Closed form solution to $\int_0^1\arctan^2(x)\,\sqrt{x}\,dx$

Question

I need to compute this integral: $$\int_0^1\arctan^2(x)\,\sqrt{x}\,dx$$ I tried integration by parts, and also introducing a parameter $\arctan(a\,x)$ and differentiation wrt it, but these approaches did not lead to anything useful. Please help.

Answer 1

The integrand has a closed-form antiderivative in terms of elementary functions and dilogarithms. Mathematica can find it if we help it by first converting the arctangent into a combination of logarithms: $$\arctan(x)=\frac i2\ln(1-i x)-\frac i2\ln(1+ix)$$ After some simplifications it takes this form. Its correctness can be checked manually using direct differentiation. Plugging the integration limits and performing some simplifications, we get the same result as given by Cleo.

Answer 2

$$\frac{\pi^2}{24}-\frac{2\pi}3+\frac1{36\sqrt{2}}\left[5\pi^2+12\left(4+\ln\left(\frac{1+\sqrt2}2\right)\right)\left(\pi-2\ln\left(1+\sqrt2\right)\right)-48\operatorname{Li}_2\left(\sqrt2-1\right)\right]$$

Answer 3

The integration range is small: $[0, 1]$ so one of the method could be to use Taylor series for the $\arctan(x)$ function:

$$\arctan^2(x) = \sum_{n = 0}^{+\infty}\sum_{m = 0}^{+\infty}\frac{(-1)^n (-1)^m}{(2n+1)(2m+1)}x^{2n+1}x^{2m+1}$$

And write $\sqrt{x} = x^{\frac{1}{2}}$ to obtain:

$$I = \int_0^1\sum_{n = 0}^{+\infty} \sum_{m = 0}^{+\infty}\frac{(-1)^n (-1)^m}{(2n+1)(2m+1)}x^{2n+1}x^{2m+1} x^{\frac{1}{2}}$$

namely

$$I = \sum_{n = 0}^{+\infty} \sum_{m = 0}^{+\infty}\frac{(-1)^n (-1)^m}{(2n+1)(2m+1)}\int_0^1x^{2n+1}x^{2m+1} x^{\frac{1}{2}}$$

That will lead you to (after a trivial integration)

$$I = \sum_{n = 0}^{+\infty} \sum_{m = 0}^{+\infty}\frac{(-1)^n (-1)^m}{(2n+1)(2m+1)}\frac{1}{\left(2(n+m) + \frac{7}{2}\right)}$$