Question

$$\Xi_1(2,m,p)=\int_0^\infty\:\frac{\ln^2(x)}{\sqrt[p]{1+x^m}}$$

İntegralini çözün.

Answer 1

İntegralimiz :

$$\Xi_1(2,m,p)=\int_0^\infty\:\frac{\ln^2(x)}{\sqrt[p]{1+x^m}}$$

Buradaki eşitlikte $n$ yerine $2$ verelim.Eşitlik :

$$\Xi_1(n,m,p)=\int_0^\infty\:\frac{\ln^n(x)}{\sqrt[p]{1+x^m}}\:dx\\=\lim\limits_{s\to0}\frac{\partial^n}{\partial{s}^n}\,\frac{1}{m\,\Gamma(p^{-1})}\Gamma\bigg(\frac{1}{m}+\frac{s}{m}\bigg)\Gamma\bigg(\frac{1}{p}-\frac{1}{m}-\frac{s}{m}\bigg)$$

$$\Xi_1(2,m,p)=\int_0^\infty\:\frac{\ln^2(x)}{\sqrt[p]{1+x^m}}\:dx\\=\lim\limits_{s\to0}\frac{\partial^2}{\partial{s}^2}\,\frac{1}{m\,\Gamma(p^{-1})}\Gamma\bigg(\frac{1}{m}+\frac{s}{m}\bigg)\Gamma\bigg(\frac{1}{p}-\frac{1}{m}-\frac{s}{m}\bigg)$$

Türevi bulalım.

$$\lim\limits_{s\to0}\,\frac{1}{m^3\,\Gamma(p^{-1})}\Bigg[\Gamma(\color{red}{{A}})\Gamma(\color{blue}{{B}})\bigg(\psi(\color{red}{{A}})-\psi(\color{blue}{{B}})\bigg)+\Gamma(\color{red}{{A}})\Gamma(\color{blue}{{B}})\bigg(\psi_1(\color{red}{{A}})+\psi(\color{blue}{{B}})\bigg)\Bigg]$$

$$\color{red}{\large{A}=\normalsize{\frac{1}{m}+\frac{s}{m}}}$$

$$\color{blue}{\large{B}=\normalsize{\frac{1}{p}-\frac{1}{m}-\frac{s}{m}}}$$

Limiti alalım.

$$\color{#A00000}{\boxed{\Xi_1(2,m,p)=\int_0^\infty\:\frac{\ln^2(x)}{\sqrt[p]{1+x^m}}=\\\frac{1}{m^3\,\Gamma(p^{-1})}\Bigg[\Gamma\Big(\frac{1}{m}\Big)\Gamma\Big(\frac{1}{p}-\frac{1}{m}\Big)\bigg(\psi\Big(\frac{1}{m}\Big)-\psi\Big(\frac{1}{p}-\frac{1}{m}\Big)\bigg)\\+\Gamma\Big(\frac{1}{m}\Big)\Gamma\Big(\frac{1}{p}-\frac{1}{m}\Big)\bigg(\psi_1\Big(\frac{1}{m}\Big)+\psi_1\Big(\frac{1}{p}-\frac{1}{m}\Big)\bigg)\Bigg]}}$$