J_{2n}=(-1)^n\left(\frac{\pi}{4}-\sum_{k=1}^n\frac{(-1)^{k-1}}{2k-1}\right)$$

A similar process leads to:

$$J_{2n+1}=(-1)^n\left(\frac{\ln 2}{2}-\sum_{k=1}^n\frac{(-1)^{k-1}}{2k}\right)$$

Plugging these results back into the expression at the end of the first line, you have a neat expression for $I_n$.

Answer 2

$\displaystyle I_n=\int_0^1 t^n\arctan t dt$

$\begin{align} I_n&=\int_0^1 t^n\left(\sum_{k=0}^{\infty}(-1)^k\dfrac{t^{2k+1}}{2k+1}\right)dt\\ &=\int_0^1 \left(\sum_{k=0}^{+\infty} (-1)^k\dfrac{t^{2k+1+n}}{2k+1}\right)dt\\ &=\sum_{k=0}^{+\infty} \dfrac{(-1)^k}{(2k+1)(2k+2+n)}\\ &=\dfrac{1}{n+1}\sum_{k=0}^{+\infty}\dfrac{(-1)^k}{2k+1}+\dfrac{1}{n+1}\sum_{k=1}^{+\infty}\dfrac{(-1)^k}{n+2k}\\ &=\dfrac{\pi}{4(n+1)}+\dfrac{1}{n+1}\sum_{k=1}^{+\infty}\dfrac{(-1)^k}{n+2k} \end{align}$