This question motivated me to look for a closed form of the generalisation of $I_3:$
$$I_n = \int_0^1 \arctan^n x\,dx.$$
Question: Does $I_n, n\geq 0$ have a closed form solution.
My attempt so far:
I was hoping to deduce a recurrence relationship for $I_n$ (similar to the answer here), that could help to find a closed form for it. So I started using IBP twice.
First:
- $u = \arctan^n x \to du = n \arctan^{n-1} x \cdot \frac{1}{1+x^2}\,dx$
- $dv = dx \to v = x$
Hence $I_n = \left(\frac{\pi}{4}\right)^n - n K_{n-1}$ with $K_{n-1} = \int_0^1 \frac{x \arctan^{n-1} x}{1+x^2}\,dx$.
Now applying IBP for $K_{n-1}$, we choose:
$u = \arctan^{n-1} x \to du = (n-1) \arctan^{n-2} x \cdot \frac{1}{1+x^2}\,dx$
$dv = \frac{x}{1+x^2}\,dx \to v = \int \frac{x}{1+x^2}\,dx = \frac{1}{2} \ln(1+x^2)$
And we get $$K_{n-1} = \left[\frac{1}{2} \ln(1+x^2) \arctan^{n-1} x\right]_0^1 - \int_0^1 \frac{1}{2} \ln(1+x^2) \cdot (n-1) \arctan^{n-2} x \cdot \frac{1}{1+x^2}\,dx \\ = \frac{1}{2} \ln 2 \left(\frac{\pi}{4}\right)^{n-1} - 0 - \frac{n-1}{2} \int_0^1 \frac{\ln(1+x^2) \arctan^{n-2} x}{1+x^2}\,dx$$
This then gives $I_n$: $$I_n = \left(\frac{\pi}{4}\right)^n - \frac{n}{2} \ln 2 \left(\frac{\pi}{4}\right)^{n-1} + \frac{n(n-1)}{2} \int_0^1 \frac{\ln(1+x^2) \arctan^{n-2} x}{1+x^2}\,dx$$
Let $L_m = \int_0^1 \frac{\ln(1+x^2) \arctan^m x}{1+x^2}\,dx$. Then $I_n$ can be written as: $$I_n = \left(\frac{\pi}{4}\right)^n - \frac{n}{2} \ln 2 \left(\frac{\pi}{4}\right)^{n-1} + \frac{n(n-1)}{2} L_{n-2}$$
Although the exponent of $\arctan$ has been reduced, $L_n$ still seems to be even more complicated, then $I_n.$
