İntegralimiz :
$$\Xi_1(2,4,2)=\int_0^\infty\:\frac{\ln^2(x)}{\sqrt{1+x^4}}\:dx$$
Buradaki eşitlikte $m$ yerine $4$ , $p$ yerine $2$ koyalım.Eşitlik :
$$\Xi_1(2,m,p)=\int_0^\infty\:\frac{\ln^2(x)}{\sqrt[p]{1+x^m}}=\\\frac{1}{m^3\,\Gamma(p^{-1})}\Bigg[\Gamma\Big(\frac{1}{m}\Big)\Gamma\Big(\frac{1}{p}-\frac{1}{m}\Big)\bigg(\psi\Big(\frac{1}{m}\Big)-\psi\Big(\frac{1}{p}-\frac{1}{m}\Big)\bigg)\\+\Gamma\Big(\frac{1}{m}\Big)\Gamma\Big(\frac{1}{p}-\frac{1}{m}\Big)\bigg(\psi_1\Big(\frac{1}{m}\Big)+\psi\Big(\frac{1}{p}-\frac{1}{m}\Big)\bigg)\Bigg]$$
$$\Xi_1(2,4,2)=\int_0^\infty\:\frac{\ln^2(x)}{\sqrt{1+x^4}}=\\\frac{1}{64\,\Gamma(2^{-1})}\Bigg[\Gamma\Big(\frac{1}{4}\Big)\Gamma\Big(\frac{1}{2}-\frac{1}{4}\Big)\bigg(\psi\Big(\frac{1}{4}\Big)-\psi\Big(\frac{1}{2}-\frac{1}{4}\Big)\bigg)\\+\Gamma\Big(\frac{1}{4}\Big)\Gamma\Big(\frac{1}{2}-\frac{1}{4}\Big)\bigg(\psi_1\Big(\frac{1}{4}\Big)+\psi\Big(\frac{1}{2}-\frac{1}{4}\Big)\bigg)\Bigg]$$
Sadeleştirelim.
$$\frac{2}{64\,\sqrt{\pi}}\Gamma^2\bigg(\frac{1}{4}\bigg)\psi_1\bigg(\frac{1}{4}\bigg)$$
$\psi_1\Big(\frac{1}{4}\Big)=\pi^2+8C$ olduğunu biliyoruz.($C\to$ catalan sabiti)
$$\color{#A00000}{{\boxed{\Xi_1(2,4,2)=\int_0^\infty\:\frac{\ln^2(x)}{\sqrt{1+x^4}}\:dx=\frac{\Gamma^2\Big(\frac{1}{4}\Big)(\pi^2+8C)}{32\,\sqrt{\pi}}}}}$$