How to Evaluate $\int_{0}^{1}\frac{\tan^{-1}(x)}{1+x^4}dx$

Question

Is there a closed form for $$\int_{0}^{1}\frac{\tan^{-1}(x)}{1+x^4}dx \approx 0.349446 $$

It may also be represented as

$$ \sum_{n=1}^{\infty}\sum_{k=1}^{\infty} \frac{(-1)^n \,(-1)^k}{(2n-1)(2n+4k-4)} $$

Using Mathematica the result is

$$ \frac{\pi \ln(1+\sqrt{2})}{8\sqrt{2}} -\frac{\tanh^{-1}(\sqrt{2})\ln(2)}{8\sqrt{2}} + \frac{\phi(2,2,1/2)}{16}+\frac{i}{4\sqrt{2}}\left(\,\operatorname{Li}_{2}\,(i\sqrt2-i)-\operatorname{Li}_{2}\,(i-i\sqrt{2})\,\right)$$

where $\phi(z,s,a)$ is the Lerch transcendent https://mathworld.wolfram.com/LerchTranscendent.html

$Li_n(x) $ is a polylogarithm https://en.wikipedia.org/wiki/Polylogarithm

This result was highly unsatisfactory to me as it was not what i would expect based on the below integrals which have a nice closed form :

$$ \int_{0}^{1} \frac{x^3 \tan^{-1}(x)}{1+x^4} dx = \frac{C}{2} - \frac{\pi \ln(1+\sqrt{2})}{8} $$ $$ \int_{0}^{1} \frac{x \tan^{-1}(x)}{1+x^4} dx = \frac{\pi^2}{32} - \frac{\ln(1+\sqrt{2})^2}{8} $$

Where $C$ is Catalan's Constant

I tried to use integral techniques (by hand) to see if I could arrive at a different closed form hopefully without the special functions but was not successful.

Q = Is there a closed form for the above integral? $\vdash$ Could the above closed form be simplified to get rid of the special functions?

*** If duplicate question please do direct me to it ****

Thank you very much for your help and time.

Answer 1

The following closed-form & the way to go in large steps is proposed by Cornel I. Valean

$$\int_0^1\frac{\arctan(x)}{1+x^4}\textrm{d}x$$ $$=\frac{1}{4}G+\frac{1}{384}(12+11\sqrt{2})\pi^2-\frac{3}{32}\sqrt{2}\log(\sqrt{2}-1)\pi-\frac{1}{8}\sqrt{2}\operatorname{Li}_2(1-\sqrt{2})-\frac{1}{64}\psi^{(1)}\left(\frac{1}{8}\right)$$

The process is easy and boring, starting from the simple transformation $$\int_0^1\frac{\arctan(x)}{1+x^4}\textrm{d}x$$ $$=-\frac{1}{32}\sqrt{2}\pi^2-\frac{1}{16}\sqrt{2}\log(\sqrt{2}-1)\pi+\frac{1}{2} \int_0^1\frac{1+x^2}{1+x^4}\arctan(x)\textrm{d}x$$$$+\frac{1}{2}\int_0^{\infty}\frac{x^2\arctan(1/x)}{1+x^4}\textrm{d}x,$$ where both integrals are easy to calculate (I would even call trivial).

After integrating by parts, rearranging, and applying a known variable change all efforts of calculating $\displaystyle \int_0^1\frac{1+x^2}{1+x^4}\arctan(x)\textrm{d}x$ reduces to calculating $\displaystyle \int_0^{\pi/2}\arctan\left(\sqrt{2}/2\tan(x)\right)\textrm{d}x$, which works great with differentiation under the integral sign, that is I mean to consider $\displaystyle I(a)= \int_0^{\pi/2}\arctan(a\tan(x))\textrm{d}x$ and differentiate with respect to $a$, well-known in the literature.

For the last one? Use the identity $\displaystyle \arctan(x)+\arctan(1/x)=\pi/2$, and all the efforts get reduced to calculating $\displaystyle \int_0^{\infty}\frac{x^2\arctan(x)}{1+x^4}\textrm{d}x$, and here we may consider $\displaystyle J(a)= \int_0^{\infty}\frac{x^2\arctan(a x)}{1+x^4}\textrm{d}x$, and then differentiate with respect to $a$.

Note: I saw more posts alike around where the closed-forms may be simplified a lot.

Best wishes ;)