Calculating in closed form $\int_0^{\infty} \frac{\text{PolyLog}^{(1,0)}(1,-x)}{1+x^2} \, dx$

199 Views Asked by At

Can you confirm the following result? Mathematica and other computational stuff I used seem
unable to do anything about this result. Maybe to confirm it numerically? $$\int_0^{\infty} \frac{\text{PolyLog}^{(1,0)}(1,-x)}{1+x^2} \, dx$$ $$=\frac{3}{8} \pi \log ^2(2)-\frac{1}{4} \gamma \pi \log (2)-\frac{1}{16} \text{StieltjesGamma}^{(0,1)}\left(1,\frac{1}{4}\right)+\frac{1}{16} \text{StieltjesGamma}^{(0,1)}\left(1,\frac{3}{4}\right)+2\log (2) G $$

Let's go a bit further and see a more advanced version

$$\int_0^{\infty} \frac{\text{PolyLog}^{(2,0)}(1,-x)}{1+x^2} \, dx$$ $$=\frac{1}{48} \left(3 \zeta ^{(2,0)}\left(2,\frac{3}{4}\right)-3 \zeta ^{(2,0)}\left(2,\frac{1}{4}\right)+6 (\gamma -1+\log (4)) \zeta ^{(1,0)}\left(2,\frac{1}{4}\right)-6 (\gamma -1+\log (4)) \zeta ^{(1,0)}\left(2,\frac{3}{4}\right)-6 \gamma \text{StieltjesGamma}^{(0,1)}\left(1,\frac{1}{4}\right)+6 \gamma \text{StieltjesGamma}^{(0,1)}\left(1,\frac{3}{4}\right)+96 G \left(\gamma -2 \log ^2(2)+\log (4)\right)+24 \pi \gamma _1 \log (2)-28 \pi \log ^3(2)+36 \gamma \pi \log ^2(2)\right)$$

Both integrals come from my research and it's for the first time I calculated in closed forms integrals with partial derivatives of the polylog. I'm also curious if you have other example of such integrals with $\operatorname{Li}_s^{(n,0)}(x), \ n\ge 1$ that are evaluated in closed forms?

If it's of any importance, I can find the closed forms for the whole family

$$\int_0^{\infty} \frac{\text{PolyLog}^{(n,0)}(1,-x)}{1+x^2} \, dx \ , n\ge1.$$